Matrices, Vectors & Fourier analysis Spring 23-24

Chapter-4
Eigenvalues and Eigenvectors
Let 𝐴𝐴 = (𝑎𝑎𝑖𝑖𝑖𝑖 )𝑛𝑛×𝑛𝑛 is a square matrix. A non-zero vector 𝑽𝑽 in ℝ𝑛𝑛 is called an eigenvector of 𝐴𝐴 if 𝐴𝐴𝑽𝑽 is
a scalar multiple of 𝑽𝑽; that is 𝐴𝐴𝑽𝑽 = 𝜆𝜆𝑽𝑽 for some scalar 𝜆𝜆. The scalar 𝜆𝜆 is called an eigenvalue of 𝐴𝐴 and
𝑽𝑽 is called the eigenvector of 𝐴𝐴 corresponding to 𝜆𝜆.

1 3 0
Example: The vector 𝑽𝑽 = � � is an eigenvector of 𝐴𝐴 = � � corresponding to the eigenvalue
2 8 −1
𝜆𝜆 = 3.
3 0 1 3 1
𝐴𝐴𝑽𝑽 = � � � � = � � = 3 � � = 𝜆𝜆𝑽𝑽.
8 −1 2 6 2

Characteristic matrix:
Provided that 𝐴𝐴 is a square matrix of order 𝑛𝑛 × 𝑛𝑛. Then the matrix 𝐴𝐴 − 𝜆𝜆𝜆𝜆 is called the characteristic
matrix where 𝜆𝜆 is scalar and 𝐼𝐼 is the unit matrix.

Example:
2 2 1
𝐴𝐴 = �1 3 1�
1 2 2
Characteristic matrix is
2 2 1 1 0 0
𝐴𝐴 − 𝜆𝜆𝜆𝜆 = �1 3 1� − 𝜆𝜆 �0 1 0�
1 2 2 0 0 1
2 − 𝜆𝜆 2 1
=� 1 3 − 𝜆𝜆 1 �
1 2 2 − 𝜆𝜆

Characteristic polynomial:
The determinant of characteristic matrix (𝐴𝐴 − 𝜆𝜆𝜆𝜆) is the polynomial of 𝜆𝜆. Characteristic polynomial is
2 − 𝜆𝜆 2 1
|𝐴𝐴 − 𝜆𝜆𝜆𝜆| = � 1 3 − 𝜆𝜆 1 �
1 2 2 − 𝜆𝜆
= (2 − 𝜆𝜆)(6 − 5𝜆𝜆 + 𝜆𝜆2 ) − 2(1 − 𝜆𝜆) + 1(2 − 3 + 𝜆𝜆)
= 𝜆𝜆3 − 7𝜆𝜆2 + 11𝜆𝜆 − 5.

Characteristic equation:
The equation |𝐴𝐴 − 𝜆𝜆𝜆𝜆| = 0 is called characteristic equation for matrix 𝐴𝐴.
For example, 𝜆𝜆3 − 7𝜆𝜆2 + 11𝜆𝜆 − 5 = 0 is characteristic equation for the above matrix 𝐴𝐴.

Characteristic root(s) or eigenvalue(s):

The roots of the characteristic equation |𝐴𝐴 − 𝜆𝜆𝜆𝜆| = 0 are called characteristic values or eigenvalues of
matrix A.
∴ 𝜆𝜆3 − 7𝜆𝜆2 + 11𝜆𝜆 − 5 = 0

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Matrices, Vectors & Fourier analysis Spring 23-24

⟹ ( 𝜆𝜆 − 1)(𝜆𝜆 − 1)(𝜆𝜆 − 5) = 0
⟹ 𝜆𝜆 = 1,1,5
So, the characteristic values or eigenvalues are 1,1 and 5

Example:
1 2 −1
Find the eigenvalues and eigenvectors of the matrix 𝐴𝐴 = �0 −2 0 �
0 5 2
Solution:
The characteristic matrix of 𝐴𝐴 is,
1 2 −1 1 0 0
𝐴𝐴 − 𝜆𝜆𝜆𝜆 = �0 −2 0 � − 𝜆𝜆 �0 1 0�
0 5 2 0 0 1
1 − 𝜆𝜆 2 −1
=� 0 −2 − 𝜆𝜆 0 �
0 5 2 − 𝜆𝜆
1 − 𝜆𝜆 2 −1
The characteristic polynomial of matrix 𝐴𝐴 is − 𝜆𝜆𝜆𝜆| = � 0
|𝐴𝐴 −2 − 𝜆𝜆 0 �
0 5 2 − 𝜆𝜆
The characteristic equation of matrix 𝐴𝐴 is |𝐴𝐴 − 𝜆𝜆𝜆𝜆| = 0
1 − 𝜆𝜆 2 −1
⇒� 0 −2 − 𝜆𝜆 0 �=0
0 5 2 − 𝜆𝜆
∴ (𝜆𝜆 − 1)(𝜆𝜆 + 2)(𝜆𝜆 − 2) = 0
So, the characteristic roots or the eigenvalues of matrix 𝐴𝐴 is 𝜆𝜆 = 1, −2,2
𝑣𝑣1
Now consider 𝑽𝑽 = �𝑣𝑣2 � is an eigenvector of matrix 𝐴𝐴 corresponding to the eigenvalue 𝜆𝜆. From the
𝑣𝑣3
definition of eigenvalue and eigenvector MatLab command for finding eigenvalues and
eigenvectors:
𝐴𝐴𝑽𝑽 = 𝜆𝜆𝑽𝑽
>> A=[1 2 -1;0 -2 0;0 5 2];
1 2 −1 𝑣𝑣1 𝑣𝑣1
⇒ �0 −2 0 � �𝑣𝑣2 � = 𝜆𝜆 �𝑣𝑣2 � >> [v,d]=eig(sym(A));
0 5 2 𝑣𝑣3 𝑣𝑣3 >> eigenvalues=eig(A)'
𝑣𝑣1 + 2𝑣𝑣2 − 𝑣𝑣3 𝜆𝜆 𝑣𝑣1 eigenvalues =
So, � −2𝑣𝑣2 𝜆𝜆
� = � 2� 𝑣𝑣 1 2 -2
5𝑣𝑣2 + 2𝑣𝑣3 𝜆𝜆 𝑣𝑣3 >> eigenvectors=double(v)
From the above equal matrices, we have eigenvectors =
𝑣𝑣1 + 2 𝑣𝑣2 − 𝑣𝑣3 = 𝜆𝜆 𝑣𝑣1 1.0000 -1.0000 0.8667
−2 𝑣𝑣2 = 𝜆𝜆 𝑣𝑣2 � 0 0 -0.8000
5𝑣𝑣2 + 2 𝑣𝑣3 = 𝜆𝜆 𝑣𝑣3 0 1.0000 1.0000

(1 − 𝜆𝜆)𝑣𝑣1 + 2 𝑣𝑣2 − 𝑣𝑣3 = 0

⟹ −(2 + 𝜆𝜆) 𝑣𝑣2 = 0 � … … … … … … … … … (1)
5𝑣𝑣2 + (2 − 𝜆𝜆) 𝑣𝑣3 = 0

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Matrices, Vectors & Fourier analysis Spring 23-24

For 𝜆𝜆 = 1, system of equations (1) becomes,

2 𝑣𝑣2 − 𝑣𝑣3 = 0
−3 𝑣𝑣2 = 0
5 𝑣𝑣2 + 𝑣𝑣3 = 0

Solving we get 𝑣𝑣2 = 𝑣𝑣3 = 0

Hence 𝑣𝑣1 is a free variable. Let, 𝑣𝑣1 = 𝑎𝑎, where 𝑎𝑎 is any real number. Therefore, the eigenvector of 𝐴𝐴
𝑎𝑎
corresponding to the eigenvalue 𝜆𝜆 = 1 are the non-zero vectors of the form 𝑽𝑽𝟏𝟏 = �0�. In particular, if
0
1
𝑎𝑎 = 1, then 𝑽𝑽𝟏𝟏 = �0� is an eigenvector corresponding to the eigenvalue of 𝜆𝜆 = 1
0
Again, for 𝜆𝜆 = −2, system of equations (1) becomes,

3 𝑣𝑣1 + 2 𝑣𝑣2 − 𝑣𝑣3 = 0

0=0
5 𝑣𝑣2 + 4 𝑣𝑣3 = 0

This system has one free variable. Let 𝑣𝑣3 = 𝑏𝑏.

4𝑏𝑏 13𝑏𝑏
∴ 𝑣𝑣2 = − and ∴ 𝑣𝑣1 =
5 15
13𝑏𝑏
15
Therefore, 𝑽𝑽𝟐𝟐 = �− 4𝑏𝑏�.
5
𝑏𝑏
In particular, let 𝑏𝑏 = −15
−13
So, 𝑽𝑽𝟐𝟐 = � 12 � is an eigenvector corresponding to the eigenvalue 𝜆𝜆 = −2
−15

For 𝜆𝜆 = 2, system of equations (1) becomes,

−𝑣𝑣1 + 2 𝑣𝑣2 − 𝑣𝑣3 = 0
−4 𝑣𝑣2 = 0
5 𝑣𝑣2 = 0

Hence, 𝑣𝑣2 = 0 and 𝑣𝑣3 is free variable. Let 𝑣𝑣3 = 𝑐𝑐 then we have 𝑣𝑣1 = −𝑐𝑐
−𝑐𝑐
Therefore, 𝑽𝑽𝟑𝟑 = � 0 �
𝑐𝑐
−1
In particular, if 𝑐𝑐 = 1 we have 𝑽𝑽𝟑𝟑 = � 0 �.
1
1 −13 −1
Therefore, eigenvectors are �0�, � 12 � and � 0 � corresponding to the eigenvalues 𝜆𝜆 = 1, −2, 2
0 −15 1
respectively.

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Matrices, Vectors & Fourier analysis Spring 23-24

Example: Solve the following system of differential equation using eigenvalues and eigenvectors.
𝑥𝑥1̇ (𝑡𝑡) = −1.5𝑥𝑥1 (𝑡𝑡) + 0.5𝑥𝑥2 (𝑡𝑡)
� with 𝑥𝑥1 (0) = 5, 𝑥𝑥2 (0) = 4.
𝑥𝑥2̇ (𝑡𝑡) = 𝑥𝑥1 (𝑡𝑡) − 𝑥𝑥2 (𝑡𝑡)
𝑑𝑑𝑥𝑥1 𝑑𝑑𝑥𝑥2
where 𝑥𝑥1̇ (𝑡𝑡) = and 𝑥𝑥2̇ (𝑡𝑡) = .
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

Solution:
𝑥𝑥1 (𝑡𝑡) 𝑥𝑥 ̇ (𝑡𝑡)
Let, 𝑋𝑋(𝑡𝑡) = � � and 𝑋𝑋(̇ 𝑡𝑡) = � 1 �
𝑥𝑥2 (𝑡𝑡) 𝑥𝑥2̇ (𝑡𝑡)
𝑥𝑥 (0) 5
So, 𝑋𝑋(0) = � 1 � = � �.
𝑥𝑥2 (0) 4
−1.5 0.5
We write, 𝐴𝐴 = � �.
1 −1
Now the system of differential equation can be written as
𝑋𝑋(̇ 𝑡𝑡) = 𝐴𝐴𝐴𝐴(𝑡𝑡)
Let 𝜆𝜆 and 𝑽𝑽 be the eigenvalue and eigenvector of 𝐴𝐴 respectively and 𝐶𝐶 is an integral constant then we
have the solution of the form,
𝑋𝑋(𝑡𝑡) = 𝐶𝐶𝐶𝐶𝑒𝑒 𝜆𝜆𝜆𝜆 .
−1.5 0.5 1 0
The characteristic matrix of 𝐴𝐴 is (𝐴𝐴 − 𝜆𝜆𝜆𝜆) = � � − 𝜆𝜆 � �
1 −1 0 1
−1.5 − 𝜆𝜆 0.5
=� �
1 −1 − 𝜆𝜆
−1.5 − 𝜆𝜆 0.5
The characteristic polynomial of 𝐴𝐴 is |𝐴𝐴 − 𝜆𝜆𝜆𝜆| = � �
1 −1 − 𝜆𝜆
The characteristic equation of 𝐴𝐴 is |𝐴𝐴 − 𝜆𝜆𝜆𝜆| = 0
−1.5 − 𝜆𝜆 0.5
⟹� �
1 −1 − 𝜆𝜆
⟹ (𝜆𝜆 + 1.5)(𝜆𝜆 + 1) − 0.5 = 0
⟹ (𝜆𝜆2 + 2.5𝜆𝜆 + 1) = 0
⟹ (𝜆𝜆 + 0.5)(𝜆𝜆 + 2) = 0
So the characteristic roots or the eigenvalues of 𝐴𝐴 is 𝜆𝜆 = −0.5, −2
𝑣𝑣1
Now, consider 𝑽𝑽 = �𝑣𝑣 � is the eigenvector of 𝐴𝐴 corresponding to the eigenvalue . From the definition
2
of eigenvalue and eigenvector, 𝑽𝑽 = 𝜆𝜆𝑽𝑽 .
−1.5 0.5 𝑣𝑣1 𝑣𝑣1
So, � � �𝑣𝑣 � = 𝜆𝜆 � 𝑣𝑣 �
1 −1 2 2
−1.5 𝑣𝑣1 + 0.5 𝑣𝑣2 𝜆𝜆 𝑣𝑣
⇒ � � = � 1�
𝑣𝑣1 − 𝑣𝑣2 𝜆𝜆 𝑣𝑣2

From the above equal matrices, we have

−1.5 𝑣𝑣1 + 0.5 𝑣𝑣2 = 𝜆𝜆 𝑣𝑣1
�
𝑣𝑣1 − 𝑣𝑣2 = 𝜆𝜆 𝑣𝑣2

−(1.5 + 𝜆𝜆 ) 𝑣𝑣1 + 0.5 𝑣𝑣2 = 0

⟹ � … … … … … … … … … (1)
𝑣𝑣1 − (1 + 𝜆𝜆 ) 𝑣𝑣2 = 0

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Matrices, Vectors & Fourier analysis Spring 23-24

For 𝜆𝜆 = −0.5, system of equations (1) becomes,

−𝑣𝑣1 + 0.5 𝑣𝑣2 = 0
𝑣𝑣1 − 0.5 𝑣𝑣2 = 0
Solving the above system, we get −𝑣𝑣1 + 0.5𝑣𝑣2 = 0.
Here 𝑣𝑣2 is a free variable. Let 𝑣𝑣2 = 2.
Thus, we get 𝑣𝑣1 = 1 and 𝑣𝑣2 = 2.
Therefore, the eigenvector of 𝐴𝐴 corresponding to the eigenvalue 𝜆𝜆 = −0.5 are the non-zero vectors of
1
the form 𝑽𝑽1 = � � .
2
Again, for 𝜆𝜆 = −2, system of equations (1) becomes,
0.5𝑣𝑣1 + 0.5𝑣𝑣2 = 0
𝑣𝑣1 + 𝑣𝑣2 = 0
Solving the above system, we get 0.5𝑣𝑣1 + 0.5𝑣𝑣2 = 0.
Here 𝑣𝑣2 is a free variable. Let 𝑣𝑣2 = 2.
Thus, we get 𝑣𝑣1 = −2 and 𝑣𝑣2 = 2.
Therefore, the eigenvector of 𝐴𝐴 corresponding to the eigenvalue 𝜆𝜆 = −2 are the non-zero vectors of
−2
the form 𝑽𝑽2 = � � .
2
So, the solution of the system of differential equation can be written as,
𝑋𝑋(𝑡𝑡) = 𝐶𝐶1 𝑉𝑉1 𝑒𝑒 𝜆𝜆1 𝑡𝑡 + 𝐶𝐶2 𝑉𝑉2 𝑒𝑒 𝜆𝜆2 𝑡𝑡
𝑥𝑥 (𝑡𝑡) 1 −2
⟹ � 1 � = 𝐶𝐶1 � � 𝑒𝑒 −0.5𝑡𝑡 + 𝐶𝐶2 � � 𝑒𝑒 −2𝑡𝑡
𝑥𝑥2 (𝑡𝑡) 2 2
𝑥𝑥 (0) 1 −2
⟹ � 1 � = 𝐶𝐶1 � � + 𝐶𝐶2 � �
𝑥𝑥2 (0) 2 2
5 1 −2
⟹ � � = 𝐶𝐶1 � � + 𝐶𝐶2 � �
4 2 2
We can write,
𝐶𝐶1 − 2𝐶𝐶2 = 5
2𝐶𝐶1 + 2𝐶𝐶2 = 4
Solving the system for 𝐶𝐶1 and 𝐶𝐶2 , we have, 𝐶𝐶1 = 3 and 𝐶𝐶2 = −1. Now,
𝑥𝑥 (𝑡𝑡) 1 −2
� 1 � = 3 � � 𝑒𝑒 −0.5𝑡𝑡 − � � 𝑒𝑒 −2𝑡𝑡
𝑥𝑥2 (𝑡𝑡) 2 2
𝑥𝑥 (𝑡𝑡) −0.5𝑡𝑡 −2𝑡𝑡
⇒ � 1 � = �3𝑒𝑒 −0.5𝑡𝑡 � − �−2𝑒𝑒−2𝑡𝑡 �
𝑥𝑥2 (𝑡𝑡) 6𝑒𝑒 2𝑒𝑒

Therefore,
𝑥𝑥1 (𝑡𝑡) = 3𝑒𝑒 −0.5𝑡𝑡 + 2𝑒𝑒 −2𝑡𝑡
𝑥𝑥2 (𝑡𝑡) = 6𝑒𝑒 −0.5𝑡𝑡 − 2𝑒𝑒 −2𝑡𝑡

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Matrices, Vectors & Fourier analysis Spring 23-24

Sample Exercise-4.1

1. Find the eigenvalues and eigenvectors of the following matrices

1 2 Ans: 𝜆𝜆1 = −1, 𝜆𝜆2 = 3, 𝑽𝑽1 = (−𝑎𝑎, 𝑎𝑎)𝑇𝑇 , 𝑽𝑽2 = (𝑏𝑏, 𝑏𝑏)𝑇𝑇
a. � �
2 1 𝑽𝑽1 |𝑎𝑎=1 = (−1, 1)𝑇𝑇 , 𝑽𝑽2 |𝑏𝑏=1 = (1, 1)𝑇𝑇

−1 0 Ans: 𝜆𝜆1 = 2, 𝜆𝜆2 = −1, 𝑽𝑽1 = (−𝑎𝑎, 𝑎𝑎)𝑇𝑇 , 𝑽𝑽2 = (0, 𝑏𝑏)𝑇𝑇
b. � �
3 2 𝑽𝑽1 |𝑎𝑎=1 = (−1, 1)𝑇𝑇 , 𝑽𝑽2 |𝑏𝑏=1 = (0, 1)𝑇𝑇
Ans: 𝜆𝜆1 = −1, 𝜆𝜆2 = 5, 𝑽𝑽1 = (−2𝑎𝑎, 𝑎𝑎)𝑇𝑇 , 𝑽𝑽2 = (𝑏𝑏, 𝑏𝑏)𝑇𝑇
1 4
c. � � 𝑽𝑽1 |𝑎𝑎=1 = (−2, 1)𝑇𝑇 , 𝑽𝑽2 |𝑏𝑏=1 = (1, 1)𝑇𝑇
2 3

Ans: 𝜆𝜆1 = 1, 𝜆𝜆2 = 2, 𝜆𝜆3 = −2

1 2 −1 𝑽𝑽1 = (𝑎𝑎, 0, 0)𝑇𝑇 , 𝑽𝑽2 = (−𝑏𝑏, 0, 𝑏𝑏)𝑇𝑇 , 𝑽𝑽3 = (−0.2𝑐𝑐, 0.8𝑐𝑐, 𝑐𝑐)𝑇𝑇
d. �0 −2 0�
𝑽𝑽1 |𝑎𝑎=1 = (1,0,0)𝑇𝑇 , 𝑽𝑽2 |𝑏𝑏=1 = (−1, 0,1)𝑇𝑇 , 𝑽𝑽2 |𝑐𝑐=1 = (−0.2, 0.8,1)𝑇𝑇
0 −5 2

Ans: 𝜆𝜆1 = 2, 𝜆𝜆2 = 1, 𝜆𝜆3 = −1

2 3 3 2𝑐𝑐 𝑐𝑐 𝑇𝑇
e. � 1 3 2� 𝑽𝑽1 = (0, −𝑎𝑎, 𝑎𝑎)𝑇𝑇 , 𝑽𝑽2 = (−𝑏𝑏, −𝑏𝑏, 𝑏𝑏)𝑇𝑇 , 𝑽𝑽3 = �− , − , 𝑐𝑐�
3 3
−1 −4 −3 2 1 𝑇𝑇
𝑽𝑽1 |𝑎𝑎=1 = (0, −1,1)𝑇𝑇 , 𝑽𝑽2 |𝑏𝑏=1 = (−1, −1,1)𝑇𝑇 , 𝑽𝑽2 |𝑐𝑐=1 = �− , − , 1�
3 3

2. Solve the following system of differential equations using eigenvalue and eigenvector where
𝑑𝑑𝑥𝑥1 𝑑𝑑𝑥𝑥2
𝑥𝑥1̇ (𝑡𝑡) = and 𝑥𝑥2̇ (𝑡𝑡) = .
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
Ans.
𝑥𝑥1̇ (𝑡𝑡) = 𝑥𝑥1 (𝑡𝑡) + 2𝑥𝑥2 (𝑡𝑡) 𝑥𝑥1 (𝑡𝑡) 8 −𝑎𝑎 4 4𝑡𝑡 2𝑏𝑏
a. � � � = − 𝑒𝑒 −𝑡𝑡 � � − 𝑒𝑒 � �
𝑥𝑥2̇ (𝑡𝑡) = 3𝑥𝑥1 (𝑡𝑡) + 2𝑥𝑥2 (𝑡𝑡) 𝑥𝑥2 (𝑡𝑡) 5𝑎𝑎 𝑎𝑎 5𝑏𝑏 3𝑏𝑏
with 𝑥𝑥1 (0) = 0, 𝑥𝑥2 (0) = −4. 𝑥𝑥1 (𝑡𝑡) 8 −1 4 2
� � = − 𝑒𝑒 −𝑡𝑡 � � − 𝑒𝑒 4𝑡𝑡 � �.
𝑥𝑥2 (𝑡𝑡) 5 1 5 3
𝑥𝑥1̇ (𝑡𝑡) = −5𝑥𝑥1 (𝑡𝑡) + 𝑥𝑥2 (𝑡𝑡) Ans.
b. �
𝑥𝑥2̇ (𝑡𝑡) = 4𝑥𝑥1 (𝑡𝑡) − 2𝑥𝑥2 (𝑡𝑡) 𝑥𝑥1 (𝑡𝑡) 3 1 2 −1
� � = 𝑒𝑒 −𝑡𝑡 � � − 𝑒𝑒 −6𝑡𝑡 � �
with 𝑥𝑥1 (0) = 1, 𝑥𝑥2 (0) = 2. 𝑥𝑥2 (𝑡𝑡) 5 4 5 1

𝑥𝑥1̇ (𝑡𝑡) = 𝑥𝑥2 (𝑡𝑡) Ans.

c. �
𝑥𝑥2̇ (𝑡𝑡) = 1.5𝑥𝑥1 (𝑡𝑡) − 2.5𝑥𝑥2 (𝑡𝑡) 𝑥𝑥1 (𝑡𝑡) 22 1 3 2
� � = − 𝑒𝑒 −3𝑡𝑡 � � − 𝑒𝑒 0.5𝑡𝑡 � �
with 𝑥𝑥1 (0) = −4, 𝑥𝑥2 (0) = 9. 𝑥𝑥2 (𝑡𝑡) 7 −3 7 1

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Matrices, Vectors & Fourier analysis Spring 23-24

Cayley-Hamilton Theorem:
Every square matrix is a zero of its characteristic polynomial.
Or,
Every square matrix satisfies its characteristic equation

𝑖𝑖. 𝑒𝑒 𝐴𝐴𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝐴𝐴𝑛𝑛−1 + 𝑎𝑎𝑛𝑛−2 𝐴𝐴𝑛𝑛−2 + ⋯ … . . + + 𝑎𝑎1 𝐴𝐴 + 𝑎𝑎0 𝐼𝐼 = 0

1 2 3
Example: Verify the Cayley-Hamilton theorem for the matrix 𝐴𝐴 = �2 −1 1�
3 1 1
1 2 3 1 0 0
Solution: The characteristic matrix of 𝐴𝐴 is 𝐴𝐴 − 𝜆𝜆𝜆𝜆 = �2 −1 1� − 𝜆𝜆 �0 1 0�
3 1 1 0 0 1
1 − 𝜆𝜆 2 3
=� 2 −1 − 𝜆𝜆 1 �
3 1 1 − 𝜆𝜆

Therefore, the characteristic equation of the matrix 𝐴𝐴 is |𝐴𝐴 − 𝜆𝜆𝜆𝜆| = 0

1 − 𝜆𝜆 2 3
� 2 −1 − 𝜆𝜆 1 �=0
3 1 1 − 𝜆𝜆
⟹ 𝜆𝜆3 − 𝜆𝜆2 − 15𝜆𝜆 − 15 = 0

Now in order to verify Cayley –Hamilton theorem we have to show that

𝐴𝐴3 − 𝐴𝐴2 − 15𝐴𝐴 − 15𝐼𝐼 = 0
1 2 3
So, 𝐴𝐴 = �2 −1 1�
3 1 1
1 2 3 1 2 3 14 3 8
𝐴𝐴2 = �2 −1 1� �2 −1 1� = � 3 6 6 �
3 1 1 3 1 1 8 6 11
14 3 8 1 2 3 44 33 53
𝐴𝐴3 = 𝐴𝐴2 𝐴𝐴 = � 3 6 6 � �2 −1 1� = �33 6 21�
8 6 11 3 1 1 53 21 41

∴ 𝐴𝐴3 − 𝐴𝐴2 − 15𝐴𝐴 − 15𝐼𝐼

44 33 53 14 3 8 1 2 3 1 0 0
= �33 6 21� − � 3 6 6 � − 15 �2 −1 1� − 15 �0 1 0�
53 21 41 8 6 11 3 1 1 0 0 1
0 0 0
= �0 0 0� = 0
0 0 0
3 2
∴ 𝐴𝐴 − 𝐴𝐴 − 15𝐴𝐴 − 15𝐼𝐼 = 0
Hence Cayley-Hamilton theorem is verified.

