INDIAN INSTITUTE OF TECHNOLOGY (BHU), VARANASI

Odd Semester 2020-2021

Engineering Mathematics I (MA 101)
Solutions to Tutorial Sheet 10

Topics Covered: Multiple integral integrals (Unit VI)

1. Evaluate the following integrals.

RR
(i) R ydxdy, where R is the region in the first quadrant bounded by the parabolas
y 2 = 4x and x2 = 4y.
RR
(ii) R rdrdθ, where R is the area bounded by {(r, θ) : r2 = 4 cos 2θ}.

Solution:

(i) Solving y 2 = 4x and x2 = 4y we get y = 0, 4 and x = 0, 4. Therefore, the points

of intersection are (0, 0) and (4, 4). The region of integration R is denoted in the
figure below:

√
Thus, R = {(x, y) : 0 ≤ x ≤ 4, x2 /4 ≤ y ≤ 2 x}
Therefore,
√
4 2 y 4 4
√ y3
ZZ Z Z Z Z  
2 y 3/2 48
ydxdy = ydxdy = y[x]y2 /4 dy = 2y − dy =
R 0 y 2 /4 0 0 4 5
 (ii) To determine the limit of integration we graph the function and see from the sym-
metry of the region that the total area is 4 times the first-quadrant portion.

√
To integrate over the shaded region, we run r from 0 to 4 cos 2θ and θ from 0 to
π/4.
Therefore,

π/4
√
4 cos 2θ π/4 r=√4 cos 2θ
r2
ZZ Z Z Z 
rdrdθ = 4 rdrdθ = 4 dθ
R 0 0 0 2 r=0
Z π/4 π/4
=4 2 cos 2θdθ = 4 sin 2θ = 4.
0 0

2. By changing the order of integration, evaluate

RaRy y
(i) 0
√
y 2 /a (a−x)
dxdy
ax−y 2
R∞R∞ e−y
(ii) 0 x y
dydx

Solution:

(i) The given limit shows that the region of integration is bounded by the curves
x = y, x = y 2 /a where 0 ≤ x ≤ a which is shown by R in the figure below.

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 To change the integration and integrate first with respect to y, the limit of integra-
√
tion of y are from x to ax and of x are from 0 to a, which can be verified from
the above figure. Thus, the changed integration becomes
√ √
Z a Z ax Z a Z ax
y 1 y
p dydx = p dydx
0 x (a − x) ax − y 2 0 (a − x) x ax − y 2
Substituting ax − y 2 = t we get
Z a Z 0
1 −1
= √ dtdx
0 (a − x) ax−x2 2 t
Z a √ 0

1
= (− t) dx
0 (a − x) ax−x2
Z a
1 √ √
= x a − x dx
0 (a − x)
Z ar
x
= dx
0 (a − x)
√
Substituting u = x we get
√
a a
u2
Z r Z
x
dx = 2 √ du
0 (a − x) 0 a − u2
√
Substituting u = a sin v we get
Z π/2 Z π/2
2 1 − cos 2v π
= 2a sin (v)dv = 2a dv = a.
0 0 2 2

(ii) The given limit shows that the region of integration is bounded by the curves
y = x, y = ∞, x = 0, x = ∞.
To change the order of integration and integrate first with respect to x, consider a

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 strip parallel to the x-axis.

Now the limit of integration of x are from 0 to y and of y are from 0 to ∞. Thus,
the changed integration becomes
∞ y ∞
e−y e−y y
Z Z Z
dxdy = [x]0 dy
0 0 y y
Z0 ∞ Z ∞
e−y
= · ydy = e−y dy = [−e−y ]∞
0 = 1.
0 y 0

R 2 R √2x−x2
3. Evaluate 0 0
√xdydx by changing into polar co-ordinates.
2
x +y 2

Solution:
The region of integration is the top half of the disk centered at (1,0) with radius 1. To
√
see this, observe that the upper bound on y is 2x − x2 :
√
y= 2x − x2 ⇐⇒ y 2 = 2x − x2 ⇐⇒ x2 − 2x + 1 + y 2 = 1 ⇐⇒ (x − 1)2 + y 2 = 1.

Further, we have

x2 +y 2 = 2x ⇐⇒ r2 = 2r cos θ ⇐⇒ r = 2 cos θ (Substituting x = r cos θ and y = r sin θ).

