1. Given P(4,2,1) and APQ=2ax +4ay +6az, find the point Q.
APQ = 2 ax + 4 ay + 6 az = (Qx-Px)ax + (Qy-Py)ay+(Qz-Pz)az
Qx-Px=Qx-4=2; Qx=6
Qy-Py=Qy-2=4; Qy=6
Qz-Pz=Qz-1=6; Qz=7
Ans: Q(6,6,7)
2. Given the points P(4,1,0)m and Q(1,3,0)m, fill in the table and make a sketch of the
vectors found in (a) through (f).
Vector Mag Unit Vector
a. Find the vector A AOP = 4 ax + 1 ay 4.12 AOP = 0.97 ax + 0.24 ay
from the origin to P
b. Find the vector B BOQ = 1 ax + 3 ay 3.16 aOQ = 0.32 ax + 0.95 ay
from the origin to Q
c. Find the vector C CPQ = -3 ax + 2 ay 3.61 aPQ = -0.83 ax + 0.55 ay
from P to Q
d. Find A + B A + B = 5 ax + 4 ay 6.4 a = 0.78 ax + 0.62 ay
e. Find C – A C - A = -7 ax + 1 ay 7.07 a = -0.99 ax + 0.14 ay
f. Find B - A B - A = -3 ax + 2 ay 3.6 a = -0.83 ax + 0.55 ay
a. AOP = (4-0)ax + (1-0)ay + (0-0)az = 4 ax + 1 ay.
AOP = 42 + 12 = 17 = 4.12
4 1
aOP = ax + a y = 0.97a x + 0.24a y
17 17
(see Figure 1)
b. BOQ =(1-0)ax + (3-0)ay + (0-0)az = 1 ax + 3 ay.
BOQ = 12 + 32 = 10 = 3.16 Fig. 1
1 3
aOQ = ax + a y = 0.32a x + 0.95a y
10 10
(see Figure 1)
c. CPQ = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + 2 ay.
C PQ = 32 + 2 2 = 13 = 3.61
-3 2
a PQ = ax + a y = -0.83ax + 0.55a y
13 13 Fig. 2
(see Figure 2)
d. A + B = (4+1)ax + (1+3)ay + (0-0)az = 5 ax + 4 ay.
A + B = 52 + 42 = 41 = 6.4
5 4
a= ax + a y = 0.78ax + 0.62a y
41 41
(see Figure 2)
e. C - A = (-3-4)ax + (2-1)ay + (0-0)az = -7 ax + 1 ay.
C - A = 72 + 12 = 50 = 7.07
-7 1 Fig 3
a= ax + a y = -0.99a x + 0.14a y
50 50
(see Figure 3)
f. B - A = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + 2 ay.
B - A = 32 + 2 2 = 13 = 3.6
-3 2
a= ax + a y = -0.83a x + 0.55a y
13 13
(see Figure 3)
3. Convert the following points from Cartesian to Spherical coordinates:
a. P(6.0, 2.0, 6.0)
b. P(0.0, -4.0, 3.0)
c. P(-5.0,-1.0, -4.0)
-1 �6 � -1 � 2� o
(a) r = 6 + 2 + 6 = 8.7, q = cos � �= 47 , f = tan � �= 18
2 2 2 o
�8.7 � �6 �
-1 �3� o -1 � -4 �
(b) r = 0 + 4 + 3 = 5, q = cos � �= 53 , f = tan � �= -90
2 2 2 o
�5 � �0 �
-1 � -1 � -1 �-1 �
(c) r = 5 + 1 + 4 = 6.5, q = cos � �= 130 , f = tan � �= 190
2 2 2 o o
�6.5 � �-5 �
4. Convert the following points from Spherical to Cartesian coordinates:
a. P(3.0, 30., 45.)
b. P(5.0, /4, 3/2)
c. P(10., 135, 180)
(a)
x = r sin q cos f = 3sin 30o cos 45o = 1.06
y = r sin q sin f = 3sin 30o sin 45o = 1.06
z = r cos q = 3cos 30o = 2.6
so P(1.1,1.1, 2.6).