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Matrices, Vectors & Fourier analysis Spring 23-24

1 2 3
Example: Using Cayley-Hamilton theorem find the inverse of the matrix 𝐴𝐴 = �2 −1 1�
3 1 1
1 2 3 1 0 0
Solution: The characteristic matrix of A is 𝐴𝐴 − 𝜆𝜆𝜆𝜆 = �2 −1 1� − 𝜆𝜆 �0 1 0�
3 1 1 0 0 1

1 − 𝜆𝜆 2 3
=� 2 −1 − 𝜆𝜆 1 �
3 1 1 − 𝜆𝜆

Therefore, the characteristic equation of the matrix A is |𝐴𝐴 − 𝜆𝜆𝜆𝜆| = 0

1 − 𝜆𝜆 2 3
� 2 −1 − 𝜆𝜆 1 �=0
3 1 1 − 𝜆𝜆
⟹ 𝜆𝜆3 − 𝜆𝜆2 − 15𝜆𝜆 − 15 = 0

Now according to the Cayley –Hamilton theorem, we have

𝐴𝐴3 − 𝐴𝐴2 − 15𝐴𝐴 − 15𝐼𝐼 = 0
⟹ 𝐴𝐴2 − 𝐴𝐴 − 15𝐼𝐼 − 15𝐴𝐴−1 = 0
1
⟹ 𝐴𝐴−1 = [𝐴𝐴2 − 𝐴𝐴 − 15𝐼𝐼]
15
Here,
1 2 3 1 2 3 14 3 8
𝐴𝐴2 = �2 −1 1� �2 −1 1 � = � 3 6 6�
3 1 1 3 1 1 8 6 11

14 3 8 1 2 3 1 0 0
1
∴ 𝐴𝐴−1 = �� 3 6 6 � − �2 −1 1� − 15 �0 1 0��
15
8 6 11 3 1 1 0 0 1
−2 1 5
1
= � 1 −8 5 �.
15
5 5 −5
1 2 2
Example: Using Cayley-Hamilton theorem find the inverse of the matrix 𝐴𝐴 = �3 1 0�
1 1 1
1 2 2 1 0 0
The characteristic matrix of 𝐴𝐴 is 𝐴𝐴 − 𝜆𝜆𝜆𝜆 = �3 1 0� − 𝜆𝜆 �0 1 0�
1 1 1 0 0 1
1 − 𝜆𝜆 2 2
=� 3 1 − 𝜆𝜆 0 �.
1 1 1 − 𝜆𝜆
Therefore, the characteristic equation of the matrix 𝐴𝐴 is |𝐴𝐴 − 𝜆𝜆𝜆𝜆| = 0
1 − 𝜆𝜆 2 2
� 3 1 − 𝜆𝜆 0 �=0
1 1 1 − 𝜆𝜆
⟹ 𝜆𝜆3 − 3𝜆𝜆2 − 5𝜆𝜆 + 1 = 0
Now according to the Cayley –Hamilton theorem, we have
𝐴𝐴3 − 3𝐴𝐴2 − 5𝐴𝐴 + 𝐼𝐼 = 0
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Matrices, Vectors & Fourier analysis Spring 23-24

⟹ 𝐴𝐴2 − 3𝐴𝐴 − 5𝐼𝐼 + 𝐴𝐴−1 = 0

𝐴𝐴−1 = 3𝐴𝐴 + 5𝐼𝐼 − 𝐴𝐴2
1 2 2
𝐴𝐴 = �3 1 0�
1 1 1
1 2 2 1 2 2 9 6 4
𝐴𝐴2 = �3 1 0� �3 1 0� = �6 7 6�
1 1 1 1 1 1 3 4 3
1 2 2 1 0 0 9 6 4 −1 0 2
−1
∴ 𝐴𝐴 = 3 �3 1 0� + 5 �0 1 0� − �6 7 6� = � 3 1 −6�
1 1 1 0 0 1 3 4 3 −2 −1 5

Sample Exercise-4.2
State the Cayley-Hamilton theorem. Hence find the inverse of the following matrices using Cayley –
Hamilton theorem and verify your result.

2 5 3 −5
Ans. 𝐴𝐴−1 = � �
a. A=  −1 2
1 3

1 2 2  −1 0 2
b. A = 3 1 0  Ans. 𝐴𝐴−1 = � 3 1 −6�
1 1 1  −2 −1 5

2 0 1 −2 5 −3
c. 𝐴𝐴 = �−2 3 4� Ans. 𝐴𝐴−1 = �−8 17 −10�
−5 5 6 5 −10 6

1 2 3 −40 16 9
d. 𝐴𝐴 = �2 5 3� Ans. 𝐴𝐴−1 = � 13 −5 −3�
1 0 8 5 −2 −1

1 −1 1 0 1 −1
e. 𝐴𝐴 = �2 −1 0� Ans. 𝐴𝐴−1 = �0 1 −2�
1 −1 0 1 0 −1

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Matrices, Vectors & Fourier analysis Spring 23-24

Vector
Vector Spaces
Vectors in ℝ𝟐𝟐 :
The set of all ordered pairs of real numbers is called two-dimensional vector space and is denoted by
ℝ2 .
ℝ2 = {(𝑥𝑥, 𝑦𝑦): 𝑥𝑥, 𝑦𝑦 ∈ ℝ}, where ℝ2 = ℝ × ℝ
Example: (1,2); (2, −1); (3,4); �√3, 5�; (0, e) ∈ ℝ2
Vectors in ℝ𝟑𝟑 :
The set of all ordered triplets of real numbers is called three-dimensional vector space and is denoted
by ℝ3
ℝ3 = {(𝑥𝑥, 𝑦𝑦, 𝑧𝑧): 𝑥𝑥, 𝑦𝑦, 𝑧𝑧 ∈ ℝ}, where ℝ3 = ℝ × ℝ × ℝ
Vectors in ℝ𝐧𝐧 :
If 𝑛𝑛 is a positive integer then the set of all ordered n triplets of real numbers is called vector 𝑛𝑛-space
and is denoted by ℝ𝑛𝑛 and if 𝒖𝒖 ∈ ℝn ; 𝑢𝑢 = (𝒖𝒖1 , 𝒖𝒖2 , … … , 𝒖𝒖𝑛𝑛 ), then 𝒖𝒖 is called a 𝑛𝑛-dimensional vector
in ℝ𝑛𝑛 . A particular n triplets in ℝ𝐧𝐧 is called co-ordinates of point.
Addition of two vectors in ℝ𝟐𝟐 :
If a= (𝑎𝑎1 , 𝑎𝑎2 ) and 𝒃𝒃 = (𝑏𝑏1 , 𝑏𝑏2 ) be two vectors in ℝ𝟐𝟐 then 𝒂𝒂 + 𝒃𝒃 = (𝑎𝑎1 + 𝑏𝑏1 , 𝑎𝑎2 + 𝑏𝑏2 )

Fig: addition of two vectors.

Subtraction of two vectors in ℝ : 𝟐𝟐

If a= (𝑎𝑎1 , 𝑎𝑎2 ) and 𝒃𝒃 = (𝑏𝑏1 , 𝑏𝑏2 ) be two vectors in ℝ𝟐𝟐 then 𝒂𝒂 − 𝒃𝒃 = (𝑎𝑎1 − 𝑏𝑏1 , 𝑎𝑎2 − 𝑏𝑏2 )

Fig: Subtraction of two vectors.

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Matrices, Vectors & Fourier analysis Spring 23-24

Scalar multiplication of vectors in ℝ𝟐𝟐 :

Let 𝑘𝑘 ∈ ℝ be a scalar and 𝒖𝒖 ∈ ℝ2 where 𝒖𝒖 = (𝑢𝑢1 , 𝑢𝑢2 ) is a vector in ℝ2 then
𝑘𝑘𝒖𝒖 = 𝑘𝑘(𝒖𝒖1 , 𝒖𝒖2 ) = (𝑘𝑘𝒖𝒖1 , 𝑘𝑘𝒖𝒖2 )
Zero Vector:
The vector whose components are all zero is called the zero vector and is denoted by 0 and defined by
𝟎𝟎 = (0,0, … . . ,0)
Dot or Inner product of two non-zero vectors:
Let 𝒖𝒖(𝑢𝑢1 , 𝑢𝑢2 ) and 𝒗𝒗(𝑣𝑣1 , 𝑣𝑣2 ) are two non-zero vectors and 𝜃𝜃 be the angle between. The dot product of
𝒖𝒖 and 𝒗𝒗 is denoted by 𝒖𝒖, 𝒗𝒗 and defined by 𝒖𝒖. 𝒗𝒗 = ‖𝒖𝒖‖‖𝒗𝒗‖cosθ = 𝑢𝑢1 𝑣𝑣1 + 𝑢𝑢2 𝑣𝑣2
Parallel vectors in ℝ𝟐𝟐 :
Let 𝒖𝒖, 𝒗𝒗𝜖𝜖ℝ2 , then the vector 𝑢𝑢 is called parallel vector of 𝑣𝑣 if 𝒖𝒖 = 𝑘𝑘𝒗𝒗. If the non-zero scalar 𝑘𝑘 > 0
then they are in the same directed and if 𝑘𝑘 < 0 then they are in opposite directed.
Perpendicular vectors in ℝ𝟐𝟐 :
Let 𝒖𝒖, 𝒗𝒗 ∈ ℝ2 then 𝒖𝒖 and 𝒗𝒗 are said to be perpendicular (or orthogonal) if 𝒖𝒖. 𝒗𝒗 = 0
Example: Let 𝒖𝒖, 𝒗𝒗 ∈ ℝ2 where 𝒖𝒖 = (3, −4), 𝒗𝒗 = (8,6); 𝒖𝒖 ∙ 𝒗𝒗 = 3 ∙ 8 + (−4) ∙ 6 = 0
Hence 𝒖𝒖 and 𝒗𝒗 are perpendicular vector.
Distance between two vectors in ℝ𝟐𝟐 :
Let 𝒖𝒖, 𝒗𝒗 ∈ ℝ2 where 𝒖𝒖(𝑢𝑢1 , 𝑢𝑢2 ) and 𝒗𝒗(𝑣𝑣1 , 𝑣𝑣2 ), then the distance between 𝒖𝒖 and 𝒗𝒗, denoted by 𝑑𝑑(𝒖𝒖, 𝒗𝒗)
are defined by 𝑑𝑑(𝒖𝒖, 𝒗𝒗) = �(𝑢𝑢1 − 𝑣𝑣1 )2 + (𝑢𝑢2 − 𝑣𝑣2 )2
Example: If 𝒖𝒖 = (5,4), 𝒗𝒗 = (1,1) then 𝑑𝑑(𝒖𝒖, 𝒗𝒗) = �(5 − 1)2 + (4 − 1)2 = √16 + 9 = 5
Norm or Length in ℝ𝟐𝟐 :
Let 𝒖𝒖 ∈ ℝ2 , where 𝒖𝒖 = (𝑢𝑢1 , 𝑢𝑢2 ) then the norm or length or modulus of 𝒖𝒖 denoted by ‖𝒖𝒖‖ and is
defined by ‖𝒖𝒖‖ = √𝒖𝒖 ∙ 𝒖𝒖 = �𝑢𝑢12 +𝑢𝑢22
Example: Let 𝒖𝒖, 𝒗𝒗 ∈ ℝ3 ; where 𝒖𝒖 = (1,0, −1), 𝒗𝒗 = (2, −3,1) then find
(𝑖𝑖)2𝒖𝒖 + 3𝒗𝒗 (𝑖𝑖𝑖𝑖)𝒖𝒖 ∙ 𝒗𝒗 (iii) ‖𝒗𝒗‖ (𝑖𝑖𝑖𝑖) ‖𝒖𝒖 − 𝒗𝒗‖ (𝑣𝑣)𝑑𝑑(𝒖𝒖, 𝒗𝒗)
Solution:
(𝑖𝑖) 2𝒖𝒖 + 3𝒗𝒗 = 2(1,0, −1) + 3(2, −3,1) = (2,0, −2) + (6, −9,3) = (8, −9,1)
(𝑖𝑖𝑖𝑖) 𝒖𝒖 ∙ 𝒗𝒗 = 1 ∙ 2 + 0 ∙ (−3) + (−1) ∙ 1 = 2 − 1 = 1
(𝑖𝑖𝑖𝑖𝑖𝑖) ‖𝒗𝒗‖ = √4 + 9 + 1 = √14
(𝑖𝑖𝑖𝑖) 𝒖𝒖 − 𝒗𝒗 = (1 − 2,0 + 3, −1 − 1) = (−1,3, −2)
∴ ‖𝒖𝒖 − 𝒗𝒗‖ = �(−1)2 + 32 + (−2)2 = √14
(𝑣𝑣)𝑑𝑑(𝒖𝒖, 𝒗𝒗) = �(1 − 2)2 + (0 + 3)2 + (−1 − 1)2 = √14 .
Vector Spaces:
Let 𝐹𝐹 be a field of scalars and 𝑽𝑽 be a non-empty set of vectors. If 𝑽𝑽 contains the following rules of
vector addition and scalar multiplication, then 𝑽𝑽 is called vector space over 𝐹𝐹.
In vector addition:
𝐴𝐴1 : 𝒖𝒖, 𝒗𝒗 ∈ 𝑉𝑉 ⇒ (𝒖𝒖 + 𝒗𝒗) ∈ 𝑽𝑽
𝐴𝐴2 : 𝒖𝒖, 𝒗𝒗, 𝒘𝒘 ∈ 𝑽𝑽 ⇒ (𝒖𝒖 + 𝒗𝒗) + 𝒘𝒘 = 𝒖𝒖 + (𝒗𝒗 + 𝒘𝒘)

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Matrices, Vectors & Fourier analysis Spring 23-24

𝐴𝐴3 : 𝒖𝒖, 𝟎𝟎 ∈ 𝑽𝑽 ⇒ (𝒖𝒖 + 𝟎𝟎) = (𝟎𝟎 + 𝒖𝒖) = 𝒖𝒖, here 𝟎𝟎 is the zero vector
𝐴𝐴4 : 𝒖𝒖, 𝒗𝒗 ∈ 𝑽𝑽 ⇒ 𝒖𝒖 + 𝒗𝒗 = 𝒗𝒗 + 𝒖𝒖
𝐴𝐴5 : 𝒖𝒖 ∈ 𝑽𝑽 ⇒ −𝒖𝒖 ∈ 𝑽𝑽 ⇒ 𝒖𝒖 + (−𝒖𝒖) = (−𝒖𝒖) + 𝒖𝒖 = 0
In scalar multiplication:
𝑀𝑀1 : 𝑎𝑎 ∈ 𝐹𝐹 and 𝒖𝒖 ∈ 𝑽𝑽 ⇒ 𝑎𝑎𝒖𝒖 ∈ 𝑽𝑽
𝑀𝑀2 : 𝑎𝑎 ∈ 𝐹𝐹 and 𝒖𝒖, 𝒗𝒗 ∈ 𝑽𝑽 ⇒ 𝑎𝑎(𝒖𝒖 + 𝒗𝒗) = 𝑎𝑎𝒖𝒖 + 𝑎𝑎𝒗𝒗
𝑀𝑀3 : 𝑎𝑎, 𝑏𝑏 ∈ 𝐹𝐹 and 𝒖𝒖 ∈ 𝑽𝑽 ⇒ (𝑎𝑎 + 𝑏𝑏)𝒖𝒖 = 𝑎𝑎𝒖𝒖 + 𝑏𝑏𝒖𝒖
𝑀𝑀4 : 𝑎𝑎, 𝑏𝑏 ∈ 𝐹𝐹 and 𝒖𝒖 ∈ 𝑽𝑽 ⇒ (𝑎𝑎𝑎𝑎)𝒖𝒖 = 𝑎𝑎(𝑏𝑏𝒖𝒖)
𝑀𝑀3 : 1 ∈ 𝐹𝐹 ⇒ 1. 𝒖𝒖 = 𝒖𝒖. 1 = 𝒖𝒖; 𝒖𝒖 ∈ 𝐹𝐹
Subspace:
Let 𝑾𝑾 be a non empty subset of a vector space 𝑽𝑽 over the field 𝐹𝐹. We call 𝑾𝑾 a subspace of 𝑽𝑽 if and
only if 𝑾𝑾 is a vector space over the field 𝐹𝐹 under the laws of vector addition and scalar multiplication
defined on 𝑽𝑽, or equivalently, 𝑾𝑾 is a subspace of 𝑽𝑽 wherever 𝑤𝑤1 , 𝑤𝑤2 ∈ 𝑾𝑾 and 𝛼𝛼, 𝛽𝛽 ∈ 𝐹𝐹 implies that
𝛼𝛼𝑤𝑤1 + 𝛽𝛽𝑤𝑤2 ∈ 𝑾𝑾.

Example: Show that 𝑆𝑆 = {(𝑎𝑎, 𝑏𝑏, 𝑐𝑐) ∣ 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 ∊ ℝ and 𝑎𝑎 + 𝑏𝑏 + 𝑐𝑐 = 0 } is a subspace of the vector space
ℝ𝟑𝟑 .

Solution: Here 𝟎𝟎 ∊ ℝ𝟑𝟑 , 𝟎𝟎 = (0,0,0) ∊ 𝑆𝑆 ∴0+0+0=0

Hence 𝑆𝑆 is non empty set. Again, let 𝒖𝒖 = (𝑎𝑎1 , 𝑏𝑏1 , 𝑐𝑐1 ) ∊ 𝑆𝑆 and 𝒗𝒗 = (𝑎𝑎2 , 𝑏𝑏2 , 𝑐𝑐2 ) ∊ 𝑆𝑆
𝑎𝑎1 + 𝑏𝑏1 + 𝑐𝑐1 = 0 and 𝑎𝑎2 + 𝑏𝑏2 + 𝑐𝑐2 = 0
For any scalars α, β we get 𝛼𝛼𝒖𝒖 + 𝛽𝛽𝒗𝒗 = 𝛼𝛼(𝑎𝑎1 , 𝑏𝑏1 , 𝑐𝑐1 ) + 𝛽𝛽(𝑎𝑎2 , 𝑏𝑏2 , 𝑐𝑐2 )
= (𝛼𝛼𝑎𝑎1 + 𝛽𝛽𝑎𝑎2 , 𝛼𝛼𝑏𝑏1 + 𝛽𝛽𝑏𝑏2 , 𝛼𝛼𝑐𝑐1 + 𝛽𝛽𝑐𝑐2 )
Now, 𝑎𝑎 + 𝑏𝑏 + 𝑐𝑐 = (𝛼𝛼𝑎𝑎1 + 𝛽𝛽𝑎𝑎2 + 𝛼𝛼𝑏𝑏1 + 𝛽𝛽𝑏𝑏2 + 𝛼𝛼𝑐𝑐1 + 𝛽𝛽𝑐𝑐2 )
= 𝛼𝛼(𝑎𝑎1 + 𝑏𝑏1 + 𝑐𝑐1 ) + 𝛽𝛽(𝑎𝑎2 + 𝑏𝑏2 + 𝑐𝑐2 ) = 𝛼𝛼. 0 + 𝛽𝛽. 0 = 0
∴ 𝛼𝛼𝒖𝒖 + 𝛽𝛽𝒗𝒗 ∈ 𝑆𝑆. Hence 𝑆𝑆 is a subspace of ℝ𝟑𝟑 .

Linear combination, Dependency and Independency of Vector

Linear combination:
Let 𝑽𝑽(𝐹𝐹) be a vector space, where 𝒗𝒗1 , 𝒗𝒗2 , 𝒗𝒗3 … 𝒗𝒗𝑛𝑛 ∊ 𝑽𝑽 and 𝛼𝛼1 , 𝛼𝛼2 , 𝛼𝛼3 … 𝛼𝛼𝑛𝑛 ∊ 𝐹𝐹. If 𝛼𝛼1 𝒗𝒗1 + 𝛼𝛼2 𝒗𝒗2 +
𝛼𝛼3 𝒗𝒗3 + ⋯ + 𝛼𝛼𝑛𝑛 𝒗𝒗𝑛𝑛 = 𝒖𝒖 ∊ 𝑽𝑽, then 𝒖𝒖 is called linear combination of 𝒗𝒗1 , 𝒗𝒗2 , 𝒗𝒗3 … 𝒗𝒗𝑛𝑛 .

Linear dependency of vectors:

Let 𝑽𝑽(𝐹𝐹) be a vector space, where 𝒗𝒗1 , 𝒗𝒗2 , 𝒗𝒗3 … 𝒗𝒗𝑛𝑛 ∊ 𝑽𝑽 and 𝛼𝛼1 , 𝛼𝛼2 , 𝛼𝛼3 … 𝛼𝛼𝑛𝑛 ∊ 𝐹𝐹. If 𝛼𝛼1 𝒗𝒗1 + 𝛼𝛼2 𝒗𝒗2 +
𝛼𝛼3 𝒗𝒗3 + ⋯ + 𝛼𝛼𝑛𝑛 𝒗𝒗𝑛𝑛 = 𝟎𝟎, and at least one of the element of the set {𝛼𝛼1 , 𝛼𝛼2 , 𝛼𝛼3 … 𝛼𝛼𝑛𝑛 } is not zero, then
the vectors 𝒗𝒗1 , 𝒗𝒗2 , 𝒗𝒗3 … 𝒗𝒗𝑛𝑛 are linearly dependent.

Linear independency of vectors:

Let 𝑽𝑽(𝐹𝐹) be a vector space and 𝒗𝒗1 , 𝒗𝒗2 , 𝒗𝒗3 , … 𝒗𝒗𝑛𝑛 ∊ 𝑽𝑽 and 𝛼𝛼1 , 𝛼𝛼2 , 𝛼𝛼3 , … 𝛼𝛼𝑛𝑛 ∊ 𝐹𝐹. If 𝛼𝛼1 𝒗𝒗1 + 𝛼𝛼2 𝒗𝒗2 +
𝛼𝛼3 𝒗𝒗3 + ⋯ + 𝛼𝛼𝑛𝑛 𝒗𝒗𝑛𝑛 = 𝟎𝟎, and all of the elements of the set {𝛼𝛼1 , 𝛼𝛼2 , 𝛼𝛼3 … 𝛼𝛼𝑛𝑛 } are zero , then the vectors
𝒗𝒗1 , 𝒗𝒗2 , 𝒗𝒗3 … 𝒗𝒗𝑛𝑛 are linearly independent.

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Matrices, Vectors & Fourier analysis Spring 23-24

Example: Write the vector 𝒖𝒖 = (1, −2,5) as a linear combination of the vectors
𝒖𝒖𝟏𝟏 = (1,1,1), 𝒖𝒖𝟐𝟐 = (1,2,3) and 𝒖𝒖𝟑𝟑 = (2, −1,1).
Solution: Let 𝛼𝛼1 𝒖𝒖𝟏𝟏 + 𝛼𝛼2 𝒖𝒖𝟐𝟐 + 𝛼𝛼3 𝒖𝒖𝟑𝟑 = 𝒖𝒖 ; where 𝛼𝛼1 , 𝛼𝛼2 , 𝛼𝛼3 are scalars.
𝛼𝛼1 (1,1,1) + 𝛼𝛼2 (1,2,3) + 𝛼𝛼3 (2, −1,1) = (1, −2,5)
(𝛼𝛼1 + 𝛼𝛼2 + 2𝛼𝛼3 , 𝛼𝛼1 + 2𝛼𝛼2 − 𝛼𝛼3 , 𝛼𝛼1 + 3𝛼𝛼2 + 𝛼𝛼3 ) = (1, −2,5)
Equating corresponding components
𝛼𝛼1 + 𝛼𝛼2 + 2𝛼𝛼3 = 1
𝛼𝛼1 + 2𝛼𝛼2 − 𝛼𝛼3 = −2
𝛼𝛼1 + 3𝛼𝛼2 + 𝛼𝛼3 = 5
Solving using elementary row operation, we get 𝛼𝛼1 = −6, 𝛼𝛼2 = 3 and 𝛼𝛼3 = 2.
∴−6 𝒖𝒖𝟏𝟏 + 3𝒖𝒖𝟐𝟐 + 2𝒖𝒖𝟑𝟑 = 𝒖𝒖
Hence 𝒖𝒖 is a linear combination of the vectors 𝒖𝒖𝟏𝟏 , 𝒖𝒖𝟐𝟐 and 𝒖𝒖𝟑𝟑 .

Example: Test whether the vectors 𝒖𝒖𝟏𝟏 = (1, 0, 1), 𝒖𝒖𝟐𝟐 = (−3, 2, 6) & 𝒖𝒖𝟑𝟑 = (4, 5, −2) are linearly
dependent or independent.
Solution: Let 𝛼𝛼1 𝒖𝒖𝟏𝟏 + 𝛼𝛼2 𝒖𝒖𝟐𝟐 + 𝛼𝛼3 𝒖𝒖𝟑𝟑 = 𝟎𝟎 ; where 𝛼𝛼1 , 𝛼𝛼2 , 𝛼𝛼3 are scalars.
𝛼𝛼1 (1, 0, 1) + 𝛼𝛼2 (−3, 2, 6) + 𝛼𝛼3 (4, 5, −2) = (0, 0, 0)
(𝛼𝛼1 − 3𝛼𝛼2 + 4𝛼𝛼3 , 0 + 2𝛼𝛼2 + 5𝛼𝛼3 , 𝛼𝛼1 + 6𝛼𝛼2 − 2𝛼𝛼3 ) = (0, 0, 0)

We can write from above equation

𝛼𝛼1 − 3𝛼𝛼2 + 4𝛼𝛼3 = 0
0 + 2𝛼𝛼2 + 5𝛼𝛼3 = 0
𝛼𝛼1 + 6𝛼𝛼2 − 2𝛼𝛼3 = 0
Solving the above linear system, we get 𝛼𝛼1 = 0, 𝛼𝛼2 = 0 and 𝛼𝛼3 = 0.
Hence the vectors 𝒖𝒖𝟏𝟏 , 𝒖𝒖𝟐𝟐 and 𝒖𝒖𝟑𝟑 are linearly independent

Sample Exercise-4.3
Write the vector 𝒖𝒖 as a linear combination of the vectors 𝒖𝒖𝟏𝟏 , 𝒖𝒖𝟐𝟐 and 𝒖𝒖𝟑𝟑 , where
1. 𝒖𝒖 = (5, 9, 5) , 𝒖𝒖𝟏𝟏 = (1, −1, 3), 𝒖𝒖𝟐𝟐 = (2, 1, 4), 𝒖𝒖𝟑𝟑 = (3, 2, 5).
2. 𝒖𝒖 = (6, 20, 2) , 𝒖𝒖𝟏𝟏 = (1, 2, 3), 𝒖𝒖𝟐𝟐 = (1, 3, −2), 𝒖𝒖𝟑𝟑 = (1, 4,1).
3. 𝒖𝒖 = (4, 2, 1, 0) , 𝒖𝒖𝟏𝟏 = (3, 1, 0, 1), 𝒖𝒖𝟐𝟐 = (1, 2, 3, 1), 𝒖𝒖𝟑𝟑 = (0, 3, 6, 6).

Test whether the following vectors are linearly independent or dependent.

4. 𝒖𝒖𝟏𝟏 = (2, −1, 4), 𝒖𝒖𝟐𝟐 = (3, 6, 2), 𝒖𝒖𝟑𝟑 = (2, 10, −4)
5. 𝒖𝒖𝟏𝟏 = (3, 0, 1, −1), 𝒖𝒖𝟐𝟐 = (2, −1, 0, 1), 𝒖𝒖𝟑𝟑 = (1, 1, 1, −2)
6. 𝒖𝒖𝟏𝟏 = (1, −1, 2), 𝒖𝒖𝟐𝟐 = (3, −5, 1), 𝒖𝒖𝟑𝟑 = (2, 7,8), 𝒖𝒖𝟒𝟒 = (−1, 1, 1)
7. 𝒖𝒖𝟏𝟏 = (2, 1, 2), 𝒖𝒖𝟐𝟐 = (0, 1, −1), 𝒖𝒖𝟑𝟑 = (4, 3, 3) .
8. 𝒖𝒖𝟏𝟏 = (2, 0, −1), 𝒖𝒖𝟐𝟐 = (1, 1 ,0), 𝒖𝒖𝟑𝟑 = (0, −1, 1)

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Matrices, Vectors & Fourier Analysis Spring 23-24

Chapter-5
Coordinate systems

Cartesian coordinates:

Figure: Differential length, area, and volume in Cartesian coordinates.

Cartesian variables x, y, z

ˆ x + yA
Vector representation, A = xA ˆ y + zA
ˆ z
 
Magnitude of A is , A = Ax 2 + Ay 2 + Az 2

Position vector, �����⃗

𝑂𝑂𝑂𝑂 = 𝑥𝑥�𝑥𝑥1 + 𝑦𝑦�𝑦𝑦1 + 𝑧𝑧̂ 𝑧𝑧1 for P = ( x1 , y1 , z1 )

Base vector properties: xˆ ⋅ xˆ = yˆ ⋅ yˆ = zˆ ⋅ zˆ = 1 and xˆ ⋅ yˆ = yˆ ⋅ xˆ = zˆ ⋅ xˆ = 0

xˆ × yˆ= zˆ , yˆ × zˆ= xˆ , zˆ × xˆ= yˆ
 
Dot product A ⋅ B = Ax Bx + Ay By + Az Bz

xˆ yˆ zˆ
 
Cross product A=
×B Ax Ay Az
= xˆ ( Ay Bz − Az By ) − yˆ ( Ax Bz − Az Bx ) + zˆ ( Ax By − Ay Bx )
Bx By Bz

Differential length, 𝑑𝑑𝑙𝑙⃗ = 𝑥𝑥� 𝑑𝑑𝑑𝑑 + 𝑦𝑦� 𝑑𝑑𝑑𝑑 + 𝑧𝑧̂ 𝑑𝑑𝑑𝑑

Differential surface area, 𝑑𝑑𝑠𝑠⃗𝑥𝑥 = 𝑥𝑥� 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑, 𝑑𝑑𝑠𝑠⃗𝑦𝑦 = 𝑦𝑦� 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑, 𝑑𝑑𝑠𝑠⃗𝑧𝑧 = 𝑧𝑧̂ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
Differential volume, 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

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 Matrices, Vectors & Fourier Analysis Spring 23-24

Cylindrical coordinates:

Figure: Differential length, area, and Figure: Point 𝑃𝑃(𝑟𝑟1 , 𝜙𝜙1 , 𝑧𝑧1 ) in cylindrical coordinates; 𝑟𝑟1 is the
volume in cylindrical coordinates. radial distance from the origin in the 𝑥𝑥𝑥𝑥 plane, 𝜙𝜙1 is the angle
in 𝑥𝑥𝑥𝑥 plane measured from the 𝑥𝑥 axis toward the 𝑦𝑦 axis, and
𝑧𝑧1 is the vertical distance from the 𝑥𝑥𝑥𝑥 plane.

Cylindrical variables r , φ , z

Vector representation, A =rA ˆ r + φˆ Aφ + zA
ˆ z
 
Magnitude of A is , A = Ar 2 + Aφ 2 + Az 2

Position vector, �����⃗

𝑂𝑂𝑂𝑂 = 𝑟𝑟̂ 𝑟𝑟1 + 𝑧𝑧̂ 𝑧𝑧1 for P = (r1 , φ1 , z1 )

Base vector properties: rˆ ⋅ rˆ = φˆ ⋅ φˆ = zˆ ⋅ zˆ = 1 and rˆ ⋅ φˆ = φˆ ⋅ zˆ = zˆ ⋅ rˆ = 0

𝑟𝑟̂ × 𝜙𝜙� = 𝑧𝑧̂ , 𝜙𝜙� × 𝑧𝑧̂ = 𝑟𝑟̂ , 𝑧𝑧̂ × 𝑟𝑟̂ = 𝜙𝜙�

 
Dot product A ⋅ B= Ar Br + Aφ Bφ + Az Bz

Cross product
𝑟𝑟̂ 𝜙𝜙� 𝑧𝑧̂
⃑ �⃑
𝐴𝐴 × 𝐵𝐵 = �𝐴𝐴𝑟𝑟 𝐴𝐴𝜙𝜙 𝐴𝐴𝑧𝑧 � = 𝑟𝑟̂ �𝐴𝐴𝜙𝜙 𝐵𝐵𝑧𝑧 − 𝐴𝐴𝑧𝑧 𝐵𝐵𝜙𝜙 � − 𝜙𝜙� (𝐴𝐴𝑟𝑟 𝐵𝐵𝑧𝑧 − 𝐴𝐴𝑧𝑧 𝐵𝐵𝑟𝑟 ) + 𝑧𝑧̂ �𝐴𝐴𝑟𝑟 𝐵𝐵𝜙𝜙 − 𝐴𝐴𝜙𝜙 𝐵𝐵𝑟𝑟 �
𝐵𝐵𝑟𝑟 𝐵𝐵𝜙𝜙 𝐵𝐵𝑧𝑧

Differential length, 𝑑𝑑𝑙𝑙⃗ = 𝑟𝑟̂ 𝑑𝑑𝑑𝑑 + 𝜙𝜙� 𝑟𝑟𝑟𝑟𝑟𝑟 + 𝑧𝑧̂ 𝑑𝑑𝑑𝑑

Differential surface area, 𝑑𝑑𝑠𝑠⃗𝑟𝑟 = 𝑟𝑟̂ 𝑟𝑟𝑟𝑟𝑟𝑟 𝑑𝑑𝑑𝑑, 𝑑𝑑𝑠𝑠⃗𝜙𝜙 = 𝜙𝜙� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑, 𝑑𝑑𝑠𝑠⃗𝑧𝑧 = 𝑧𝑧̂ 𝑑𝑑𝑑𝑑 𝑟𝑟𝑟𝑟𝑟𝑟
Differential volume, 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 𝑟𝑟𝑟𝑟𝑟𝑟 𝑑𝑑𝑑𝑑

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Spherical coordinates:

Figure: point 𝑃𝑃(𝑅𝑅1 , 𝜃𝜃1 , 𝜙𝜙1 ) in spherical Figure: Differential volume in spherical
coordinates. coordinates.

R, θ , φ

ˆ + θˆ A + φˆ A
A = RA r θ φ

 
Magnitude of A is , A = AR 2 + Aθ 2 + Aφ 2

Position vector, �����⃗

𝑂𝑂𝑂𝑂 = 𝑅𝑅�𝑅𝑅1 for P = ( R1 , θ1 , φ1 )

Base vector properties: Rˆ ⋅ Rˆ = θˆ ⋅ θˆ = φˆ ⋅ φˆ = 1 and Rˆ ⋅ θˆ = θˆ ⋅ φˆ = φˆ ⋅ Rˆ = 0

Rˆ × θˆ= φˆ, θˆ ×φˆ= Rˆ , φˆ × Rˆ= θˆ

 
Dot product A ⋅ = B AR BR + Aθ Bθ + Aφ Bφ

Cross product

Rˆ θˆ φˆ
 
A=
×B AR Aθ Aφ
= Rˆ ( Aθ Bφ − Aφ Bθ ) − θˆ ( AR Bφ − Aφ BR ) + φˆ ( AR Bθ − Aθ BR )
BR Bθ Bφ

Differential length, 𝑑𝑑𝑙𝑙⃗ = 𝑅𝑅� 𝑑𝑑𝑑𝑑 + 𝜃𝜃� 𝑅𝑅 𝑑𝑑𝑑𝑑 + 𝜙𝜙� 𝑅𝑅 sin 𝜃𝜃 𝑑𝑑𝑑𝑑
Differential surface area, 𝑑𝑑𝑠𝑠⃗𝑅𝑅 = 𝑅𝑅� 𝑅𝑅𝑅𝑅𝑅𝑅 𝑅𝑅 sin 𝜃𝜃 𝑑𝑑𝑑𝑑, 𝑑𝑑𝑠𝑠⃗𝜃𝜃 = 𝜃𝜃� 𝑑𝑑𝑑𝑑 𝑅𝑅 sin 𝜃𝜃 𝑑𝑑𝑑𝑑,
𝑑𝑑𝑠𝑠⃗𝜙𝜙 = 𝜙𝜙� 𝑑𝑑𝑑𝑑 𝑅𝑅𝑅𝑅𝑅𝑅
Differential volume, 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 𝑅𝑅 𝑑𝑑𝑑𝑑 𝑅𝑅 sin 𝜃𝜃 𝑑𝑑𝑑𝑑

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Transformation (Cartesian coordinates to cylindrical and vice versa):

Figure: Interrelationships between Figure: Interrelationships between base vectors

Cartesian coordinates (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) and (𝑥𝑥�, 𝑦𝑦�) and (𝑟𝑟̂ , 𝜙𝜙�).
cylindrical coordinates(𝑟𝑟, 𝜙𝜙, 𝑧𝑧).

We know,
= cos φ , y r sin φ and z = z
x r=

 y
r = x 2 + y 2 , φ =arctan   , z =z ; 0 ≤ r < ∞ , 0 ≤ φ < 2π , − ∞ < z < ∞
x
𝐴𝐴𝑟𝑟 𝐴𝐴𝜙𝜙 𝐴𝐴𝑧𝑧
𝑟𝑟̂ 𝜙𝜙� 𝑧𝑧̂
𝐴𝐴𝑥𝑥 𝑥𝑥� cos 𝜙𝜙 − sin 𝜙𝜙 0
𝐴𝐴𝑦𝑦 𝑦𝑦� sin 𝜙𝜙 cos 𝜙𝜙 0
𝐴𝐴𝑧𝑧 𝑧𝑧̂ 0 0 1

xˆ cos φ + yˆ sin φ , φˆ =
That is, rˆ = − xˆ sin φ + yˆ cos φ and zˆ = z.
ˆ

Ax cos φ + Ay sin φ , Aφ =
Ar = − Ax sin φ + Ay cos φ and Az =
Az .

𝑥𝑥� = r� cos 𝜙𝜙 − 𝜙𝜙� sin 𝜙𝜙, 𝑦𝑦� = 𝑟𝑟̂ sin 𝜙𝜙 + 𝜙𝜙� cos 𝜙𝜙, 𝑧𝑧̂ = 𝑧𝑧̂
𝐴𝐴𝑥𝑥 = 𝐴𝐴𝑟𝑟 cos 𝜙𝜙 − 𝐴𝐴𝜑𝜑 sin 𝜙𝜙, 𝐴𝐴𝑦𝑦 = 𝐴𝐴𝑟𝑟 sin 𝜙𝜙 + 𝐴𝐴𝜙𝜙 cos 𝜙𝜙, 𝐴𝐴𝑧𝑧 = 𝐴𝐴𝑧𝑧

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Transformation (Spherical coordinates to cylindrical coordinates and vice versa):

We know,
= sin θ , z R cos θ and φ = φ
r R=

r
R = r 2 + z 2 , θ =arctan   , φ =φ ; 0 ≤ r < ∞ , 0 ≤ φ < 2π , − ∞ < z < ∞
z
𝐴𝐴𝑅𝑅 𝐴𝐴𝜃𝜃 𝐴𝐴𝜙𝜙

𝑅𝑅� 𝜃𝜃� 𝜙𝜙�

𝐴𝐴𝑟𝑟 𝑟𝑟̂ sin 𝜃𝜃 cos 𝜃𝜃 0
𝐴𝐴𝑧𝑧 𝑧𝑧̂ cos 𝜃𝜃 −sin 𝜃𝜃 0
𝐴𝐴𝜙𝜙 𝜙𝜙� 0 0 1

Rˆ sin θ + θˆ cos θ , zˆ =
That is, rˆ = Rˆ cos θ − θˆ sin θ and φˆ =
φˆ .
AR sin θ + Aθ cos θ , Az =
Ar = AR cos θ − Aθ sin θ and Aφ =
Aφ .

𝑅𝑅� = r� sin 𝜃𝜃 − 𝑧𝑧̂ cos 𝜃𝜃, 𝜃𝜃� = 𝑟𝑟̂ cos 𝜃𝜃 − 𝑧𝑧̂ sin 𝜃𝜃, 𝜙𝜙� = 𝜙𝜙
𝐴𝐴𝑅𝑅 = 𝐴𝐴𝑟𝑟 sin 𝜃𝜃 − 𝐴𝐴𝑧𝑧 cos 𝜃𝜃, 𝐴𝐴𝜃𝜃 = 𝐴𝐴𝑟𝑟 cos 𝜃𝜃 − 𝐴𝐴𝑧𝑧 sin 𝜃𝜃, 𝐴𝐴𝜙𝜙 = 𝐴𝐴𝜙𝜙

Transformation (spherical coordinates to Cartesian coordinates and vice versa):

We know, x R=
= sin θ cos φ , y R sin θ sin φ and z = R cos θ ;
�𝑥𝑥 2 +𝑦𝑦2 𝑦𝑦
𝑅𝑅 = �𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 , 𝜃𝜃 = arctan � � , 𝜙𝜙 = 𝑎𝑎𝑎𝑎𝑎𝑎 𝑡𝑡𝑡𝑡𝑡𝑡 �𝑥𝑥 � ;
𝑧𝑧

0 ≤ r < ∞ , 0 ≤ φ < 2π , − ∞ < z < ∞

𝐴𝐴𝑅𝑅 𝐴𝐴𝜃𝜃 𝐴𝐴𝜙𝜙

𝑅𝑅� 𝜃𝜃� 𝜙𝜙�
𝐴𝐴𝑥𝑥 𝑥𝑥� sin 𝜃𝜃 cos 𝜙𝜙 cos 𝜃𝜃 cos 𝜙𝜙 − sin 𝜙𝜙
𝐴𝐴𝑦𝑦 𝑦𝑦� sin 𝜃𝜃 sin 𝜙𝜙 cos 𝜃𝜃 sin 𝜙𝜙 cos 𝜙𝜙
𝐴𝐴𝑧𝑧 𝑧𝑧̂ cos 𝜃𝜃 − sin 𝜃𝜃 0
That is,
xˆ = Rˆ sin θ cos φ + θˆ cos θ cos φ − φˆ sin φ , yˆ = Rˆ sin θ sin φ + θˆ cos θ sin φ + φˆ cos φ
and zˆ Rˆ cos θ − θˆ sin θ .
=
Ax = AR sin θ cos φ + Aθ cos θ cos φ − Aφ sin φ ,
Ay = AR sin θ sin φ + Aθ cos θ sin φ + Aφ cos φ
and Az AR cos θ − Aθ sin θ ..
=

𝑅𝑅� = 𝑥𝑥� sin 𝜃𝜃 cos 𝜙𝜙 + 𝑦𝑦� sin 𝜃𝜃 sin 𝜙𝜙 + 𝑧𝑧̂ cos 𝜃𝜃

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𝜃𝜃� = 𝑥𝑥� cos 𝜃𝜃 cos 𝜙𝜙 + 𝑦𝑦� cos 𝜃𝜃 sin 𝜙𝜙 − 𝑧𝑧̂ sin 𝜃𝜃

𝜙𝜙� = −𝑥𝑥� sin 𝜙𝜙 + 𝑦𝑦� cos 𝜙𝜙
𝐴𝐴𝑅𝑅 = 𝐴𝐴𝑥𝑥 sin 𝜃𝜃 cos 𝜙𝜙 + 𝐴𝐴𝑦𝑦 sin 𝜃𝜃 sin 𝜙𝜙 + 𝐴𝐴𝑧𝑧 cos 𝜃𝜃
𝐴𝐴𝜃𝜃 = 𝐴𝐴𝑥𝑥 cos 𝜃𝜃 cos 𝜙𝜙 + 𝐴𝐴𝑦𝑦 cos 𝜃𝜃 sin 𝜙𝜙 − 𝐴𝐴𝑧𝑧 sin 𝜃𝜃
𝐴𝐴𝜙𝜙 = −𝐴𝐴𝑥𝑥 sin 𝜙𝜙 + 𝐴𝐴𝑦𝑦 cos 𝜙𝜙

Figure: Interrelationships between (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) and (𝑅𝑅, 𝜃𝜃, 𝜙𝜙)

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Matrices, Vectors & Fourier Analysis Spring 23-24

3𝜋𝜋
Example 1: Transform �√2, , 3� from cylindrical coordinates to Cartesian coordinates.
4
3𝜋𝜋
Solution: 𝑥𝑥 = 𝑟𝑟 cos 𝜙𝜙 = √2 cos = −1,
4
3𝜋𝜋
𝑦𝑦 = 𝑟𝑟 sin 𝜙𝜙 = √2 sin = 1,
4

𝑧𝑧 = 𝑧𝑧 = 3
So, Cartesian point (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = (−1,1,3)

Example 2: Transform (−1,1,3) from Cartesian coordinate to cylindrical coordinate.