(b)
x = r sin q cos f = 5sin 45o cos 270 o = 0
y = r sin q sin f = 5sin 45o sin 270o = -3.5
z = r cos q = 5cos 45o = 3.5
so P(0, -3.5,3.5).
(c)
x = r sin q cos f = 10sin135o cos180o = -7.1
y = r sin q sin f = 10sin135o sin180o = 0
z = r cos q = 10 cos135o = -7.1
so P(-7.1, 0, -7.1).
5. Convert the following points from cylindrical to Cartesian coordinates:
a. P(2.83, 45.0, 2.00)
b. P(6.00, 120., -3.00)
c. P(10.0, -90.0, 6.00)
(a)
x = r cos f = 2.83cos 45o = 2.00
y = r sin f = 2.83sin 45o = 2.00
z = z = 2.00
so P (2.00, 2.00, 2.00).
(b)
x = r cos f = 6.00 cos120o = -3.00
y = r sin f = 6.00sin120o = 5.20
z = z = -3.00
so P( -3.00,5.20, -3.00).
(c)
x = r cos f = 10.0 cos(-90.0o ) = 0
y = r sin f = 10.0sin(-90.0o ) = -10.0
z = z = 6.00
so P (0, -10.0, 6.00).
6. A 20.0 cm long section of copper pipe has a 1.00 cm thick wall and outer diameter of
6.00 cm.
a. Determine the total surface area (this could actually be useful if, say, you needed
to do an electroplating step on this piece of pipe)
b. Determine the weight of the pipe given the density of copper is 8.96 g/cm3
Fig. P2.14
(a) The top area, Stop, is equal to the bottom area. We must also find the inner area, Sinner,
and the outer area, Souter.
3 2
Stop = �
�r d r df = �
rd r �
df = 5 cm 2 .
2 0
Sbottom = Stop .
2 20
Souter = �
�r df dz = 3 �
df �
dz = 120 cm 2
0 0
2 20
Sinner == �
�r df dz = 2 �
df �
dz = 80 cm 2
0 0
The total area, then, is 210 cm2, or Stot = 660 cm2.
(b) Determining the weight of the pipe requires the volume:
V =�
�
�r d r df dz
3 2 20
rd r �
=� df �
dz = 100 cm3 .
2 0 0
� g �
M pipe = �
�
8.96 3 �
cm
(
�
100 cm3 )
= 2815 g.
So Mpipe = 2820g.
7. Sketch the following volumes and find the total charge for each given a volume charge
density of rv = 1nC/m3. Units (other than degrees) are meters.
(a) 0 ≤ x ≤ 4, 0 ≤ y ≤ 5, 0 ≤ z ≤ 6
(b) 1 ≤ r ≤ 5, 0 ≤ q≤ 60
(c) 1 ≤ r≤ 5, 0 ≤ f≤ 90, 0 ≤ z ≤ 5
4 5 6
r v dv = rv ���
(a) Q = � dx dy dz = 120nC
0 0 0
(b)
Q=� r v dv
5 60o 2
= rv � sin q dq
r dr �
2
df = 130nC
�
1 0 0
Fig. P2.24a
(c)
Q=�r v dv
5 2 5
= rv �
r dr �df �
dz = 94nC
1 0 0
Fig. P2.24b
Fig. P2.24c
8. You have a cylinder of 4.00 inch diameter and 5.00 inch length (imagine a can of
tomatoes) that has a charge distribution that varies with radius as rv = (6 r) nC/in3 where
r is in inches. (It may help you with the units to think of this as rv (nC/in3)= 6 (nC/in4)
r(in)) Find the total charge contained in this cylinder.
2 2 5
rv dv = �
Q=� �
�( 6r ) r d r df dz = 6�
r dr �
df �
2
dz = 160 nC = 503nC
0 0 0
9. Verify Stokes’s theorem for the vector field B= by evaluating:
B dl
C
over the semicircular contour shown in Fig. below.
10. Evaluate the line integral of E = x x – y y along the segment P1 to P2
of the circular path shown in the figure.