Solution: 𝑟𝑟 = �𝑥𝑥 2 + 𝑦𝑦 2 = �(−1)2 + 12 = √2
1 3𝜋𝜋
𝜙𝜙 = tan−1 � � = tan−1 (−1) =
−1 4
𝑧𝑧 = 𝑧𝑧 = 3

Example 3: Transform �1,0, √3� from Cartesian coordinate to spherical coordinate.

Solution: 𝑅𝑅 = �𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 = �12 + 02 + (√3)2 = √4 = 2

�𝑥𝑥 2 + 𝑦𝑦 2 √12 + 02 𝜋𝜋
𝜃𝜃 = tan−1 � � = 𝑡𝑡𝑡𝑡𝑛𝑛−1 � �=
𝑧𝑧 (√3)2 6
𝑦𝑦 0
𝜙𝜙 = tan−1 = tan−1 = 0
𝑥𝑥 1
𝜋𝜋
So, the spherical point (𝑅𝑅, 𝜃𝜃, 𝜙𝜙) = �2, , 0�
6

𝜋𝜋
Example 4: Transform �2, , 0� from spherical coordinate to Cartesian coordinate.
6
𝜋𝜋
Solution: 𝑥𝑥 = 𝑅𝑅 sin 𝜃𝜃 cos 𝜙𝜙 = 2 sin cos 0 = 1
6
𝜋𝜋
𝑦𝑦 = 𝑅𝑅 sin 𝜃𝜃 sin 𝜙𝜙 = 2 sin sin 0 = 0
6
𝜋𝜋
𝑧𝑧 = 𝑅𝑅 cos 𝜃𝜃 = 2 cos = √3
6
So, the Cartesian point (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = �1,0, √3�

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𝜋𝜋
Example 5: Transform �2, , 0� from spherical coordinate to cylindrical coordinate.
6
𝜋𝜋
Solution: 𝑟𝑟 = 𝑅𝑅 sin 𝜃𝜃 = 2 sin = 1
6
𝜋𝜋
𝑧𝑧 = 𝑅𝑅 cos 𝜃𝜃 = 2 cos = √3
6
𝜙𝜙 = 𝜙𝜙 = 0
So, the cylindrical pint (𝑟𝑟, 𝑧𝑧, 𝜙𝜙) = �1, √3, 0�
Example 6: Transform �1,0, √3� from cylindrical coordinate to spherical coordinate.

2
Solution: 𝑅𝑅 = √𝑟𝑟 2 + 𝑧𝑧 2 = �12 + �√3� = 2

𝑟𝑟 1 𝜋𝜋
𝜃𝜃 = tan−1 � � = 𝑡𝑡𝑡𝑡𝑛𝑛−1 � � =
𝑧𝑧 √3 6
𝜙𝜙 = 𝜙𝜙 = 0
𝜋𝜋
So, the spherical point (𝑅𝑅, 𝜃𝜃, 𝜙𝜙) = �2, , 0�
6

Example 7: Express vector 𝐴𝐴⃗ = 𝑥𝑥�(𝑥𝑥 + 𝑦𝑦) + 𝑦𝑦�(𝑦𝑦 − 𝑥𝑥) + 𝑧𝑧̂ 𝑧𝑧 in spherical coordinate.
Solution: We know in spherical coordinate
𝐴𝐴⃗ = 𝑅𝑅� 𝐴𝐴𝑅𝑅 + 𝜃𝜃�𝐴𝐴𝜃𝜃 + 𝜙𝜙�𝐴𝐴𝜙𝜙
𝐴𝐴𝑅𝑅 = 𝐴𝐴𝑥𝑥 sin 𝜃𝜃 cos 𝜙𝜙 + 𝐴𝐴𝑦𝑦 sin 𝜃𝜃 sin 𝜙𝜙 + 𝐴𝐴𝑧𝑧 cos 𝜃𝜃
= (𝑥𝑥 + 𝑦𝑦) sin 𝜃𝜃 cos 𝜙𝜙 + (𝑦𝑦 − 𝑥𝑥) sin 𝜃𝜃 sin 𝜙𝜙 + 𝑧𝑧 cos 𝜃𝜃
= (𝑅𝑅 sin 𝜃𝜃 cos 𝜙𝜙 + 𝑅𝑅 sin 𝜃𝜃 sin 𝜙𝜙) sin 𝜃𝜃 cos 𝜙𝜙
+ ( 𝑅𝑅 sin 𝜃𝜃 sin 𝜙𝜙 − 𝑅𝑅 sin 𝜃𝜃 cos 𝜙𝜙) sin 𝜃𝜃 sin 𝜙𝜙 + 𝑅𝑅 cos 𝜃𝜃 cos 𝜃𝜃
= 𝑅𝑅 sin2 𝜃𝜃 cos2 𝜙𝜙 + 𝑅𝑅 sin2 𝜃𝜃 sin 𝜙𝜙 cos 𝜙𝜙 + 𝑅𝑅 sin2 𝜃𝜃 sin2 𝜙𝜙 − 𝑅𝑅 sin2 𝜃𝜃 sin 𝜙𝜙 cos 𝜙𝜙
+ 𝑅𝑅 cos 2 𝜃𝜃
= 𝑅𝑅 sin2 𝜃𝜃 (cos 2 𝜙𝜙 + sin2 𝜙𝜙) + 𝑅𝑅 cos2 𝜃𝜃
= 𝑅𝑅 sin2 𝜃𝜃 + 𝑅𝑅 cos 2 𝜃𝜃 = 𝑅𝑅
𝐴𝐴𝜃𝜃 = 𝐴𝐴𝑥𝑥 cos 𝜃𝜃 cos 𝜙𝜙 + 𝐴𝐴𝑦𝑦 cos 𝜃𝜃 sin 𝜙𝜙 − 𝐴𝐴𝑧𝑧 sin 𝜃𝜃
= (𝑥𝑥 + 𝑦𝑦) cos 𝜃𝜃 cos 𝜙𝜙 + (𝑦𝑦 − 𝑥𝑥) cos 𝜃𝜃 sin 𝜙𝜙 − 𝑧𝑧 sin 𝜃𝜃
= (𝑅𝑅 sin 𝜃𝜃 cos 𝜙𝜙 + 𝑅𝑅 sin 𝜃𝜃 sin 𝜙𝜙) cos 𝜃𝜃 cos 𝜙𝜙
+ ( 𝑅𝑅 sin 𝜃𝜃 sin 𝜙𝜙 − 𝑅𝑅 sin 𝜃𝜃 cos 𝜙𝜙) cos 𝜃𝜃 sin 𝜙𝜙 − 𝑅𝑅 sin 𝜃𝜃 cos 𝜃𝜃
= 𝑅𝑅 sin 𝜃𝜃 cos 𝜃𝜃 cos 2 𝜙𝜙 + 𝑅𝑅 sin 𝜃𝜃 cos 𝜃𝜃 sin 𝜙𝜙 cos 𝜙𝜙 + 𝑅𝑅 sin 𝜃𝜃 cos 𝜃𝜃 sin2 𝜙𝜙
− 𝑅𝑅 sin 𝜃𝜃 cos 𝜃𝜃 sin 𝜙𝜙 cos 𝜙𝜙 − 𝑅𝑅 sin 𝜃𝜃 cos 𝜃𝜃
= 𝑅𝑅 sin 𝜃𝜃 cos 𝜃𝜃 (sin2 𝜙𝜙 + cos2 𝜙𝜙) − 𝑅𝑅 sin 𝜃𝜃 cos 𝜃𝜃
=0

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𝐴𝐴𝜙𝜙 = −𝐴𝐴𝑥𝑥 sin 𝜙𝜙 + 𝐴𝐴𝑦𝑦 cos 𝜙𝜙

= −(𝑥𝑥 + 𝑦𝑦) sin 𝜙𝜙 + (𝑦𝑦 − 𝑥𝑥) cos 𝜙𝜙
= −(𝑅𝑅 sin 𝜃𝜃 cos 𝜙𝜙 + 𝑅𝑅 sin 𝜃𝜃 sin 𝜙𝜙) sin 𝜙𝜙 + ( 𝑅𝑅 sin 𝜃𝜃 sin 𝜙𝜙 − 𝑅𝑅 sin 𝜃𝜃 cos 𝜙𝜙) cos 𝜙𝜙
= −𝑅𝑅 sin 𝜃𝜃 sin 𝜙𝜙 cos 𝜙𝜙 − 𝑅𝑅 sin 𝜃𝜃 sin2 𝜙𝜙 + 𝑅𝑅 sin 𝜃𝜃 sin 𝜙𝜙 cos 𝜙𝜙 − 𝑅𝑅 sin 𝜃𝜃 cos2 𝜙𝜙
= −𝑅𝑅 sin 𝜃𝜃(sin2 𝜙𝜙 + cos2 𝜙𝜙) = −𝑅𝑅 sin 𝜃𝜃
∴ 𝐴𝐴⃗ = 𝑅𝑅� 𝑅𝑅 − 𝜙𝜙� 𝑅𝑅 sin 𝜃𝜃
Example 8: Express vector 𝐴𝐴⃗ = 𝑅𝑅� 𝑅𝑅 − 𝜙𝜙� 𝑅𝑅 sin 𝜃𝜃 in Cartesian form.
Solution: We know in Cartesian coordinate 𝐴𝐴⃗ = 𝑥𝑥�𝐴𝐴𝑥𝑥 + 𝑦𝑦�𝐴𝐴𝑦𝑦 + 𝑧𝑧̂ 𝐴𝐴𝑧𝑧
𝐴𝐴𝑥𝑥 = 𝐴𝐴𝑅𝑅 sin 𝜃𝜃 cos 𝜙𝜙 + 𝐴𝐴𝜃𝜃 cos 𝜃𝜃 cos 𝜙𝜙 – 𝐴𝐴𝜙𝜙 sin 𝜙𝜙
= 𝑅𝑅 sin 𝜃𝜃 cos 𝜙𝜙 + 0. cos 𝜃𝜃 cos 𝜙𝜙 + 𝑅𝑅 sin 𝜃𝜃 sin 𝜙𝜙
= 𝑥𝑥 + 𝑦𝑦
𝐴𝐴𝑦𝑦 = 𝐴𝐴𝑅𝑅 sin 𝜃𝜃 sin 𝜙𝜙 + 𝐴𝐴𝜃𝜃 cos 𝜃𝜃 sin 𝜙𝜙 + 𝐴𝐴𝜙𝜙 cos 𝜙𝜙
= 𝑅𝑅 sin 𝜃𝜃 sin 𝜙𝜙 + 0. cos 𝜃𝜃 sin 𝜙𝜙 – 𝑅𝑅 sin 𝜃𝜃 cos 𝜙𝜙
= 𝑦𝑦 − 𝑥𝑥
𝐴𝐴𝑧𝑧 = 𝐴𝐴𝑅𝑅 cos 𝜃𝜃 − 𝐴𝐴𝜃𝜃 sin 𝜃𝜃
= 𝑅𝑅 cos 𝜃𝜃 − 0. sin 𝜃𝜃
= 𝑧𝑧
∴ 𝐴𝐴⃗ = 𝑥𝑥�(𝑥𝑥 + 𝑦𝑦) + 𝑦𝑦�(𝑦𝑦 − 𝑥𝑥) + 𝑧𝑧̂ 𝑧𝑧
Example 9: Transform vector 𝐴𝐴⃗ = 𝑥𝑥�(𝑥𝑥 + 𝑦𝑦) + 𝑦𝑦�(𝑦𝑦 − 𝑥𝑥) + 𝑧𝑧̂ 𝑧𝑧 to cylindrical coordinates.
Solution: We know, cylindrical coordinate 𝐴𝐴⃗ = 𝑟𝑟̂ 𝐴𝐴𝑟𝑟 + 𝜙𝜙�𝐴𝐴𝜙𝜙 + 𝑧𝑧̂ 𝐴𝐴𝑧𝑧
𝐴𝐴𝑟𝑟 = 𝐴𝐴𝑥𝑥 cos 𝜙𝜙 + 𝐴𝐴𝑦𝑦 sin 𝜙𝜙 = (𝑥𝑥 + 𝑦𝑦) cos 𝜙𝜙 + (𝑦𝑦 − 𝑥𝑥) sin 𝜙𝜙
= (𝑟𝑟 cos 𝜙𝜙 + 𝑟𝑟 sin 𝜙𝜙) cos 𝜙𝜙 + (𝑟𝑟 sin 𝜙𝜙 − 𝑟𝑟 cos 𝜙𝜙) sin 𝜙𝜙
= 𝑟𝑟 cos2 𝜙𝜙 + 𝑟𝑟 sin 𝜙𝜙 cos 𝜙𝜙 + 𝑟𝑟 sin2 𝜙𝜙 − 𝑟𝑟 sin 𝜙𝜙 cos 𝜙𝜙
= 𝑟𝑟(cos 2 𝜙𝜙 + sin2 𝜙𝜙) = 𝑟𝑟
𝐴𝐴𝜙𝜙 = −𝐴𝐴𝑥𝑥 sin 𝜙𝜙 + 𝐴𝐴𝑦𝑦 cos 𝜙𝜙
= −(𝑥𝑥 + 𝑦𝑦) sin 𝜙𝜙 + (𝑦𝑦 − 𝑥𝑥) cos 𝜙𝜙
= −(𝑟𝑟 cos 𝜙𝜙 + 𝑟𝑟 sin 𝜙𝜙) sin 𝜙𝜙 + (𝑟𝑟 sin 𝜙𝜙 − 𝑟𝑟 cos 𝜙𝜙) cos 𝜙𝜙
= −𝑟𝑟 cos 𝜙𝜙 sin 𝜙𝜙 − 𝑟𝑟 sin2 𝜙𝜙 + 𝑟𝑟 cos 𝜙𝜙 sin 𝜙𝜙 − 𝑟𝑟 𝑐𝑐𝑐𝑐𝑠𝑠 2 𝜙𝜙
= −𝑟𝑟(sin2 𝜙𝜙 + 𝑐𝑐𝑐𝑐𝑠𝑠 2 𝜙𝜙)
= −𝑟𝑟
𝐴𝐴𝑧𝑧 = 𝑧𝑧
∴ 𝐴𝐴⃗ = 𝑟𝑟̂ 𝑟𝑟 − 𝜙𝜙�𝑟𝑟 + 𝑧𝑧̂ 𝑧𝑧

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Matrices, Vectors & Fourier Analysis Spring 23-24

Example 10: Express vector 𝐴𝐴⃗ = r� 𝑟𝑟 − ϕ� r + z� 𝑧𝑧 in Cartesian coordinate.

Solution: We know in Cartesian coordinate 𝐴𝐴⃗ = 𝑥𝑥�𝐴𝐴𝑥𝑥 + 𝑦𝑦�𝐴𝐴𝑦𝑦 + 𝑧𝑧̂ 𝐴𝐴𝑧𝑧
𝐴𝐴𝑥𝑥 = 𝐴𝐴𝑟𝑟 cos 𝜑𝜑 − 𝐴𝐴𝜑𝜑 sin 𝜑𝜑
= 𝑟𝑟 cos 𝜑𝜑 + r sin 𝜑𝜑
= 𝑥𝑥 + 𝑦𝑦
𝐴𝐴𝑦𝑦 = 𝐴𝐴𝑟𝑟 sin 𝜑𝜑 + 𝐴𝐴𝜑𝜑 cos 𝜑𝜑
= 𝑟𝑟 sin 𝜑𝜑 − 𝑟𝑟 cos 𝜑𝜑
= 𝑦𝑦 − 𝑥𝑥
𝐴𝐴𝑧𝑧 = 𝑧𝑧
Hence, ∴ 𝐴𝐴⃗ = x� (𝑥𝑥 + 𝑦𝑦) + y� (𝑦𝑦 − 𝑥𝑥 ) + z� 𝑧𝑧
Example 11: Transform the vector 𝐴𝐴⃗ = 𝑟𝑟̂ 3 cos 𝜙𝜙 − 𝜙𝜙�2𝑟𝑟 + 𝑧𝑧̂ 𝑧𝑧 to spherical coordinate.
Solution: We know in spherical coordinate 𝐴𝐴⃗ = 𝑅𝑅� 𝐴𝐴𝑅𝑅 + 𝜃𝜃�𝐴𝐴𝜃𝜃 + 𝜙𝜙�𝐴𝐴𝜙𝜙
𝐴𝐴𝑅𝑅 = 𝐴𝐴𝑟𝑟 sin 𝜃𝜃 + 𝐴𝐴𝑧𝑧 cos 𝜃𝜃
= 3 cos 𝜙𝜙 sin 𝜃𝜃 + 𝑧𝑧 cos 𝜃𝜃 = 3 cos 𝜙𝜙 sin 𝜃𝜃 + 𝑅𝑅 cos 𝜃𝜃 cos 𝜃𝜃
𝐴𝐴𝜃𝜃 = 𝐴𝐴𝑟𝑟 cos 𝜃𝜃 − 𝐴𝐴𝑧𝑧 sin 𝜃𝜃
= 3 cos 𝜙𝜙 cos 𝜃𝜃 − 𝑧𝑧 sin 𝜃𝜃 = 3 cos 𝜙𝜙 cos 𝜃𝜃 − 𝑅𝑅 cos 𝜃𝜃 sin 𝜃𝜃
𝐴𝐴𝜙𝜙 = 𝐴𝐴𝜙𝜙 = −2𝑟𝑟 = −2𝑅𝑅 sin 𝜃𝜃

∴ 𝐴𝐴⃗ = 𝑅𝑅� (3 cos 𝜙𝜙 sin 𝜃𝜃 + 𝑅𝑅 cos 2 𝜃𝜃) − 𝜃𝜃�2𝑅𝑅 sin 𝜃𝜃 + 𝜙𝜙�(3 cos 𝜙𝜙 cos 𝜃𝜃 − 𝑅𝑅 cos 𝜃𝜃 sin 𝜃𝜃)
 π 
Example 12: Transform the vector Aˆ = Rˆ cos φ + θˆ sin φ + φˆ sin 2 θ at the point  3, , π  to
 2 
cylindrical coordinates.
 π 
Solution: Given that, Aˆ = Rˆ cos φ + θˆ sin φ + φˆ sin 2 θ at the point  3, , π  .
 2 
π
Here,= θ
R 3, = φ π then r =Rsin
,= = θ 3,=
φ π and z =Rcos
= θ 0
2
 π 
We get,  3, , π  → ( 3, π , 0 )
 2 
Now, from given equation 𝐴𝐴𝑅𝑅 = cos 𝜑𝜑 = −1, 𝐴𝐴𝜃𝜃 = sin 𝜑𝜑 = 0, 𝐴𝐴𝜑𝜑 = sin2 𝜃𝜃 = 1
We know,
Aˆ = Ar rˆ + Aφφˆ + Az zˆ
AR sin θ + Aθ cos θ =
∴ Ar = −1, Aφ = AR cos θ − Aθ sin θ =
1 and Az =
Aφ = 0

Hence, Aˆ =−rˆ + φˆ .

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 Matrices, Vectors & Fourier Analysis Spring 23-24

  π 
Example 13: Transform the vector A = rˆ cos φ − φˆ sin φ + zˆ sin φ cos φ at the point  2, , 2  .
 4 
to spherical coordinates.
  π 
Solution: Given that , A = rˆ cos φ − φˆ sin φ + zˆ sin φ cos φ at the point  2, , 2  .
 4 
π
cos φ , Aφ =
Here, Ar = − sin φ , and Az =
cos φ sin φ and r =2, φ = , z=
2
4
r 2 π π
We get, R = r2 + z2 = 4 + 4= 8 , θ = tan −1  = tan −1   = , φ=
z 2 4 4

 π π
∴ ( R, θ , φ ) = 8, , 
 4 4

We know, A = AR Rˆ + Aθ θˆ + Aφφˆ

Ar sin θ + Az cos θ , Aθ =
Here, AR = Ar cos θ − Az sin θ and Aφ =
Aφ .

AR = cos φ sin θ + cos φ sin φ cos θ , Aθ = Ar cos φ cos θ − cos φ sin φ sin θ and Aφ = − sin φ .

∴ A ( cos φ sin θ + cos φ sin φ cos θ ) Rˆ + ( Ar cos φ cos θ − cos φ sin φ sin θ ) θˆ + ( − sin φ ) φˆ
=

 π π  1 ˆ
At the point  8, ,  A=.853Rˆ + 0.1464θˆ − φ.
 4 4 2

Sample exercise-5.1
1. Convert the coordinates of the following points from Cartesian to cylindrical and
spherical coordinates:
2𝜋𝜋 5𝜋𝜋 2𝜋𝜋
a. 𝑝𝑝1 = (−1, √3, −2√3) Ans: �2, , −2√3� and �4, , �.
3 6 3
3𝜋𝜋
b. 𝑝𝑝2 = (4, 0, −4) Ans: (4, 0, −4) and �4√2, , 0�.
4
7𝜋𝜋 𝜋𝜋 7𝜋𝜋
c. 𝑝𝑝3 = (√8, −√8, 4) Ans: �4, , 4� and �4√2, , �.
4 4 4

2. Convert the coordinates of the following points from cylindrical to Cartesian and
spherical coordinates:
2𝜋𝜋 𝜋𝜋 2𝜋𝜋
a. 𝑝𝑝1 = (2, , 2√3) Ans: �−1, √3, 2√3� and �4, , �.
3 6 3
2𝜋𝜋
b. 𝑝𝑝2 = (√3, 00 , −1) Ans: (√3, 0, −1) and �2, , 0�.
3
2𝜋𝜋
c. 𝑝𝑝3 = (4√3, 𝜋𝜋, −4) Ans: (−4√3, 0, −4) and �8, , 𝜋𝜋�.
3

Page 11 of 12
Matrices, Vectors & Fourier Analysis Spring 23-24

3. Convert the coordinates of the following points from spherical to Cartesian and
cylindrical coordinates:
3𝜋𝜋 5𝜋𝜋 3√2 3√6 −3 3 5𝜋𝜋 −3
a. 𝑝𝑝1 = (3, , ) Ans: � ,− , � and � , , �.
4 3 4 4 √2 √2 3 √2
𝜋𝜋 𝜋𝜋 𝜋𝜋
b. 𝑝𝑝2 = (4, , ) Ans: (2√2, 2√2, 0) and �4, , 0�.
2 4 4
𝜋𝜋 𝜋𝜋 1 1 1 𝜋𝜋 1
c. 𝑝𝑝3 = (1, , ) Ans: �0, , � and � , , �.
4 2 √2 √2 √2 2 √2

4. Transform the following vectors into cylindrical coordinates at the indicated

points:
� − 4 z�.
a. 𝑨𝑨 = x� (𝑥𝑥 + 𝑦𝑦) + y� (𝑥𝑥 + 𝑦𝑦) + z� z at 𝑝𝑝 = (4, 0, −4) Ans: 𝑨𝑨 = 4 r� + 4 ϕ
b. 𝑩𝑩 = R � cos 2 𝜙𝜙 at 𝑝𝑝 = (2, 𝜋𝜋 , 𝜋𝜋) Ans: 𝑩𝑩 = r� + 1 ϕ
� sin 𝜃𝜃 + θ� cos 𝜃𝜃 + ϕ �.
2 4 2

5. Transform the following vectors into spherical coordinates at the indicated

points:
a. 𝑨𝑨 = x� 𝑦𝑦 − y� 𝑥𝑥 at 𝑝𝑝 = (1, −1, 0) �.
Ans: 𝑨𝑨 = −√2 ϕ
𝜋𝜋 1
b. 𝑩𝑩 = z� sin 𝜙𝜙 at 𝑝𝑝 = (2, , 2) Ans: 𝑩𝑩 = R� − 1 θ.
�
4 2 2

6. Transform the following vectors into cartesian coordinates at the indicated

points:
� 𝑅𝑅 sin 𝜃𝜃 at 𝑝𝑝 = (1, 𝜋𝜋 , 0)
a. 𝑨𝑨 = −ϕ Ans: 𝑨𝑨 = −y� .
2
2𝜋𝜋 1 √3
� sin 𝜙𝜙 + z� z at 𝑝𝑝 = (2,
b. 𝑩𝑩 = −r� cos 𝜙𝜙 − ϕ , 2√3) Ans: 𝑩𝑩 = x� + y� + 2√3 z� .
3 2 2

Page 12 of 12
Matrices, Vectors & Fourier Analysis Spring 23-24

Chapter - 6
Gradient, Divergence and Curl
1. The Gradient Vector: grad

Consider a room in which the temperature is given by a scalar field, 𝑇𝑇(𝑢𝑢1 , 𝑢𝑢2 , 𝑢𝑢3 ) at each
point (𝑢𝑢1 , 𝑢𝑢2 , 𝑢𝑢3 ) (assume that the temperature does not change over time.). At each point in the
room, the gradient of 𝑇𝑇 at that point will show the direction in which the temperature rises most
quickly. The magnitude of the gradient will determine how fast the temperature rises in that
direction.

A vector field, called the gradient, written: gradT or 𝛁𝛁 T, where 𝛁𝛁 is called ‘del’, can be associated
with a scalar field 𝑇𝑇.

At every point, the direction of the vector field (𝛁𝛁 T) is

 orthogonal to the scalar field contour (𝐶𝐶) and

 in the direction of the maximum rate of change of 𝑇𝑇.𝑢𝑢�3

The gradient of a scalar function 𝑇𝑇(𝑢𝑢1 , 𝑢𝑢2 , 𝑢𝑢3 ) is given by

𝑢𝑢�1 𝜕𝜕𝜕𝜕 𝑢𝑢�2 𝜕𝜕𝜕𝜕 u� 3 𝜕𝜕𝜕𝜕

gradT = 𝛁𝛁 T = + + ∙
ℎ1 𝜕𝜕𝑢𝑢1 ℎ2 𝜕𝜕𝑢𝑢2 ℎ3 𝜕𝜕𝑢𝑢3

where ℎ𝑖𝑖 are scale factors and 𝑢𝑢�𝑖𝑖 are the unit vectors along 𝑢𝑢𝑖𝑖 , (𝑖𝑖 = 1, 2, 3). For cartesian
coordinates 𝑢𝑢1 = 𝑥𝑥, 𝑢𝑢2 = 𝑦𝑦, 𝑢𝑢3 = 𝑧𝑧, ℎ1 = ℎ2 = ℎ3 = 1, for cylindrical coordinates 𝑢𝑢1 = 𝑟𝑟, 𝑢𝑢2 =
𝜑𝜑, 𝑢𝑢3 = 𝑧𝑧, ℎ1 = ℎ3 = 1 and ℎ2 = 𝑟𝑟 and for spherical coordinates 𝑢𝑢1 = 𝑅𝑅, 𝑢𝑢2 = 𝜃𝜃, 𝑢𝑢3 = 𝜑𝜑, ℎ1 =
1, ℎ2 = 𝑅𝑅& ℎ3 = 𝑅𝑅 sin 𝜃𝜃.

1
Matrices, Vectors & Fourier Analysis Spring 23-24

Example 1.1. Find the gradient of the following scalar functions:

𝑥𝑥𝑥𝑥𝑥𝑥
(a) 𝑇𝑇(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = (𝑥𝑥 2 at the point (1, 1,1).
+𝑦𝑦 2 +𝑧𝑧 2 )
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
For Cartesian coordinates, gradT = 𝛁𝛁 𝑇𝑇 = x� + y� + z� ∙
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

𝜕𝜕 𝑥𝑥𝑥𝑥𝑥𝑥 𝜕𝜕 𝑥𝑥𝑥𝑥𝑥𝑥 𝜕𝜕 𝑥𝑥𝑥𝑥𝑥𝑥

= x� � 2 � + y
� � � + z
� � �
𝜕𝜕𝜕𝜕 (𝑥𝑥 + 𝑦𝑦 2 +𝑧𝑧 2 ) 𝜕𝜕𝜕𝜕 (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 ) 𝜕𝜕𝜕𝜕 (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )

(𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )𝑦𝑦𝑦𝑦 − 2𝑥𝑥 2 𝑦𝑦𝑦𝑦 (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )𝑥𝑥𝑥𝑥 − 2𝑦𝑦 2 𝑥𝑥𝑥𝑥 (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )𝑥𝑥𝑥𝑥 − 2𝑧𝑧 2 𝑥𝑥𝑥𝑥
= x� + y
� + z
�
(𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )2 (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )2 (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )2

1
At the point (1, 1,1), 𝛁𝛁 𝑇𝑇 = (x� + y� + z�) ∙
9
𝑧𝑧 cos ϕ
(b) 𝑇𝑇(𝑟𝑟, ϕ, 𝑧𝑧) = (1+𝑟𝑟 2) , at the point (1, 𝜋𝜋, 2)
r� 𝜕𝜕𝜕𝜕 � 𝜕𝜕𝜕𝜕
ϕ z� 𝜕𝜕𝜕𝜕
For cylindrical coordinates, gradT = 𝛁𝛁 𝑇𝑇 = + +
1 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕ϕ 1 𝜕𝜕𝜕𝜕

𝜕𝜕 𝑧𝑧 cos ϕ 1 𝜕𝜕 𝑧𝑧 cos ϕ 𝜕𝜕 𝑧𝑧 cos ϕ

= r� � �+ϕ� � � + z� � �
𝜕𝜕𝜕𝜕 (1 + 𝑟𝑟 )
2 r 𝜕𝜕ϕ (1 + 𝑟𝑟 )
2 𝜕𝜕𝜕𝜕 (1 + 𝑟𝑟 2 )
2𝑧𝑧𝑧𝑧 cos ϕ 1 𝑧𝑧 sin ϕ cos ϕ
= −r� �
−ϕ + z� ∙
(1 + 𝑟𝑟 )
2 2 r (1 + 𝑟𝑟 )
2 (1 + 𝑟𝑟 2 )
1
At the point (1, 𝜋𝜋, 2), 𝛁𝛁 𝑇𝑇 = r� − z� ∙
2
𝜋𝜋 𝜋𝜋
(c) 𝑇𝑇(𝑅𝑅, 𝜃𝜃, ϕ) = 𝑅𝑅 cos 𝜃𝜃 sin ϕ , at the point �2, , � ∙
2 4
� 𝜕𝜕𝜕𝜕
R � 𝜕𝜕𝜕𝜕
θ �
ϕ 𝜕𝜕𝜕𝜕
For cylindrical coordinates, gradT = 𝛁𝛁 𝑇𝑇 = + +
1 𝜕𝜕𝜕𝜕 𝑅𝑅 𝜕𝜕θ 𝑅𝑅 sin θ 𝜕𝜕ϕ

𝜕𝜕 θ� 𝜕𝜕 �
ϕ 𝜕𝜕
�
=R (𝑅𝑅 cos 𝜃𝜃 sin ϕ) + (𝑅𝑅 cos 𝜃𝜃 sin ϕ) + (𝑅𝑅 cos 𝜃𝜃 sin ϕ)
𝜕𝜕𝜕𝜕 𝑅𝑅 𝜕𝜕θ 𝑅𝑅 sin θ 𝜕𝜕ϕ
�
ϕ
� cos 𝜃𝜃 sin ϕ − θ� sin 𝜃𝜃 sin ϕ +
=R 𝑅𝑅 cos 𝜃𝜃 cos ϕ.
𝑅𝑅 sin θ

𝜋𝜋 𝜋𝜋 1
At the point �2, , � , 𝛁𝛁 𝑇𝑇 = −θ� .
2 4 √2

2
Matrices, Vectors & Fourier Analysis Spring 23-24

2. Directional Derivatives:

Consider the temperature 𝑇𝑇 at various points of a heated metal plate. Some contours for 𝑇𝑇 are

shown in the following diagram.

We are interested in how 𝑇𝑇 changes from one point to another. The rate of change of 𝑇𝑇 in the
direction specified by 𝐴𝐴𝐴𝐴 is given by (20 − 15)/𝐴𝐴𝐴𝐴 = 5/𝐴𝐴𝐴𝐴, examples of directional
derivative.

In general, for a given function 𝑇𝑇 = 𝑇𝑇(𝑢𝑢1 , 𝑢𝑢2 ), the directional derivative in the direction of a
unit vector is the gradient vector at a point 𝐴𝐴

 magnitude = the largest directional derivative, and

 pointing in the direction in which this largest directional derivative occurs, is known as
the gradient vector.

Hence, the component of 𝛁𝛁 𝑇𝑇 in the direction of a vector𝐝𝐝 is equal to𝛁𝛁 𝑇𝑇 ∙ 𝐝𝐝 and it is called the
directional derivative of 𝑇𝑇 in the direction of 𝐝𝐝.

Example 2.1 Find the directional derivative of 𝑇𝑇(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑥𝑥𝑦𝑦 2 − 𝑧𝑧 2 at the point (1, −1, 4 )in the
direction 𝐝𝐝 = x� − y� + 4 z� .
𝜕𝜕 𝜕𝜕 𝜕𝜕
Solution: 𝛁𝛁 T = x� (𝑥𝑥𝑦𝑦 2 − 𝑧𝑧 2 ) + y� (𝑥𝑥𝑦𝑦 2 − 𝑧𝑧 2 ) + z� (𝑥𝑥𝑦𝑦 2 − 𝑧𝑧 2 )
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

= x� 𝑦𝑦 2 + y� 2𝑥𝑥𝑥𝑥 + z�(−2𝑧𝑧)
3
Matrices, Vectors & Fourier Analysis Spring 23-24

At the point (1, −1, 4 ), 𝛁𝛁 𝑇𝑇 = x� − 2 y� − 8 z�

x�−y
�+4 z� 1 1 4
Now, the unit vector in the direction of x� − y� + 4 z� is a� = = x� − y� + z�
√1+1+16 √18 √18 √18
1 1 4
Then the required directional derivative is,𝛁𝛁 𝑇𝑇 ∙ a� = (x� − 2 y� − 8 z�) ∙ � x� − y� + z��
√18 √18 √18

1 2 32 29
= + − =− .
√18 √18 √18 √18

𝑟𝑟
1 𝜋𝜋
Example 2.2 Find the directional derivative of 𝑇𝑇(𝑟𝑟, ϕ, 𝑧𝑧) = 𝑒𝑒 −5 cos ϕ at the point �2, , 3� in
2 4

the direction r�.

𝜕𝜕 1
Solution: 𝛁𝛁 𝑇𝑇 = r� �1
� 𝑒𝑒 −𝑟𝑟/5 cos ϕ� + ϕ
𝜕𝜕 1
� 𝑒𝑒 −𝑟𝑟/5 cos ϕ� + z�
𝜕𝜕 1
� 𝑒𝑒 −𝑟𝑟/5 cos ϕ�
𝜕𝜕𝜕𝜕 2 r 𝜕𝜕ϕ 2 𝜕𝜕𝜕𝜕 2

1 −𝑟𝑟/5 1
= −r� 𝑒𝑒 � 𝑒𝑒 −𝑟𝑟/5 sin ϕ
cos ϕ − ϕ
10 2r
𝜋𝜋 1 1
At the point �2, , 3 �, 𝛁𝛁 𝑇𝑇 = −r� �
𝑒𝑒 −2/5 − ϕ 𝑒𝑒 −2/5
4 10√2 4√2
1
Then the required directional derivative is, 𝛁𝛁 𝑇𝑇 ∙ r� = − 𝑒𝑒 −2/5 .
10√2

1 𝜋𝜋 𝜋𝜋
Example 2.3 Find the directional derivative of 𝑇𝑇(𝑅𝑅, 𝜃𝜃, ϕ) = sin2 𝜃𝜃 at the point �5, , � in the
𝑅𝑅 4 2

�.
direction R
𝜕𝜕 1 � 𝜕𝜕
θ 1 �
ϕ 𝜕𝜕 1
�
Solution: 𝛁𝛁 𝑇𝑇 = R � sin2 𝜃𝜃� + � sin2 𝜃𝜃� + � sin2 𝜃𝜃�
𝜕𝜕𝜕𝜕 𝑅𝑅 𝑅𝑅 𝜕𝜕θ 𝑅𝑅 𝑅𝑅 sin θ 𝜕𝜕ϕ 𝑅𝑅

1 1
�
= −R 2
sin2 𝜃𝜃 + θ� 2 2 sin 𝜃𝜃 cos 𝜃𝜃
𝑅𝑅 𝑅𝑅
𝜋𝜋 𝜋𝜋 1 1
�
At the point �5, , �, 𝛁𝛁 𝑇𝑇 = −R + θ�
4 2 50 25

� = − 1.
Then the required directional derivative is, 𝛁𝛁 𝑇𝑇 ∙ R
50

3. The divergence and curl of a vector function

Consider air as it is heated or cooled. The velocity of the air at each point defines a vector field.
While air is heated in a region, it expands in all directions, and thus the velocity field points
outward from that region. The divergence of the velocity field in that region would thus have a
positive value. While the air is cooled and thus contracting, the divergence of the velocity has a
negative value.
The divergence of a vector field is relatively easy to understand intuitively. Imagine that the vector
field A pictured below gives the velocity of some fluid flow. It appears that the fluid is exploding
outward from the origin.

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Matrices, Vectors & Fourier Analysis Spring 23-24

𝛁𝛁 ∙ 𝐀𝐀 > 0
This expansion of fluid flowing with velocity field 𝑨𝑨 is captured by the divergence of 𝑨𝑨, which we
denote div 𝑨𝑨 or mathematically 𝛁𝛁 ∙ 𝐀𝐀. The divergence of the above vector field is positive since
the flow is expanding.
In contrast, the below vector field represents fluid flowing so that it compresses as it moves toward
the origin. Since this compression of fluid is the opposite of expansion, the divergence of this
vector field is negative.

𝛁𝛁 ∙ 𝐀𝐀 < 0
Lastly, a solenoidal vector field (also known as an incompressible vector field, a divergence-
free vector field) is a vector field 𝑨𝑨 with divergence zero at all points in the field. That is
𝛁𝛁 ∙ 𝐀𝐀 = 𝟎𝟎

Hence the illustration of the divergence of a vector field at any point 𝑃𝑃 is given below:

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Matrices, Vectors & Fourier Analysis Spring 23-24

The curl of a vector field captures the idea of how a flow may rotate. Imagine that the below
vector field F represents fluid flow. The vector field indicates that the fluid is circulating around
a central axis. This rotation of fluid flowing with velocity field 𝑨𝑨 is captured by the curl of 𝑨𝑨,
which we denote curl𝐀𝐀 or mathematically 𝛁𝛁 × 𝐀𝐀.

If curl𝐀𝐀 = 0 then 𝐀𝐀 is called conservative or irrotational.

Now, let 𝐀𝐀 = u� 1 𝐴𝐴1 (𝑢𝑢1 , 𝑢𝑢2 , 𝑢𝑢3 ) + u� 2 𝐴𝐴2 (𝑢𝑢1 , 𝑢𝑢2 , 𝑢𝑢3 ) + u� 3 𝐴𝐴3 (𝑢𝑢1 , 𝑢𝑢2 , 𝑢𝑢3 ) is a vector. The
divergence and the curl is given by

1 𝜕𝜕 𝜕𝜕 𝜕𝜕
div𝐀𝐀 = 𝛁𝛁 ∙ 𝐀𝐀 = � (𝐴𝐴1 ℎ2 ℎ3 ) + (ℎ1 𝐴𝐴2 ℎ3 ) + (ℎ ℎ 𝐴𝐴 )� ∙
ℎ1 ℎ2 ℎ3 𝜕𝜕𝑢𝑢1 𝜕𝜕𝑢𝑢2 𝜕𝜕𝑢𝑢3 1 2 3

6
Matrices, Vectors & Fourier Analysis Spring 23-24

u� 1 ℎ1 u� 2 ℎ2 𝑢𝑢�3 ℎ3
1 𝜕𝜕 𝜕𝜕 𝜕𝜕
and curl𝐀𝐀 = 𝛁𝛁 × 𝐀𝐀 = � 𝜕𝜕𝑢𝑢1 𝜕𝜕𝑢𝑢2 𝜕𝜕𝑢𝑢3
�,
ℎ1 ℎ2 ℎ3
ℎ1 𝐴𝐴1 ℎ2 𝐴𝐴2 ℎ3 𝐴𝐴3

• The derivation can be found in advanced calculus books.

Example 3.1 Determine divergence and curl. Also check each of the following vector fields
solenoidal, conservative or both.

(a) 𝐀𝐀 = x� 𝑥𝑥 2 + y� 2𝑥𝑥𝑥𝑥
𝜕𝜕 2 𝜕𝜕
div𝐀𝐀 = 𝛁𝛁 ∙ 𝐀𝐀 = � (𝑥𝑥 ) + (2𝑥𝑥𝑥𝑥)� = 2𝑥𝑥 + 2𝑥𝑥 = 4𝑥𝑥.
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

x� y� z�
𝜕𝜕 𝜕𝜕 𝜕𝜕
curl𝐀𝐀 = 𝛁𝛁 × 𝐀𝐀 = � 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
� = z� 2𝑦𝑦.
𝑥𝑥 2 2𝑥𝑥𝑥𝑥 0

∴ 𝐀𝐀 is not solenoidal or conservative.

sin ϕ cos ϕ
(b) 𝐀𝐀 = r� �
+ϕ
𝑟𝑟 2 𝑟𝑟 2

1 𝜕𝜕 𝜕𝜕 𝜕𝜕
div𝐀𝐀 = 𝛁𝛁 ∙ 𝐀𝐀 = � (𝐴𝐴1 ℎ2 ℎ3 ) + (ℎ1 𝐴𝐴2 ℎ3 ) + (ℎ ℎ 𝐴𝐴 )�
ℎ1 ℎ2 ℎ3 𝜕𝜕𝑢𝑢1 𝜕𝜕𝑢𝑢2 𝜕𝜕𝑢𝑢3 1 2 3

1 𝜕𝜕 sin ϕ 𝜕𝜕 cos ϕ 𝜕𝜕 sin ϕ sin ϕ 2sin ϕ

= � � 2 𝑟𝑟� + � 2 � + (0 ∙ 𝑟𝑟)� = − 3 − 3 = − .
𝑟𝑟 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕ϕ 𝑟𝑟 𝜕𝜕𝜕𝜕 𝑟𝑟 𝑟𝑟 𝑟𝑟 3

r� �r
ϕ z�
u� 1 ℎ1 u� 2 ℎ2 u� 3 ℎ3
𝜕𝜕 𝜕𝜕 𝜕𝜕
1 𝜕𝜕 𝜕𝜕 𝜕𝜕 1� � 2cos ϕ
curl𝐀𝐀 = 𝛁𝛁 × 𝐀𝐀 = � �= 𝜕𝜕𝜕𝜕 𝜕𝜕ϕ 𝜕𝜕𝜕𝜕 = −z� .
ℎ1 ℎ2 ℎ3 𝜕𝜕𝑢𝑢1 𝜕𝜕𝑢𝑢2 𝜕𝜕𝑢𝑢3 𝑟𝑟 � � 𝑟𝑟 2
sin ϕ cos ϕ
ℎ1 𝐴𝐴1 ℎ2 𝐴𝐴2 ℎ3 𝐴𝐴3 𝑟𝑟 0
𝑟𝑟 2 𝑟𝑟 2
∴ 𝐀𝐀 is not solenoidal or conservative.

�(𝑅𝑅𝑒𝑒 −𝑅𝑅 )
(c) 𝐀𝐀 = R
1 𝜕𝜕 −𝑅𝑅 2
𝜕𝜕 𝜕𝜕
div𝐀𝐀 = 𝛁𝛁 ∙ 𝐀𝐀 = � (𝑅𝑅𝑒𝑒 ∙ 𝑅𝑅 sin 𝜃𝜃) + (0 ∙ 𝑅𝑅 sin 𝜃𝜃) + (0 ∙ 𝑅𝑅)� = 𝑒𝑒 −𝑅𝑅 (3 − 𝑅𝑅).
𝑅𝑅2 sin 𝜃𝜃 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕ϕ

�
R θ� 𝑅𝑅 � 𝑅𝑅 sin 𝜃𝜃
ϕ
1 𝜕𝜕 𝜕𝜕 𝜕𝜕
𝛁𝛁 × 𝐀𝐀 =
𝑅𝑅2 sin 𝜃𝜃
� � = 0. ∴ 𝐀𝐀 is conservative but not solenoidal.
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕ϕ
−𝑅𝑅
𝑅𝑅𝑒𝑒 𝑅𝑅 ∙ 0 𝑅𝑅 sin 𝜃𝜃 ∙ 0
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Matrices, Vectors & Fourier Analysis Spring 23-24

Example 3.2 Test whether 𝐀𝐀 = x� (𝑦𝑦 2 cos 𝑥𝑥 + 𝑧𝑧 3 ) + y� (2𝑦𝑦 sin 𝑥𝑥 − 𝑦𝑦 ) + z�(3𝑥𝑥𝑧𝑧 2 + 2)is a
conservative force field. If conservative, find the scalar potential 𝑇𝑇such that 𝐀𝐀 = 𝛁𝛁 𝑇𝑇. Hence find
𝜋𝜋
the work done in moving an object in this field from (0, 1, −1) to � , −1, 2� ∙
2

Solution: We know for conservative force field curl𝐀𝐀 = 𝛁𝛁 × 𝐀𝐀 = 0

x� y� z�
𝜕𝜕 𝜕𝜕 𝜕𝜕
𝛁𝛁 × 𝐀𝐀 = �� �=0
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 �
𝑦𝑦 cos 𝑥𝑥 + 𝑧𝑧 3
2
2𝑦𝑦 sin 𝑥𝑥 − 𝑦𝑦 3𝑥𝑥𝑧𝑧 2 + 2
Hence 𝐀𝐀 is conservative force field.
Let 𝑇𝑇(𝑥𝑥, 𝑦𝑦, 𝑧𝑧)be a scalar potential of 𝐀𝐀, i.e. 𝐀𝐀 = 𝛁𝛁 𝑇𝑇 ∙∴ 𝑇𝑇(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = ∫ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥

= �[x� (𝑦𝑦 2 cos 𝑥𝑥 + 𝑧𝑧 3 ) + y� (2𝑦𝑦 sin 𝑥𝑥 − 𝑦𝑦 ) + z�(3𝑥𝑥𝑧𝑧 2 + 2)]. [x� 𝑑𝑑𝑑𝑑 + y� 𝑑𝑑𝑑𝑑 + z�𝑑𝑑𝑑𝑑]

= �[(𝑦𝑦 2 cos 𝑥𝑥 + 𝑧𝑧 3 )𝑑𝑑𝑑𝑑 + (2𝑦𝑦 sin 𝑥𝑥 − 𝑦𝑦 )𝑑𝑑𝑑𝑑 + (3𝑥𝑥𝑧𝑧 2 + 2)𝑑𝑑𝑑𝑑]

𝑦𝑦 2 𝑦𝑦 2
= ∫ 𝑑𝑑 �𝑥𝑥𝑧𝑧 3 + 𝑦𝑦 2 sin 𝑥𝑥 + 2𝑧𝑧 − � = 𝑥𝑥𝑧𝑧 3 + 𝑦𝑦 2 sin 𝑥𝑥 + 2𝑧𝑧 − + 𝑐𝑐, 𝑐𝑐 is a constant.
2 2
𝜋𝜋 𝜋𝜋
� 2 ,−1,2� � ,−1,2�
𝑦𝑦 2 2
Now, work done, = ∫(0,1,−1) 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = �𝑥𝑥𝑧𝑧 + 𝑦𝑦 sin 𝑥𝑥 + 2𝑧𝑧 −
3 2
� = 4𝜋𝜋 + 7 ∙
2 (0,1,−1)

Example 3.3 Test whether 𝐀𝐀 = r�(cos ϕ) + ϕ � (− sin ϕ)is a conservative force field. If
conservative, find the scalar potential 𝑇𝑇such that 𝐀𝐀 = 𝛁𝛁 𝑇𝑇. Hence find the work done in moving
an object in this field from (0, 𝜋𝜋, −1) to (1,2𝜋𝜋, 2) ∙
Solution: We know for conservative force field curl𝐀𝐀 = 𝛁𝛁 × 𝐀𝐀 = 0
r� �r
ϕ z�
1 𝜕𝜕 𝜕𝜕 𝜕𝜕
𝛁𝛁 × 𝐀𝐀 = �� �� = 0
𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕ϕ 𝜕𝜕𝜕𝜕
cos ϕ – 𝑟𝑟 sin ϕ 0
Hence 𝐀𝐀 is conservative force field.
Let 𝑇𝑇(𝑟𝑟, ϕ, 𝑧𝑧) be a scalar potential of 𝐀𝐀, i.e.𝐀𝐀 = 𝛁𝛁 𝑇𝑇 ∙∴ 𝑇𝑇(𝑟𝑟, ϕ, 𝑧𝑧) = ∫ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥

� (– sin ϕ)�. �r� 𝑑𝑑𝑑𝑑 + ϕ

= ��r�(cos ϕ) + ϕ � 𝑟𝑟 𝑑𝑑ϕ + z�𝑑𝑑𝑑𝑑�

= �[cos ϕ 𝑑𝑑𝑑𝑑 − rsin ϕ 𝑑𝑑ϕ]

= ∫ 𝑑𝑑(𝑟𝑟 cos ϕ) = 𝑟𝑟 cos ϕ + 𝑐𝑐, 𝑐𝑐 is a constant.

Now, work done,

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Matrices, Vectors & Fourier Analysis Spring 23-24

(1,2𝜋𝜋,2)
(1,2𝜋𝜋,2)
� 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = [𝑟𝑟 cos ϕ](0,𝜋𝜋,−1) = 1 ∙
(0,𝜋𝜋,−1)

Example 3.4 Test whether 𝐀𝐀 = R � (sin 𝜃𝜃 cos ϕ) + 𝜃𝜃�(cos 𝜃𝜃 cos ϕ) + ϕ

� (− sin ϕ)is a conservative
force field. If conservative, find the scalar potential𝑇𝑇such that 𝐀𝐀 = 𝛁𝛁 𝑇𝑇. Hence find the work
𝜋𝜋 𝜋𝜋 𝜋𝜋
done in moving an object in this field from �0, , � to �1, , 𝜋𝜋� ∙
4 3 2

Solution: We know for conservative force field curl𝐀𝐀 = 𝛁𝛁 × 𝐀𝐀 = 0

�
R θ� R ϕ� R sin θ
1 𝜕𝜕 𝜕𝜕 𝜕𝜕
𝛁𝛁 × 𝐀𝐀 = 2 �� �� = 0
𝑅𝑅 sin 𝜃𝜃 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕ϕ
sin 𝜃𝜃 cos ϕ R cos 𝜃𝜃 cos ϕ − R sin θ sin ϕ
Hence𝐀𝐀 is conservative force field.
Let𝑇𝑇(𝑅𝑅, 𝜃𝜃, ϕ)be a scalar potential of 𝐀𝐀, i.e.𝐀𝐀 = 𝛁𝛁 𝑇𝑇 ∙∴ 𝑇𝑇(𝑅𝑅, 𝜃𝜃, ϕ) = ∫ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥

�(sin 𝜃𝜃 cos ϕ) + 𝜃𝜃�(cos 𝜃𝜃 cos ϕ) + ϕ

= ��R � (− sin ϕ)�. �R
� 𝑑𝑑𝑑𝑑 + 𝜃𝜃� 𝑅𝑅 𝑑𝑑𝑑𝑑 + ϕ
� R sin θ 𝑑𝑑ϕ�

= �[(sin 𝜃𝜃 cos ϕ)𝑑𝑑𝑑𝑑 + (cos 𝜃𝜃 cos ϕ)𝑅𝑅 𝑑𝑑𝑑𝑑 + (− sin ϕ)R sin θ 𝑑𝑑ϕ]

= ∫ 𝑑𝑑(𝑅𝑅 sin 𝜃𝜃 cos ϕ) = 𝑅𝑅 sin 𝜃𝜃 cos ϕ + 𝑐𝑐, 𝑐𝑐 is a constant.

𝜋𝜋 𝜋𝜋
�1, 2 ,𝜋𝜋� �1, ,𝜋𝜋�
Now, work done, ∫ 𝜋𝜋 𝜋𝜋 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = [𝑅𝑅 sin 𝜃𝜃 cos ϕ] 2
𝜋𝜋 𝜋𝜋 = −1 ∙
�0, 4 , 3 � �0, 4 , 3 �

4. Laplacian operator
The Laplacian of a scalar function is defined as the divergence of the gradient of that function.

The differential operator 𝛁𝛁𝟐𝟐 = 𝛁𝛁 ∙ 𝛁𝛁 is called Laplacian operator, and

𝛁𝛁 ∙ (𝛁𝛁T) = 𝛁𝛁𝟐𝟐 𝑇𝑇(𝑢𝑢1 , 𝑢𝑢2 , 𝑢𝑢3 ) = 0 is called Laplace Equation.

The Laplacian of a scalar function 𝑇𝑇 in different coordinate system are defined as follows:
𝜕𝜕2 𝑇𝑇 𝜕𝜕2 𝑇𝑇 𝜕𝜕2 𝑇𝑇
In Cartesian coordinates 𝛁𝛁𝟐𝟐 𝑇𝑇 = + + ,
𝜕𝜕𝑥𝑥 2 𝜕𝜕𝑦𝑦 2 𝜕𝜕𝑧𝑧 2

1 𝜕𝜕 𝜕𝜕𝜕𝜕 1 𝜕𝜕2 𝑇𝑇 𝜕𝜕2 𝑇𝑇

in cylindrical coordinates 𝛁𝛁𝟐𝟐 𝑇𝑇 = �𝑟𝑟 �+ + and
𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑟𝑟 2 𝜕𝜕ϕ2 𝜕𝜕𝑧𝑧 2

1 𝜕𝜕 𝜕𝜕𝜕𝜕 1 𝜕𝜕 𝜕𝜕𝜕𝜕 1 𝜕𝜕2 𝑇𝑇

in Spherical coordinates 𝛁𝛁𝟐𝟐 𝑇𝑇 = �𝑅𝑅2 �+ �sin 𝜃𝜃 �+ � �∙
𝑅𝑅2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑅𝑅2 sin 𝜃𝜃 𝜕𝜕θ 𝜕𝜕θ 𝑅𝑅2 sin2 𝜃𝜃 𝜕𝜕ϕ2

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Matrices, Vectors & Fourier Analysis Spring 23-24

3
Example 4.1 Find the Laplacian of the scalar function T = .
x2 +y2

Solution: In Cartesian co-ordinates we know the Laplacian is

𝟐𝟐
𝜕𝜕 2 𝑇𝑇 𝜕𝜕 2 𝑇𝑇 𝜕𝜕 2 𝑇𝑇
𝛁𝛁 𝑇𝑇 = 2 + 2 + 2
𝜕𝜕𝑥𝑥 𝜕𝜕𝑦𝑦 𝜕𝜕𝑧𝑧

𝜕𝜕 𝜕𝜕 𝜕𝜕 𝜕𝜕 𝜕𝜕 𝜕𝜕
= ( (3(𝑥𝑥 2 + 𝑦𝑦 2 )−1 )) + ( (3(𝑥𝑥 2 + 𝑦𝑦 2 )−1 )) + ( (3(𝑥𝑥 2 + 𝑦𝑦 2 )−1 ))
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

𝜕𝜕 𝜕𝜕
= (−3(𝑥𝑥 2 + 𝑦𝑦 2 )−2 . 2𝑥𝑥) + (−3(𝑥𝑥 2 + 𝑦𝑦 2 )−2 . 2𝑦𝑦) + 0
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

= 24𝑥𝑥 2 (𝑥𝑥 2 + 𝑦𝑦 2 )−3 − 6(𝑥𝑥 2 + 𝑦𝑦 2 )−2 + 24𝑦𝑦 2 (𝑥𝑥 2 + 𝑦𝑦 2 )−3 − 6(𝑥𝑥 2 + 𝑦𝑦 2 )−2

12
=
(𝑥𝑥 2 + 𝑦𝑦 2 )2

Example 4.2 Find the Laplacian of the scalar function T = 5e−r cosϕ.

Solution: In Cylindrical coordinates we know the Laplacian is

1 𝜕𝜕 𝜕𝜕𝜕𝜕 1 𝜕𝜕 2 𝑇𝑇 𝜕𝜕 2 𝑇𝑇
𝛁𝛁𝟐𝟐 𝑇𝑇 = �𝑟𝑟 � + 2 2 + 2
𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕ϕ 𝜕𝜕𝑧𝑧

1 𝜕𝜕 𝜕𝜕 1 𝜕𝜕 𝜕𝜕 𝜕𝜕 𝜕𝜕
= �𝑟𝑟 (5e−r cosϕ)� + 2 � (5e−r cosϕ)� + � (5e−r cosϕ)�
𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕ϕ 𝜕𝜕ϕ 𝜕𝜕z 𝜕𝜕z

5cosϕ 𝜕𝜕 5e−r 𝜕𝜕
= (−𝑟𝑟 𝑒𝑒 −𝑟𝑟 ) + 2 (−𝑠𝑠𝑠𝑠𝑠𝑠ϕ) + 0
𝑟𝑟 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕ϕ

5cosϕ −𝑟𝑟 −𝑟𝑟 ]

5e−r
=− [−𝑒𝑒 − 𝑟𝑟 𝑒𝑒 − 2 𝑐𝑐𝑐𝑐𝑐𝑐ϕ
𝑟𝑟 𝑟𝑟
−5𝑒𝑒 −𝑟𝑟 cosϕ 5e−r
= + 5e−r cosϕ − 𝑐𝑐𝑐𝑐𝑐𝑐ϕ.
𝑟𝑟 𝑟𝑟 2

Example 4.3 Find the Laplacian of the scalar function T = 10e−R sinθ.

Solution: In Spherical coordinates we know the Laplacian is,

1 𝜕𝜕 𝜕𝜕𝜕𝜕 1 𝜕𝜕 𝜕𝜕𝜕𝜕 1 𝜕𝜕 2 𝑇𝑇
𝛁𝛁𝟐𝟐 𝑇𝑇 = �𝑅𝑅 2
� + �sin 𝜃𝜃 � + � �∙
𝑅𝑅2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑅𝑅2 sin 𝜃𝜃 𝜕𝜕θ 𝜕𝜕θ 𝑅𝑅2 sin2 𝜃𝜃 𝜕𝜕ϕ2

10
Matrices, Vectors & Fourier Analysis Spring 23-24

1 𝜕𝜕 𝜕𝜕 1 𝜕𝜕 𝜕𝜕
= 2
�𝑅𝑅2 (10e−R sinθ)� + 2 �𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (10e−R sinθ)�
𝑅𝑅 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑅𝑅 sin 𝜃𝜃 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
1 𝜕𝜕 𝜕𝜕
+ 2 2
� (10e−R sinθ)�
𝑅𝑅 sin 𝜃𝜃 𝜕𝜕ϕ 𝜕𝜕ϕ

10sinθ 𝜕𝜕 2 −𝑅𝑅
10e−R 𝜕𝜕
= (−𝑅𝑅 𝑒𝑒 ) + (𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) + 0
𝑅𝑅2 𝜕𝜕𝜕𝜕 𝑅𝑅2 sin 𝜃𝜃 𝜕𝜕𝜕𝜕
10sinθ 𝜕𝜕 2 −𝑅𝑅
10e−R 𝜕𝜕
=− (−𝑅𝑅 𝑒𝑒 ) + 2 (𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) + 0
𝑅𝑅2 𝜕𝜕𝜕𝜕 𝑅𝑅 sin 𝜃𝜃 𝜕𝜕𝜕𝜕
10sinθ −𝑅𝑅 2 −𝑅𝑅
10e−R 𝜕𝜕
=− [2𝑅𝑅𝑒𝑒 − 𝑅𝑅 𝑒𝑒 ] − 2 (𝑠𝑠𝑠𝑠𝑠𝑠2𝜃𝜃)
𝑅𝑅2 2𝑅𝑅 sin 𝜃𝜃 𝜕𝜕𝜕𝜕
10sinθ −𝑅𝑅 2 −𝑅𝑅 ]
10e−R
=− [2𝑅𝑅𝑒𝑒 − 𝑅𝑅 𝑒𝑒 − 𝑐𝑐𝑐𝑐𝑐𝑐2𝜃𝜃.
𝑅𝑅2 𝑅𝑅2 sin 𝜃𝜃

Sample exercise - 6.1

1. Find the gradient of the following scalar functions at the indicated point:
𝑦𝑦
(a) 𝑇𝑇(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 2𝑥𝑥 3 𝑦𝑦 𝑧𝑧 + y 2 𝑥𝑥 2 − 5 at the point (0, 2, −1). ∙ Ans: 𝛁𝛁T = 5 y� + 10z�
𝑧𝑧
𝑧𝑧+sin ϕ 3𝜋𝜋 1
(b) 𝑇𝑇(𝑟𝑟, ϕ, 𝑧𝑧) = , at the point �2, , 1� ∙ Ans: 𝛁𝛁T = z�
𝑟𝑟 2 2
𝜋𝜋 2𝜋𝜋
(c) 𝑇𝑇(𝑅𝑅, 𝜃𝜃, ϕ) = 𝑅𝑅2 cos ϕ sin 𝜃𝜃 , at the point �2, , �∙
4 3
1
�−
Ans: 𝛁𝛁T = −√2 R 𝜃𝜃� − √3 ϕ
�
√2

2. (a) Find the directional derivative (D. D.) of 𝑇𝑇(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑥𝑥 2 𝑦𝑦 − 𝑥𝑥𝑥𝑥 at the point (1, 0, 2 )in the
2
direction 𝐝𝐝 = x� − 2y� − 6 z� . Ans: 𝛁𝛁T = x� (2xy − z) + y� x 2 − z� 𝑥𝑥 and D. D. is ∙
√41

𝜋𝜋
(b) Find the directional derivative (D. D.) of 𝑇𝑇(𝑟𝑟, ϕ, 𝑧𝑧) = 𝑟𝑟 3 cos ϕ at the point �2, , 1� in the
4
12
direction r�. Ans: D.D. is ∙
√2

1 𝜋𝜋 𝜋𝜋
(c) Find the directional derivative (D. D.) of 𝑇𝑇(𝑅𝑅, 𝜃𝜃, ϕ) = cos 2 𝜃𝜃 at the point �1, , � in the
𝑅𝑅 4 2
1 1
� − 𝜃𝜃�. Ans: D. D. is −
direction R + ∙
2√2 √2

3. Determine divergence and curl. Also check each of the following vector fields solenoidal,
conservative or both.

11
Matrices, Vectors & Fourier Analysis Spring 23-24

(a) 𝐀𝐀 = x� 𝑧𝑧𝑧𝑧 3 + y� 2𝑦𝑦 sin(𝑥𝑥𝑥𝑥) + z� 3𝑥𝑥 2 ln 𝑧𝑧

3
Ans: 𝛁𝛁 . 𝐀𝐀 = 𝑥𝑥2 + 2 sin 𝑥𝑥𝑥𝑥 + 2𝑥𝑥𝑥𝑥 cos 𝑥𝑥𝑥𝑥 and 𝛁𝛁 × 𝐀𝐀 = y� (𝑦𝑦 3 − 6𝑥𝑥 lnz) + z� (2𝑦𝑦 2 cos 𝑥𝑥𝑥𝑥 − 3𝑦𝑦 2 𝑧𝑧) ∙
𝑧𝑧

sin ϕ cos ϕ sin ϕ cos ϕ 1 1

(b) 𝐀𝐀 = r� �
+ϕ Ans: 𝛁𝛁 . 𝐀𝐀 = − and 𝛁𝛁 × 𝐀𝐀 = −z� ( + )∙
𝑟𝑟 𝑟𝑟 2 𝑟𝑟 3 r 𝑟𝑟 2 𝑟𝑟

� cos 𝜃𝜃 + 𝜃𝜃�(𝑅𝑅 − sin 𝜃𝜃)

(c) 𝐀𝐀 = R � 2.
Ans: 𝛁𝛁 . 𝐀𝐀 = cot 𝜃𝜃 and 𝛁𝛁 × 𝐀𝐀 = ϕ
4. (a) Test whether 𝐀𝐀 = x� (sin 𝑦𝑦 + 1) + y� (2𝑦𝑦𝑦𝑦 + 𝑥𝑥 cos 𝑦𝑦 ) + z� (𝑦𝑦 2 − 3) is a conservative force
field. If conservative, find the scalar potential 𝑇𝑇such that 𝐀𝐀 = 𝛁𝛁 T. Hence find the work done in
𝜋𝜋
moving an object in this field from (1, −1, 5) to �2, , 1� ∙ Ans: 𝑇𝑇 = 𝑥𝑥 sin 𝑦𝑦 + 𝑦𝑦 2 𝑧𝑧 + 𝑥𝑥 − 3𝑧𝑧 + 𝑐𝑐
2

(b) Test whether 𝐀𝐀 = r� (2𝑟𝑟𝑟𝑟 − cos ϕ) + ϕ � (sin ϕ) + z� (𝑟𝑟 2 ) is a conservative force field. If
conservative, find the scalar potential 𝑇𝑇such that 𝐀𝐀 = 𝛁𝛁 T. Hence find the work done in moving
an object in this field from (1, 0, −1) to (1, 𝜋𝜋, 0) ∙ Ans: 𝑇𝑇 = r 2 z − r cos ϕ + 𝑐𝑐
(c) Test whether 𝐀𝐀 = R�(cos 𝜃𝜃) + 𝜃𝜃�(−sin 𝜃𝜃) is a conservative force field. If conservative, find the
scalar potential 𝑇𝑇such that 𝐀𝐀 = 𝛁𝛁 T. Hence find the work done in moving an object in this field
𝜋𝜋 𝜋𝜋 𝜋𝜋
from �0, , � to �1, , 0� ∙ Ans: 𝑇𝑇 = 𝑅𝑅 cos 𝜃𝜃 + 𝑐𝑐
4 3 2

5. Find the Laplacian of the following scalar functions:

(a) T = 4𝑦𝑦 2 𝑧𝑧 2 ∙ Ans: 8(𝑦𝑦 2 + 𝑧𝑧 2 )

(b) T = 𝑥𝑥𝑥𝑥 + 𝑧𝑧𝑧𝑧 ∙ Ans: 0 ∙

(c) T = 10r 3 cos2ϕ ∙ Ans: 50𝑟𝑟 cos 2ϕ ∙

2 2
(d) T = cosθsinϕ ∙ Ans: cos 𝜃𝜃 sin ϕ (4 − csc 2 𝜃𝜃) ∙
𝑅𝑅3 𝑅𝑅5

MATLAB for Gradient, divergence and curl

Find the gradient of the following scalar >>syms x y z

3
functions 𝑇𝑇(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = (𝑥𝑥 2 2) ∙ >>gradient(3/(x^2 + z^2), [x, y, z])
+𝑧𝑧

Determine divergence of 𝐀𝐀 = x� 𝑥𝑥 2 + y� 2𝑥𝑥𝑥𝑥 >>syms x y z

>>divergence([x^2, 2*y^2, 0], [x, y, z])

Determine curl of 𝐀𝐀 = x� 𝑥𝑥 + y� 2𝑦𝑦 2 + z� 3𝑧𝑧 3 >>syms x y z

>>curl([x, 2*y^2, 3*z^3], [x, y, z])

Find the Laplacian of T = 4xy 2 z 3 . >> syms x y z

>> laplacian(4*x*y^2*z^3)

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Matrices, Vectors & Fourier Analysis Spring 23-24

Gauss’s-Divergence Theorem and Stoke’s Theorem

1. Some Prerequisite Example

Example 1.1. For a vector function 𝐀𝐀 = r�𝑟𝑟 2 + z� 2𝑧𝑧, find surface integral for the circular
cylindrical region enclosed by 𝑟𝑟 = 5, 𝑧𝑧 = 0, 𝑧𝑧 = 4.

Solution

� 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 + � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 + � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬

𝑠𝑠 top bottom walls
face face face

i. Top face: 𝑧𝑧 = 4, 𝐀𝐀 = r�𝑟𝑟 2 + z� 8and 𝑑𝑑𝐬𝐬 = z� 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ

2𝜋𝜋 5

∴ � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 8 � � 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ = 200𝜋𝜋.

top 0 0
face

ii. Bottom face: 𝑧𝑧 = 0, 𝐀𝐀 = r�𝑟𝑟 2 and 𝑑𝑑𝒔𝒔 = −z� 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ ∴ ∫bottom 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 0
face

iii. Walls face: 𝑟𝑟 = 5, 𝐀𝐀 = r� 25 + z�2𝑧𝑧 and 𝑑𝑑𝒔𝒔 = r� 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ

2𝜋𝜋 4

∴ � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 25 � � 5𝑑𝑑𝑑𝑑𝑑𝑑ϕ = 1000𝜋𝜋.

walls 0 0
face

Total, ∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 1200𝜋𝜋

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Matrices, Vectors & Fourier Analysis Spring 2023-24

Example 1.2. Evaluate ∭𝐸𝐸 16 𝑧𝑧 𝑑𝑑𝑑𝑑 where E is the upper half of the sphere
𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 = 1.

Solution
Since we are taking the upper half of the sphere the limits for the variables are,

0 ≤ 𝑅𝑅 ≤ 1
𝜋𝜋
0 ≤ 𝜃𝜃 ≤
2
0 ≤ 𝜑𝜑 ≤ 2𝜋𝜋
The integral is then,
𝜋𝜋
2𝜋𝜋 2 1

� 16𝑧𝑧 𝑑𝑑𝑑𝑑 = � � � 𝑅𝑅2 sin 𝜃𝜃(16𝑅𝑅 cos 𝜃𝜃 ) 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝐸𝐸 0 0 0
𝜋𝜋
2 2𝜋𝜋 1

= � � � 8𝑅𝑅3 sin 2𝜃𝜃 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

0 0 0
𝜋𝜋
2 2𝜋𝜋

= � � 2 sin 2𝜃𝜃𝜃𝜃𝜃𝜃 𝑑𝑑𝑑𝑑

0 0
𝜋𝜋
2

= � 4𝜋𝜋 sin 2𝜃𝜃 𝑑𝑑𝑑𝑑

0
𝜋𝜋
= −2𝜋𝜋 cos 2𝜃𝜃 2 = 4𝜋𝜋.
0

2. Gauss’s divergence theorem

Statement: The surface integral of the normal component of a vector function 𝐀𝐀 taken arround a
closed surfce 𝑆𝑆 is equal to the integral of the divergence of 𝐀𝐀 taken over the volume 𝑉𝑉 enclosed
by the surface 𝑆𝑆.
Mathematically, ∫𝑉𝑉 𝛁𝛁 ∙ 𝐀𝐀 𝑑𝑑𝑑𝑑 = ∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬

Example 2.1. For the vector field 𝐀𝐀 = x� 𝑥𝑥𝑥𝑥 − y� 𝑦𝑦𝑧𝑧 2 − z� 𝑥𝑥𝑥𝑥, verify the divergence theorem by
computing (a) the total outward flux flowing through the surface of a cube centered at the origin
and with sides equal to 2 units each and parallel to the Cartesian axes, (b) the integral of 𝛁𝛁 ∙ 𝐀𝐀 over
the cube’s volume.

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Matrices, Vectors & Fourier Analysis Spring 2023-24

Solution
We first evaluate the surface integral over the six faces.
2 2
i. Front face: 𝑥𝑥 = 2, 𝑑𝑑𝐬𝐬 = x� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ∴ ∫front 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = ∫0 ∫0 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 = 8.
face

ii. Back face: 𝑥𝑥 = 0, 𝑑𝑑𝐬𝐬 = −x� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ∴ ∫back 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 0.

face
2 2 32
iii. Right face: 𝑦𝑦 = 2, 𝑑𝑑𝐬𝐬 = y� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ∴ ∫right 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = ∫0 ∫0 −𝑦𝑦𝑧𝑧 2 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = − .
3
face

iv. Left face: 𝑦𝑦 = 0, 𝑑𝑑𝐬𝐬 = −y� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ∴ ∫left 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 0.

face
2 2
v. Top face: 𝑧𝑧 = 2, 𝑑𝑑𝐬𝐬 = z� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ∴ ∫top 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = ∫0 ∫0 −𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 = −4.
face
2 2
vi. Bottom face: 𝑧𝑧 = 0, 𝑑𝑑𝐬𝐬 = −z� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ∴ ∫bottom 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = ∫0 ∫0 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 = 4.
face
32 8
Adding the above six values, we have∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 8 + 0 − +0−4+4=− ∙
3 3
𝜕𝜕 𝜕𝜕 𝜕𝜕
Now the divergence of 𝐀𝐀 is (𝑥𝑥𝑥𝑥) + (−𝑦𝑦𝑧𝑧 2 ) + (−𝑥𝑥𝑥𝑥) = 𝑧𝑧 − 𝑧𝑧 2 .
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
2 2 2 8
Hence ∫𝑉𝑉 𝛁𝛁 ∙ 𝐀𝐀 𝑑𝑑𝑑𝑑 = ∫0 ∫0 ∫0 (𝑧𝑧 − 𝑧𝑧 2 )𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = − ∙
3

which is the same as the result of the closed surface integral. The divergence theorem is therefore
verified.

Example 2.2. Given 𝐴𝐴 = x� 𝑥𝑥 2 + y� 𝑥𝑥𝑥𝑥 + z� 𝑦𝑦𝑦𝑦,verify the divergence theorem over a cubeone unit
on each side. The cube is situated in the first octant of the Cartesian coordinate system with one
corner at the origin.

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Matrices, Vectors & Fourier Analysis Spring 2023-24

Solution
We first evaluate the surface integral over the six faces.

i. Front face: 𝑥𝑥 = 1, 𝑑𝑑𝐬𝐬 = x� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 iv. Left face: 𝑦𝑦 = 0, 𝑑𝑑𝐬𝐬 = −y� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
1 1

� 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = � � 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 1. � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 0.

front 0 0 left
face face

ii. Back face: 𝑥𝑥 = 0, 𝑑𝑑𝒔𝒔 = −x� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 v. Top face: 𝑧𝑧 = 1, 𝑑𝑑𝒔𝒔 = z�𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
1 1
1
� 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 0. � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = � � 𝑦𝑦 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = .
2
back top 0 0
face face
iii. Right face: 𝑦𝑦 = 1, 𝑑𝑑𝒔𝒔 = y� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 vi. Bottom face: 𝑧𝑧 = 0, 𝑑𝑑𝒔𝒔 = −z� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
1 1
1
� 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = � � 𝑥𝑥 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = . � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 0
2
right 0 0 bottom
face face

1 1
Adding the above six values, we have∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 1 + 0 + + 0 + + 0 = 2.
2 2
𝜕𝜕 𝜕𝜕 𝜕𝜕
Now the divergence of 𝐀𝐀 is (𝑥𝑥 2 ) + (𝑥𝑥𝑥𝑥) + (𝑦𝑦𝑦𝑦) = 3𝑥𝑥 + 𝑦𝑦.
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
1 1 1
Hence ∫𝑉𝑉 𝛁𝛁 ∙ 𝐀𝐀 𝑑𝑑𝑑𝑑 = ∫0 ∫0 ∫0 (3𝑥𝑥 + 𝑦𝑦)𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 2.
which is the same as the result of the closed surface integral. The divergence theorem is therefore
verified.

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Matrices, Vectors & Fourier Analysis Spring 2023-24

Example 2.3. For a vector function 𝐀𝐀 = r�𝑟𝑟 2 + z� 2𝑧𝑧, verify for the circular cylindrical region
enclosed by 𝑟𝑟 = 5, 𝑧𝑧 = 0, 𝑧𝑧 = 4.

Solution
1 𝜕𝜕 1 𝜕𝜕 1 𝜕𝜕
𝛁𝛁 ∙ 𝐀𝐀 = (𝑟𝑟𝐴𝐴𝑟𝑟 ) + �𝐴𝐴ϕ � + (𝑟𝑟𝐴𝐴𝑧𝑧 )
𝑟𝑟 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕ϕ 𝑟𝑟 𝜕𝜕𝜕𝜕
1 𝜕𝜕 1 𝜕𝜕 1 𝜕𝜕 3
= (𝑟𝑟𝑟𝑟 2 ) + 0 + (𝑟𝑟2𝑧𝑧) = (𝑟𝑟 ) + 2 = (3𝑟𝑟 + 2)
𝑟𝑟 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕
4 2𝜋𝜋 5

� 𝛁𝛁 ∙ 𝐀𝐀 𝑑𝑑𝑑𝑑 = � � �(3𝑟𝑟 + 2) 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ𝑑𝑑𝑑𝑑 = 1200𝜋𝜋.

𝑉𝑉 𝑧𝑧=0 𝜑𝜑=0 𝑟𝑟=0

� 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 + � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 + � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬

𝑠𝑠 top bottom walls
face face face

i. Top face: 𝑧𝑧 = 4, 𝐀𝐀 = r�𝑟𝑟 2 + z� 8and 𝑑𝑑𝐬𝐬 = z� 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ

2𝜋𝜋 5

∴ � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 8 � � 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ = 200𝜋𝜋.

top 0 0
face

ii. Bottom face: 𝑧𝑧 = 0, 𝐀𝐀 = r�𝑟𝑟 2 and 𝑑𝑑𝒔𝒔 = −z� 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ ∴ ∫bottom 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 0.
face

iii. Walls face: 𝑟𝑟 = 5, 𝐀𝐀 = r� 25 + z�2𝑧𝑧 and 𝑑𝑑𝒔𝒔 = r� 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ

2𝜋𝜋 4

∴ � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 25 � � 5𝑑𝑑𝑑𝑑𝑑𝑑ϕ = 1000𝜋𝜋.

walls 0 0
face

Total, ∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 1200𝜋𝜋 = ∫𝑉𝑉 𝛁𝛁 ∙ 𝐀𝐀 𝑑𝑑𝑑𝑑 (verified).

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Matrices, Vectors & Fourier Analysis Spring 2023-24

Example 2.4 A vector field 𝐀𝐀 = r� 10𝑒𝑒 −𝑟𝑟 − z� 3𝑧𝑧, verify the divergence theorem for the cylindrical
region enclosed by 𝑟𝑟 = 2, 𝑧𝑧 = 0 and 𝑧𝑧 = 4.

Solution
1 𝜕𝜕 1 𝜕𝜕 1 𝜕𝜕
𝛁𝛁 ∙ 𝐀𝐀 = (𝑟𝑟𝐴𝐴𝑟𝑟 ) + �𝐴𝐴ϕ � + (𝑟𝑟𝐴𝐴𝑧𝑧 )
𝑟𝑟 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕ϕ 𝑟𝑟 𝜕𝜕𝜕𝜕
1 𝜕𝜕 1 𝜕𝜕 1
= (𝑟𝑟10𝑒𝑒 −𝑟𝑟 ) + 0 − (3𝑧𝑧𝑧𝑧) = 10𝑒𝑒 −𝑟𝑟 (1 − 𝑟𝑟) − 3.
𝑟𝑟 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕 𝑟𝑟

4 2𝜋𝜋 2
1
� 𝛁𝛁 ∙ 𝐀𝐀𝑑𝑑𝑑𝑑 = � � � � 10𝑒𝑒 −𝑟𝑟 (1 − 𝑟𝑟) − 3� 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ𝑑𝑑𝑑𝑑
𝑟𝑟
𝑉𝑉 0 0 0
4 2𝜋𝜋 2 4 2𝜋𝜋 2
−𝑟𝑟 (1
160𝜋𝜋
= 10 � � � 𝑒𝑒 − 𝑟𝑟)𝑑𝑑𝑑𝑑𝑑𝑑ϕ𝑑𝑑𝑑𝑑 − 3 � � � 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ𝑑𝑑𝑑𝑑 = − 48𝜋𝜋.
𝑒𝑒 2
0 0 0 0 0 0

i. Top face: 𝑧𝑧 = 4, 𝐀𝐀 = r� 10𝑒𝑒 −𝑟𝑟 − z� 12 and 𝑑𝑑𝒔𝒔 = z� 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ

2𝜋𝜋 2

∴ � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = −12 � � 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ = −48𝜋𝜋.

top 0 0
face

ii. Bottom face: 𝑧𝑧 = 0, 𝐀𝐀 = r� 10𝑒𝑒 −𝑟𝑟 and 𝑑𝑑𝒔𝒔 = z� 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ ∴ ∫bottom 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 0.
face

iii. Walls face: 𝑟𝑟 = 2, 𝐀𝐀 = r� 10𝑒𝑒 −2

− z� 3𝑧𝑧 and and 𝑑𝑑𝐬𝐬 = r� 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ
2𝜋𝜋 4
160𝜋𝜋
∴ � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 20 � � 𝑒𝑒 −2 𝑑𝑑𝑑𝑑𝑑𝑑ϕ = .
𝑒𝑒 2
walls 0 0
face

160𝜋𝜋 160𝜋𝜋
Total = + 0 − 48𝜋𝜋 = − 48𝜋𝜋.
𝑒𝑒 2 𝑒𝑒 2

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Matrices, Vectors & Fourier Analysis Spring 2023-24

Example 2.5 A vector field 𝐀𝐀 = r�𝑟𝑟 3 exists in the region between two concentric cylindrical
surfaces defined by 𝑟𝑟 = 1 and 𝑟𝑟 = 2, with both cylinders extending between 𝑧𝑧 = 0 and 𝑧𝑧 = 5.
Verify the divergence theorem by evaluating the following:(a) ∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 and (b) ∫𝑣𝑣 𝛁𝛁 ∙ 𝐀𝐀𝑑𝑑𝑑𝑑 .

Solution:
1 𝜕𝜕 1 𝜕𝜕 1 𝜕𝜕
𝛁𝛁 ∙ 𝐀𝐀 = (𝑟𝑟𝐴𝐴𝑟𝑟 ) + �𝐴𝐴ϕ � + (𝑟𝑟𝐴𝐴𝑧𝑧 )
𝑟𝑟 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕ϕ 𝑟𝑟 𝜕𝜕𝜕𝜕
1 𝜕𝜕 1 𝜕𝜕 1 𝜕𝜕
= (𝑟𝑟𝑟𝑟 3 ) + 0 + (0) = (𝑟𝑟 4 ) = 4𝑟𝑟 2 .
𝑟𝑟 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕ϕ 𝑟𝑟 𝜕𝜕𝜕𝜕

5 2𝜋𝜋 2 5 2𝜋𝜋 2 5 2𝜋𝜋

∫𝑉𝑉 𝛁𝛁 ∙ 𝐀𝐀𝑑𝑑𝑑𝑑 = ∫0 ∫0 ∫1 4𝑟𝑟 2 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = ∫0 ∫0 ∫1 4𝑟𝑟 3 𝑑𝑑𝑑𝑑𝑑𝑑ϕ𝑑𝑑𝑑𝑑 = ∫0 ∫0 [𝑟𝑟 4 ]12 𝑑𝑑ϕ𝑑𝑑𝑑𝑑 = 150𝜋𝜋.

i. Top face: 𝑧𝑧 = 5, 𝐀𝐀 = r�𝑟𝑟 3 and 𝑑𝑑𝒔𝒔 = z� 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ ∫top 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 0.

face

ii. Bottom face: 𝑧𝑧 = 0, 𝐀𝐀 = r�𝑟𝑟 3 and 𝑑𝑑𝒔𝒔 = −z�𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ ∫bottom 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 0.

face

iii. Outside surface: 𝑟𝑟 = 2, 𝐀𝐀 = r� 8 and 𝑑𝑑𝒔𝒔 = r�𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ

2𝜋𝜋 5

∴ � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 16 � � 𝑑𝑑𝑑𝑑𝑑𝑑ϕ = 160𝜋𝜋.

outside 0 0

iv. Inside surface: 𝑟𝑟 = 1, 𝐀𝐀 = r� and 𝑑𝑑𝒔𝒔 = −r�𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ

2𝜋𝜋 5

∴ � 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = − � � 𝑑𝑑𝑑𝑑𝑑𝑑ϕ = −10𝜋𝜋.

intside 0 0

Adding all surface, 160𝜋𝜋 − 10𝜋𝜋 = 150𝜋𝜋.

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Matrices, Vectors & Fourier Analysis Spring 2023-24

� 3𝑅𝑅2 , evalute both sides the divergence theorem for the

Example 2.6 For the vector field 𝐴𝐴 = R
region enclosed between spharical shells defined by 𝑅𝑅 = 1 and 𝑅𝑅 = 2.

Solution

�(𝑅𝑅2 )2 sin 𝜃𝜃 𝑑𝑑𝑑𝑑𝑑𝑑ϕ and we get 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 3(𝑅𝑅2 )4 sin 𝜃𝜃 𝑑𝑑𝑑𝑑𝑑𝑑ϕ

At the outer surface 𝑑𝑑𝐬𝐬 = R
2𝜋𝜋 𝜋𝜋
∴ ∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = ∫𝜑𝜑=0 ∫𝜃𝜃=0 3 × 24 × sin 𝜃𝜃 𝑑𝑑𝑑𝑑 𝑑𝑑ϕ = 192𝜋𝜋.
� (𝑅𝑅2 )2 sin 𝜃𝜃 𝑑𝑑𝑑𝑑𝑑𝑑ϕ and we get 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 3(𝑅𝑅2 )4 sin 𝜃𝜃 𝑑𝑑𝑑𝑑𝑑𝑑ϕ
At the inner surface 𝑑𝑑𝐬𝐬 = −R
2𝜋𝜋 𝜋𝜋
∴ ∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = − ∫𝜑𝜑=0 ∫𝜃𝜃=0 3 × 14 × sin 𝜃𝜃 𝑑𝑑𝑑𝑑 𝑑𝑑ϕ = −12𝜋𝜋.

Total = 192𝜋𝜋 − 12𝜋𝜋 = 180𝜋𝜋.

Again,
1 𝜕𝜕 𝜕𝜕 𝜕𝜕
𝛁𝛁 ∙ 𝐀𝐀 = � (3𝑅𝑅2 ∙ 𝑅𝑅2 sin 𝜃𝜃) + (0 ∙ 𝑅𝑅 sin 𝜃𝜃) + (0 ∙ 𝑅𝑅)� = 12𝑅𝑅
𝑅𝑅2 sin 𝜃𝜃 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕ϕ

Now, Outer sphere: ∫𝑉𝑉 𝛁𝛁 ∙ 𝐀𝐀 𝑑𝑑𝑑𝑑 = ∫𝑉𝑉 12𝑅𝑅 𝑑𝑑𝑑𝑑 =

𝜋𝜋 2𝜋𝜋 2

� � � 12 𝑅𝑅3 sin 𝜃𝜃 𝑑𝑑𝑑𝑑 𝑑𝑑ϕ 𝑑𝑑𝑑𝑑 = 192𝜋𝜋

𝜃𝜃=0 𝜑𝜑=0 𝑅𝑅=0

Inner sphere: ∫𝑉𝑉 𝛁𝛁 ∙ 𝐀𝐀 𝑑𝑑𝑑𝑑 = ∫𝑉𝑉 12𝑅𝑅 𝑑𝑑𝑑𝑑 =

𝜋𝜋 2𝜋𝜋 1

� � � 12 𝑅𝑅3 sin 𝜃𝜃 𝑑𝑑𝑑𝑑 𝑑𝑑ϕ 𝑑𝑑𝑑𝑑 = 12𝜋𝜋

𝜃𝜃=0 𝜑𝜑=0 𝑅𝑅=0

Total = 192𝜋𝜋 − 12𝜋𝜋 = 180𝜋𝜋.

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Matrices, Vectors & Fourier Analysis Spring 2023-24

Example 2.7 Find ∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 over the surface of a hemispherical region that is the top half of a
sphere of radius 3 centered at (0, 0, 0) with its flat base coinciding with the 𝑥𝑥𝑥𝑥 plane. Also verify
divergence theorem where 𝐀𝐀 = z� 𝑧𝑧.

Solution

Given that 𝑨𝑨 = z� 𝑅𝑅 cos 𝜃𝜃 (0 ≤ 𝜃𝜃 ≤ 𝜋𝜋).

� 𝑅𝑅 cos 2 𝜃𝜃 − θ� R cos 𝜃𝜃 sin 𝜃𝜃
In spherical coordinates 𝐀𝐀 = R
� 𝑅𝑅2 sin 𝜃𝜃 𝑑𝑑𝑑𝑑𝑑𝑑ϕ and we get 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = 𝑅𝑅3 cos 2 𝜃𝜃 sin 𝜃𝜃 𝑑𝑑𝑑𝑑𝑑𝑑ϕ
Over the hemisphere surface 𝑑𝑑𝐬𝐬 = R
𝜋𝜋
2𝜋𝜋 2
∴ ∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 = ∫𝜑𝜑=0 ∫𝜃𝜃=0
2 𝑅𝑅3 cos2 𝜃𝜃 sin 𝜃𝜃 𝑑𝑑𝑑𝑑 𝑑𝑑ϕ = 𝜋𝜋𝑅𝑅3 = 18𝜋𝜋. (𝑅𝑅 = 3)
3
1 𝜕𝜕 𝜕𝜕 𝜕𝜕
Now,𝛁𝛁 ∙ 𝐀𝐀 = � (𝑅𝑅 cos 2 𝜃𝜃 ∙ 𝑅𝑅2 sin 𝜃𝜃) + (−R cos 𝜃𝜃 sin 𝜃𝜃 ∙ 𝑅𝑅 sin 𝜃𝜃) + (0 ∙ 𝑅𝑅)�
𝑅𝑅2 sin 𝜃𝜃 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕ϕ

1
[3𝑅𝑅2 cos 2 𝜃𝜃 sin 𝜃𝜃 − 𝑅𝑅2 (2 sin 𝜃𝜃 cos 2 𝜃𝜃 − sin3 𝜃𝜃)] = 3cos 2 𝜃𝜃 − 2 cos 2 𝜃𝜃 + sin2 𝜃𝜃 = 1
𝑅𝑅2 sin 𝜃𝜃
and 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑙𝑙𝑅𝑅 𝑑𝑑𝑙𝑙𝜃𝜃 𝑑𝑑𝑙𝑙𝜑𝜑 = 𝑑𝑑𝑑𝑑(𝑅𝑅)𝑑𝑑𝑑𝑑(𝑅𝑅 sin 𝜃𝜃)𝑑𝑑ϕ
𝜋𝜋
2𝜋𝜋 3 2
So, ∫𝑉𝑉 𝛁𝛁 ∙ 𝐀𝐀 𝑑𝑑𝑑𝑑 = ∫𝑉𝑉 𝑑𝑑𝑑𝑑 = ∫𝜃𝜃=0
2
∫𝜑𝜑=0 ∫𝑅𝑅=0 𝑅𝑅2 sin 𝜃𝜃 𝑑𝑑𝑑𝑑 𝑑𝑑ϕ 𝑑𝑑𝑑𝑑 = 𝜋𝜋33 = 18𝜋𝜋 = ∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 ∙
3

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Matrices, Vectors & Fourier Analysis Spring 2023-24

Sample exercise - 6.2

1. For the vector field𝐀𝐀 = x� 𝑥𝑥𝑥𝑥 + y� 𝑦𝑦 2 𝑧𝑧 + z� 𝑥𝑥𝑥𝑥, verify the divergence theorem by computing
(a) the total outward flux flowing through the surface of a cube centered at the origin and
with sides equal to 2 units each and parallel to the Cartesian axes, (b) the integral of 𝛁𝛁 ∙ 𝐀𝐀
over the cube’s volume.

2. For a vector function 𝐀𝐀 = r�𝑟𝑟 2 + z� 3𝑧𝑧, verify the divergence theorem for the circular
cylindrical region enclosed by 𝑟𝑟 = 1, 𝑧𝑧 = 0, 𝑧𝑧 = 4. Ans: 20𝜋𝜋

3. A vector field 𝐀𝐀 = r� 𝑟𝑟 2 exists in the region between two concentric cylindrical surfaces
defined by 𝑟𝑟 = 2 and 𝑟𝑟 = 3, with both cylinders extending between 𝑧𝑧 = 0 and 𝑧𝑧 = 3.
Verify the divergence theorem by evaluating the following: (a) ∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 and (b)
∫𝑣𝑣 𝛁𝛁 ∙ 𝐀𝐀 𝑑𝑑𝑑𝑑. Ans: 114𝜋𝜋

4. Find ∮𝑠𝑠 𝐀𝐀 ∙ 𝑑𝑑𝐬𝐬 over the surface of a hemispherical region that is the top half of a sphere of
radius 4 centered at (0, 0, 0) with its flat base coinciding with the 𝑥𝑥𝑥𝑥 plane. Also verify
divergence theorem. where 𝐀𝐀 = R � 𝑅𝑅 cos 𝜃𝜃. Ans: 64𝜋𝜋

� 2𝑅𝑅3 , evaluate both sides of the divergence theorem for the

5. For the vector field 𝐴𝐴 = R
region enclosed between spherical shells defined by 𝑅𝑅 = 2 and 𝑅𝑅 = 3. Ans: 1688𝜋𝜋

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Matrices, Vectors & Fourier Analysis Spring 2023-24

3. Stokes’s Theorem:
Let 𝑆𝑆 be the open surface (two-sided) and 𝐶𝐶 be the closed boundary of 𝑆𝑆, the vector field 𝐀𝐀 is
continuous on 𝑆𝑆. Then

� (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 = � 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 .

𝑠𝑠 𝑐𝑐

Example 3.1. Assume that a vector field 𝐀𝐀 = x� (2𝑥𝑥 2 + 𝑦𝑦 2 ) + y� (𝑥𝑥𝑥𝑥 − 𝑦𝑦 2 ) , (a) find ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥
around the triangular contour, (b) find ∫𝑠𝑠 (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 over triangular arc, (c) verify Stokes’s
theorem and (d) can 𝐀𝐀 be expressed as gradient of a scalar? Explain.

Solution

(a) 𝑑𝑑𝒍𝒍 = x� 𝑑𝑑𝑑𝑑 + y� 𝑑𝑑𝑑𝑑 ∴ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = (2𝑥𝑥 2 + 𝑦𝑦 2 )𝑑𝑑𝑑𝑑 + (𝑥𝑥𝑥𝑥 − 𝑦𝑦 2 )𝑑𝑑𝑑𝑑

0 8
Path 𝑐𝑐1 ; 𝑥𝑥 = 0, 𝑑𝑑𝑑𝑑 = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = − ∫2 𝑦𝑦 2 𝑑𝑑𝑑𝑑 = ∙
1 3
2 16
Path 𝑐𝑐2 ; 𝑦𝑦 = 0, 𝑑𝑑𝑑𝑑 = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫0 2𝑥𝑥 2 𝑑𝑑𝑑𝑑 = ∙
2 3

Path 𝑐𝑐3 ; 𝑦𝑦 = 2 − 𝑥𝑥, 𝑑𝑑𝑑𝑑 = −𝑑𝑑𝑑𝑑,

0
28
� 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = �[{2𝑥𝑥 2 + (2 − 𝑥𝑥)2 } − {𝑥𝑥(2 − 𝑥𝑥) − (2 − 𝑥𝑥)2 }]𝑑𝑑𝑑𝑑 = − ∙
3
𝑐𝑐3 2
4
Total, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = − ∙
3
2 2−𝑥𝑥 4
Now, 𝛁𝛁 × 𝐀𝐀 = −z� 𝑦𝑦, 𝑑𝑑𝒔𝒔 = z� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ∴ ∫𝑠𝑠 (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 = − ∫0 ∫0 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 = − ∙
3

(c) Stokes’s theorem is verified. (d) No, ∵ 𝛁𝛁 × 𝐀𝐀 ≠ 0.

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Matrices, Vectors & Fourier Analysis Spring 2023-24

Example 3.2. Assume that a vector field 𝐀𝐀 = x� 𝑥𝑥𝑥𝑥 − y� (𝑥𝑥 2 + 2𝑦𝑦 2 ), (a) find ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 around the
triangular contour, (b) find ∫𝑠𝑠 (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 over triangular arc.

Solution

(a) 𝑑𝑑𝐥𝐥 = x� 𝑑𝑑𝑑𝑑 + y� 𝑑𝑑𝑑𝑑 ∴ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = 𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑 − (𝑥𝑥 2 + 2𝑦𝑦 2 ) 𝑑𝑑𝑑𝑑
1
Path 𝑐𝑐1 ; 𝑦𝑦 = 0, 𝑑𝑑𝑑𝑑 = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫0 𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑 − (𝑥𝑥 2 + 2𝑦𝑦 2 ) 𝑑𝑑𝑑𝑑 = 0 ∙
1

1
Path 𝑐𝑐2 ; 𝑥𝑥 = 1, 𝑑𝑑𝑑𝑑 = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫0 𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑 − (𝑥𝑥 2 + 2𝑦𝑦 2 ) 𝑑𝑑𝑑𝑑
2
1
5
= � −(1 + 2𝑦𝑦 2 )𝑑𝑑𝑑𝑑 = − ∙
3
0

Path 𝑐𝑐3 ; 𝑦𝑦 = 𝑥𝑥, 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑,

0 0
2
� 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = � 𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑 − (𝑥𝑥 2 + 2𝑦𝑦 2)
𝑑𝑑𝑑𝑑 = � 𝑦𝑦 2 𝑑𝑑𝑑𝑑 − (𝑦𝑦 2 + 2𝑦𝑦 2 ) 𝑑𝑑𝑑𝑑 = ∙
3
𝑐𝑐3 1 1
5 2
Total, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = − + = −1 ∙
3 3
𝑥𝑥� 𝑦𝑦� 𝑧𝑧̂
𝜕𝜕 𝜕𝜕 𝜕𝜕
Now, 𝛁𝛁 × 𝐀𝐀 = 𝛁𝛁 × 𝐀𝐀 = � 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
� = −3𝑥𝑥 z�, 𝑑𝑑𝒔𝒔 = z� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑥𝑥𝑥𝑥 −(𝑥𝑥 2 + 2𝑦𝑦 2 ) 0
1 𝑥𝑥

∴ � (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 = − � � 3𝑥𝑥 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = −1 ∙

𝑠𝑠 0 0

Stokes’s theorem is verified.

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Matrices, Vectors & Fourier Analysis Spring 2023-24

Example 3.3. Repeat Ex3.2. for the contour shown below.

Solution
(a) 𝑑𝑑𝒍𝒍 = x� 𝑑𝑑𝑑𝑑 + y� 𝑑𝑑𝑑𝑑 ∴ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = 𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑 − (𝑥𝑥 2 + 2𝑦𝑦 2 ) 𝑑𝑑𝑑𝑑
2
Path 𝑐𝑐1 ; 𝑦𝑦 = 0, 𝑑𝑑𝑑𝑑 = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫0 𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑 − (𝑥𝑥 2 + 2𝑦𝑦 2 ) 𝑑𝑑𝑑𝑑 = 0 ∙
1

Path 𝑐𝑐2 ; 𝑦𝑦 = 2 − 𝑥𝑥, 𝑑𝑑𝑑𝑑 = −𝑑𝑑𝑑𝑑,

1 1
11
� 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = � 𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑 − (𝑥𝑥 2 + 2𝑦𝑦 2)
𝑑𝑑𝑑𝑑 = � 𝑥𝑥(2 − 𝑥𝑥) 𝑑𝑑𝑑𝑑 + (𝑥𝑥 2 + 2(2 − 𝑥𝑥)2 ) 𝑑𝑑𝑑𝑑 = − ∙
3
𝑐𝑐3 0 2

Path 𝑐𝑐3 ; 𝑦𝑦 = 𝑥𝑥, 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑,

0 0
2
� 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = � 𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑 − (𝑥𝑥 2 + 2𝑦𝑦 2 ) 𝑑𝑑𝑑𝑑 = � 𝑦𝑦 2 𝑑𝑑𝑑𝑑 − (𝑦𝑦 2 + 2𝑦𝑦 2 ) 𝑑𝑑𝑑𝑑 =
3
𝑐𝑐3 1 1
11 2
Total, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = − + = −3 ∙
3 3
𝑥𝑥� 𝑦𝑦� 𝑧𝑧̂
𝜕𝜕 𝜕𝜕 𝜕𝜕
Now, 𝛁𝛁 × 𝐀𝐀 = 𝛁𝛁 × 𝐀𝐀 = � 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
� = −3𝑥𝑥 z�, 𝑑𝑑𝒔𝒔 = z� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑥𝑥𝑥𝑥 −(𝑥𝑥 2 + 2𝑦𝑦 2 ) 0
1 2−𝑦𝑦

∴ � (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 = − � � 3𝑥𝑥 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = −3 ∙

𝑠𝑠 0 𝑦𝑦

Stokes’s theorem is verified.

cosϕ
Example 3.4. For vector field 𝐀𝐀 = z� , verify Stokes’s theorem for a segment of a cylindrical
𝑟𝑟
surface defined by 𝑟𝑟 = 2, 𝜋𝜋/3 ≤ ϕ ≤ 𝜋𝜋/2, and 0 ≤ 𝑧𝑧 ≤ 3.

Solution
Stokes’s theorem states that ∫𝑠𝑠 (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝒔𝒔 = ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥

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Matrices, Vectors & Fourier Analysis Spring 2023-24

cos ϕ
LHS: With 𝐀𝐀 having only a component 𝐴𝐴𝑧𝑧 = ,use of the expression for 𝛁𝛁 × 𝐀𝐀 in cylindrical
𝑟𝑟
coordinates,
1 𝜕𝜕𝐴𝐴𝑧𝑧 𝜕𝜕𝐴𝐴ϕ 𝜕𝜕𝐴𝐴 𝜕𝜕𝐴𝐴 1 𝜕𝜕�𝑟𝑟𝐴𝐴ϕ � 𝜕𝜕𝐴𝐴𝑟𝑟
𝛁𝛁 × 𝐀𝐀 = r� � − � � 𝑟𝑟 − 𝑧𝑧 � + z� �
�+ϕ − �
𝑟𝑟 𝜕𝜕ϕ 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕ϕ
cos ϕ cos ϕ
1 𝜕𝜕 � 𝑟𝑟 � 𝜕𝜕 � � sin ϕ cos ϕ
= r� �
−ϕ 𝑟𝑟 ��
= −r� � 2 � + ϕ �
𝑟𝑟 𝜕𝜕ϕ 𝜕𝜕𝜕𝜕 𝑟𝑟 𝑟𝑟 2
The integral of 𝛁𝛁 × 𝐀𝐀 over the specified surface S is
3 𝜋𝜋/2 3 𝜋𝜋/2
sin ϕ cos ϕ sin ϕ 3
��
� (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝒔𝒔 = � � �−r� � 2 � + ϕ �� ∙ r� 𝑟𝑟𝑟𝑟ϕ𝑑𝑑𝑑𝑑 = − � � 𝑑𝑑ϕ𝑑𝑑𝑑𝑑 = − ∙
𝑟𝑟 𝑟𝑟 2 𝑟𝑟 4
𝑠𝑠 0 𝜋𝜋/3 0 𝜋𝜋/3

RHS: The surface S is bounded by contour 𝑐𝑐 = 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 shown in figure above. The direction of
𝑐𝑐 is chosen so that it is compatible with the surface normal r� by the right-hand rule. Hence,
𝑄𝑄 𝑅𝑅 𝑇𝑇 𝑃𝑃

� 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = � 𝐀𝐀 𝑃𝑃𝑃𝑃 ∙ 𝑑𝑑𝐥𝐥 + � 𝐀𝐀 𝑄𝑄𝑄𝑄 ∙ 𝑑𝑑𝐥𝐥 + � 𝐀𝐀 𝑅𝑅𝑅𝑅 ∙ 𝑑𝑑𝐥𝐥 + � 𝐀𝐀 𝑇𝑇𝑇𝑇 ∙ 𝑑𝑑𝐥𝐥

𝑐𝑐 𝑃𝑃 𝑄𝑄 𝑅𝑅 𝑇𝑇

where 𝑨𝑨𝑃𝑃𝑃𝑃 , 𝑨𝑨𝑄𝑄𝑄𝑄 , 𝑨𝑨𝑅𝑅𝑅𝑅 and 𝑨𝑨 𝑇𝑇𝑇𝑇 are the field 𝑨𝑨 along segments 𝑃𝑃𝑃𝑃, 𝑄𝑄𝑄𝑄, 𝑅𝑅𝑅𝑅 and 𝑇𝑇 𝑃𝑃respectively.
𝑐𝑐𝑐𝑐𝑐𝑐ϕ
Over segment 𝑃𝑃𝑃𝑃 the dot product of 𝑨𝑨𝑃𝑃𝑃𝑃 = z� � 𝑟𝑟𝑟𝑟ϕ is zero, and the same is true for
and 𝑑𝑑𝐥𝐥 = ϕ
𝑟𝑟
𝑐𝑐𝑐𝑐𝑐𝑐𝜋𝜋/2
segment 𝑅𝑅𝑅𝑅 . Over segment 𝑄𝑄𝑄𝑄 , ϕ = 𝜋𝜋/2 ; hence, 𝐀𝐀 𝑄𝑄𝑄𝑄 = z� = 0. For the last segment,
2
𝑐𝑐𝑐𝑐𝑐𝑐𝜋𝜋/3
𝐀𝐀 𝑇𝑇𝑇𝑇 = z� = z�/4 and 𝑑𝑑𝐥𝐥 = z�𝑑𝑑𝑑𝑑. Hence,
2

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Matrices, Vectors & Fourier Analysis Spring 2023-24

𝑃𝑃 0
1 1 3
� 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = � �z� ∙ � 𝑧𝑧̂ 𝑑𝑑𝑑𝑑 = � 𝑑𝑑𝑑𝑑 = − ∙
4 4 4
𝑐𝑐 𝑇𝑇 3

which is the same as the result obtained by evaluating the left-hand side of Stokes’s equation.

� sin ϕ, (a) find ∮ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 over the

Example 3.5. Assume that a vector field 𝐀𝐀 = r� 𝑟𝑟 cos ϕ + ϕ 𝑐𝑐
semicircular contour, and (b) find ∫𝑠𝑠 (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 over the surface of the semicircle.

Solution
� 𝑟𝑟𝑟𝑟ϕ ∴ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = 𝑟𝑟 cos ϕ 𝑑𝑑𝑑𝑑 + sin ϕ 𝑟𝑟 𝑑𝑑ϕ
(a) 𝑑𝑑𝐥𝐥 = r� 𝑑𝑑𝑑𝑑 + ϕ
2
Path 𝑐𝑐1 ; ϕ = 0, 𝑑𝑑ϕ = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫0 𝑟𝑟 cos ϕ 𝑑𝑑𝑑𝑑 = 2.
1

𝜋𝜋
Path 𝑐𝑐2 ; 𝑟𝑟 = 2, 𝑑𝑑𝑑𝑑 = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫0 sin ϕ 𝑟𝑟 𝑑𝑑ϕ = 4.
2

Path 𝑐𝑐3 ; ϕ = 𝜋𝜋, 𝑑𝑑ϕ = 0,

0 0
∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫2 𝑟𝑟 cos 𝜑𝜑 𝑑𝑑𝑑𝑑 = − ∫2 𝑟𝑟 𝑑𝑑𝑑𝑑 = 2 ∙ Total, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = 2 + 4 + 2 = 8 ∙
3

r� �
𝑟𝑟ϕ z�
1 𝜕𝜕 𝜕𝜕 𝜕𝜕 1
Now, 𝛁𝛁 × 𝐀𝐀 = 𝛁𝛁 × 𝐀𝐀 = � � = z� 𝑟𝑟 (sin ϕ + 𝑟𝑟 sin ϕ), 𝑑𝑑𝒔𝒔 = z� 𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑ϕ
𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕ϕ 𝜕𝜕𝜕𝜕
𝑟𝑟 cos ϕ 𝑟𝑟 sin ϕ 0
𝜋𝜋 2
∴ ∫𝑠𝑠 (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 = ∫0 ∫0 (sin ϕ + 𝑟𝑟 sin ϕ) 𝑑𝑑𝑑𝑑𝑑𝑑ϕ = 8 ∙ Stokes’s theorem is verified.

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Example 3.6. Repeat Ex3.5. for the contour shown below.

Solution
� 𝑟𝑟𝑟𝑟ϕ ∴ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = 𝑟𝑟 cos ϕ 𝑑𝑑𝑑𝑑 + sin ϕ 𝑟𝑟 𝑑𝑑ϕ
(a) 𝑑𝑑𝐥𝐥 = r� 𝑑𝑑𝑑𝑑 + ϕ
2 3
Path 𝑐𝑐1 ; ϕ = 0, 𝑑𝑑ϕ = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫1 𝑟𝑟 cos ϕ 𝑑𝑑𝑑𝑑 = .
1 2
𝜋𝜋
Path 𝑐𝑐2 ; 𝑟𝑟 = 2, 𝑑𝑑𝑑𝑑 = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫02 sin ϕ 𝑟𝑟 𝑑𝑑ϕ = 2.
2
𝜋𝜋 1
Path 𝑐𝑐3 ; ϕ = , 𝑑𝑑ϕ = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫2 𝑟𝑟 cos ϕ 𝑑𝑑𝑑𝑑 = 0 ∙
2 3
0
Path 𝑐𝑐4 ; 𝑟𝑟 = 1, 𝑑𝑑𝑑𝑑 = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫𝜋𝜋 sin ϕ 𝑟𝑟 𝑑𝑑ϕ = −1.
2 2
3 5
Total, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = + 0 + 2 − 1 = ∙
2 2

r� �
𝑟𝑟ϕ z�
1 𝜕𝜕 𝜕𝜕 𝜕𝜕 1
Now, 𝛁𝛁 × 𝐀𝐀 = 𝛁𝛁 × 𝐀𝐀 = � � = z� 𝑟𝑟 (sin ϕ + 𝑟𝑟 sin ϕ), 𝑑𝑑𝒔𝒔 = z� 𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑ϕ
𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕ϕ 𝜕𝜕𝜕𝜕
𝑟𝑟 cos ϕ 𝑟𝑟 sin ϕ 0
𝜋𝜋/2 2 5
∴ ∫𝑠𝑠(𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 = ∫0 ∫1 (sin ϕ + 𝑟𝑟 sin ϕ) 𝑑𝑑𝑑𝑑𝑑𝑑ϕ = 2 ∙ Stokes’s theorem is verified.

Example 3.7. Assume that a vector field 𝐀𝐀 = r� cos ϕ + ϕ � sin ϕ, (a) find ∮ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 over the path
𝑐𝑐
comprising a quarter section of a circle, and (b) find ∫𝑠𝑠(𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 over the surface of the quarter
section.

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Solution
� 𝑟𝑟𝑟𝑟ϕ ∴ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = cos ϕ 𝑑𝑑𝑑𝑑 + sin ϕ 𝑟𝑟𝑟𝑟ϕ
(a) 𝑑𝑑𝐥𝐥 = r� 𝑑𝑑𝑑𝑑 + ϕ
𝜋𝜋 3 3 𝜋𝜋
Path 𝑐𝑐1 ; ϕ = , 𝑑𝑑ϕ = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫0 cos ϕ 𝑑𝑑𝑑𝑑 = ∫0 cos 𝑑𝑑𝑑𝑑 = 0.
2 1 2
𝜋𝜋
Path 𝑐𝑐2 ; 𝑟𝑟 = 3, 𝑑𝑑𝑑𝑑 = 0, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫𝜋𝜋/2 sin ϕ 𝑟𝑟𝑟𝑟ϕ = 3.
2

0 0
Path 𝑐𝑐3 ; ϕ = 𝜋𝜋, 𝑑𝑑ϕ = 0, ∴ ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫3 cos ϕ 𝑑𝑑𝑑𝑑 = − ∫3 𝑑𝑑𝑑𝑑 = 3 ∙
3

Total, ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = 0 + 3 + 3 = 6 ∙

r� �
𝑟𝑟 ϕ z�
1 𝜕𝜕 𝜕𝜕 𝜕𝜕 1
Now, 𝛁𝛁 × 𝐀𝐀 = 𝛁𝛁 × 𝐀𝐀 = � � = z� r 2sin ϕ , 𝑑𝑑𝐬𝐬 = z� 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟ϕ
𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕ϕ 𝜕𝜕𝜕𝜕
cos ϕ 𝑟𝑟 sin ϕ 0
𝜋𝜋 3
∴ ∫𝑠𝑠 (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 = ∫𝜋𝜋/2 ∫0 2sin ϕ 𝑑𝑑𝑑𝑑𝑑𝑑ϕ = 6 ∙∴ Stokes’s theorem is verified.

� sin 𝜃𝜃 by evaluating it
� cos 𝜃𝜃 + ϕ
Example 3.8. Verify Stokes’s theorem for the vector field, 𝐀𝐀 = R
on the hemisphere of unit radius.
Solution
�
R 𝑅𝑅 θ� �
𝑅𝑅 sin 𝜃𝜃 ϕ
𝟏𝟏 𝜕𝜕 𝜕𝜕 𝜕𝜕
𝛁𝛁 × 𝐀𝐀 = 2 � �
𝑅𝑅 sin 𝜃𝜃 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕ϕ
cos 𝜃𝜃 𝑅𝑅 ∙ 0 𝑅𝑅 sin 𝜃𝜃 sin 𝜃𝜃
� 𝑅𝑅 sin2 𝜃𝜃+ϕ
� 2Rsin 𝜃𝜃 cos 𝜃𝜃−θ
R � 𝑅𝑅 sin2 𝜃𝜃
= ∙
𝑅𝑅2 sin 𝜃𝜃

and 𝑑𝑑𝐬𝐬 = �(𝑅𝑅𝑅𝑅𝑅𝑅)(𝑅𝑅 sin 𝜃𝜃 𝑑𝑑ϕ) ∴ (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 =

R
2R sin 𝜃𝜃 cos 𝜃𝜃 𝑑𝑑𝑑𝑑 𝑑𝑑ϕ , (𝑅𝑅 = 1).
𝜋𝜋/2 2𝜋𝜋

� (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 = � � 2 sin 𝜃𝜃 cos 𝜃𝜃 𝑑𝑑ϕ𝑑𝑑𝑑𝑑

𝑠𝑠 𝜃𝜃=0 𝜑𝜑=0
𝜋𝜋/2 1

= 4𝜋𝜋 � sin 𝜃𝜃 cos 𝜃𝜃 𝑑𝑑𝑑𝑑 = 4𝜋𝜋 � 𝑢𝑢 𝑑𝑑𝑑𝑑 = 2𝜋𝜋.

0 0
� 𝑅𝑅 sin 𝜃𝜃 𝑑𝑑ϕ ∴ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = 𝑅𝑅 sin 𝜃𝜃 𝑑𝑑ϕ = 𝑑𝑑ϕ, (𝑅𝑅 = 1 & 𝜃𝜃 = 𝜋𝜋/2).
Again, 𝑑𝑑𝐥𝐥 = ϕ 2

2𝜋𝜋
∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 = ∫0 𝑑𝑑ϕ = 2𝜋𝜋. ∴ Stokes’s theorem is verified.

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Sample exercise – 6.3

� 3sin ϕ by evaluating it on the

1. Verify Stokes’s theorem for the vector field 𝐀𝐀 = ϕ
2
hemisphere of radius 2. Ans: 24

2. For vector field 𝐀𝐀 = z� cos 𝜑𝜑, verify Stokes’s theorem for a segment of a cylindrical
surface defined by 𝑟𝑟 = 5, 𝜋𝜋/4 ≤ 𝜑𝜑 ≤ 𝜋𝜋/2, and 0 ≤ 𝑧𝑧 ≤ 4. Ans: −2√2

3. Assume that a vector field 𝐀𝐀 = x� (𝑥𝑥 2 − 𝑦𝑦 2 ) + y� (𝑥𝑥 2 − 𝑥𝑥𝑥𝑥), (a) find ∮𝑐𝑐 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 around the
rectangular contour, (b) find ∫𝑠𝑠 (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 over rectangular arc, (c) verify Stokes’s
theorem and (d) can 𝐀𝐀 be expressed as gradient of a scalar? Explain. Ans: 12

� cos ϕ , (a) find ∮ 𝐀𝐀 ∙ 𝑑𝑑𝐥𝐥 over the

4. Assume that a vector field, 𝐀𝐀 = r� 𝑟𝑟 sin ϕ + ϕ 𝑐𝑐
semicircular contour shown below, and (b) find ∫𝑠𝑠 (𝛁𝛁 × 𝐀𝐀) ∙ 𝑑𝑑𝐬𝐬 over the surface of the
semicircles. Ans: 0

Book: Fundamentals of Applied Electromagnetics (6th Edition)

Fawwaz T. Ulaby, Eric Michielssen, Umberto Ravaioli

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