√
This indicates that y = 2x − x2 can be represented by r = 2 cos θ in polar coordinates,
which can be shown in the figure below

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To evaluate the integral in the polar co-ordinate, first we find the Jacobian of the trans-
formation x = r cos θ and y = r sin θ, which is given by the following formula:

∂x ∂x
∂(x, y) ∂r ∂θ
= ∂y ∂y
.
∂(r, θ) ∂r ∂θ

As ∂x
∂r
= cos θ, ∂x
∂θ
= −r sin θ, ∂y
∂r
= sin θ and ∂y
∂θ
= r cos θ.
Thus,
∂(x, y) cos θ −r sin θ
= = r.
∂(r, θ) sin θ r cos θ

we know that
∂(x, y)
dxdy = drdθ.
∂(r, θ)
Therefore,
√
Z 2 Z 2x−x2 Z π/2 Z 2 cos θ
xdydx r cos θrdrdθ
p =
0 0 x2 + y 2 0 0 r
π/2 Z 2 cos θ π/2 2 cos θ
r2
Z Z
= r cos θdrdθ = cos θ dθ
0 0 0 2 0
Z π/2 Z π/2
3
=2 cos θdθ = 2 (1 − sin2 θ) cos θdθ
0 0

substituting u = sin θ we get

Z 1 1
u3

2 4
=2 (1 − u )du = 2 u − = .
u=0 3 0 3

∂(x,y)
4. Find the Jacobian ∂(u,v)
of the following transformation:
x = u cos v, y = u sin v.
Solution:

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The Jacobian of the transformation

Φ : (u, v) → (x(u, v), y(u, v))

is the 2 × 2 determinant
∂x ∂x
∂(x, y) ∂u ∂v
= ∂y ∂y
.
∂(u, v) ∂u ∂v

∂x ∂x ∂y ∂y
Since ∂u
= cos v, ∂v
= −u sin v, ∂u
= sin v, and ∂v
= u cos v, therefore, our Jacobian is

∂(x, y) cos v −u sin v

= = u.
∂(u, v) sin v u cos v

∂(x,y,z)
5. Find the Jacobian ∂(u,v,w)
of the following transformation:
x = 2u − 1, y = 3v − 4 and z = (1/2)(w − 4).
Solution:
The Jacobian of the transformation

Φ : (u, v, w) → (x(u, v, w), y(u, v, w), z(u, v, w))

is the 3 × 3 determinant
∂x ∂x ∂x
∂u ∂v ∂w
∂(x, y, z) ∂y ∂y ∂y
= ∂u ∂v ∂w
.
∂(u, v, w) ∂z ∂z ∂z
∂u ∂v ∂w

∂x ∂x ∂x ∂y ∂y ∂y ∂z ∂z ∂x
Since ∂u
= 2, ∂v
= ∂w
= 0, ∂v
= 3, ∂u
= ∂w
= 0, and ∂w
= 1/2, ∂u
= ∂v
= 0, therefore,
our Jacobian is

2 0 0
∂(x, y, z)
= 0 3 0 = 3.
∂(u, v, w)
0 0 1/2
RR
6. Evaluate R
(x + y)2 dxdy, where R is the parallelogram in the xy-plane with vertices
(1,0), (3,0), (2,2) and (0,1), using the transformation u = x + y and v = x − 2y.
Solution:
Let R is the parallelogram in xy-plane with vertices (1,0), (3,0), (2,2) and (0,1) is shown
in the figure below

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Note that R is bounded by the lines x + y = 1, y = 0, 2y − x = 2 and y + 2x = 6.
Plugging in the transformation gives:

x + y = 1 =⇒ u = 1
y = 0 =⇒ u = v
2y − x = 2 =⇒ v = −2
y + 2x = 6 =⇒ 5u + v = 18.

Therefore, after applying the transformations the parallelogram S in the uv-plane whose
sides are u = 1, u = v, v = −2 and 5u + v = 18 is shown in the figure below. The
equation of the line joining the co-ordinates (1,-2), (3,3) is 5u − 2v = 9

From the given transformation we can find the values of x and y in terms of u and v.

2u + v
x= ,
3

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 u−v
y= .
3
From which we find the Jacobian

∂(x, y) 2/3 1/3

= = −1/3,
∂(u, v) 1/3 −1/3

Therefore, the given integral in xy-plane can be transformed into the integral in uv-plane
by the following
9+2v
ZZ Z 1 Z Z 4 Z 18−5u
2
5
2 ∂(x, y) ∂(x, y)
(x + y) dxdy = u dvdu + u2 dvdu
R −2 v ∂(u, v) 1 5u−9
2
∂(u, v)
9+2v
Z 1 Z
5
Z 4 Z 18−5u
2
= u (1/3)dvdu + u2 (1/3)dvdu
5u−9
−2 v 1 2

7. Find by double integration, the area lying inside the circle r = a sin θ and outside the
cardiod r = a(1 − cos θ).

Solution:
For the given curves r = a sin θ and r = a(1−cos θ), the required area is shown by shaded
region in the figure below (we have drawn the figure for a=3)

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 The point of intersection of two curves is given by

a sin θ = a(1 − cos θ)

=⇒ sin θ + cos θ = 1
=⇒ 2 sin θ/2 cos θ/2 + (1 − 2 sin2 θ/2) = 1
=⇒ sin θ/2(sin θ/2 − cos θ/2) = 0
=⇒ sin θ/2 = 0 or (sin θ/2 − cos θ/2) = 0
=⇒ θ = 0 or tan θ/2 = 1 =⇒ θ = π/2.

Hence, in limit of integration r varies from a(1 − cos θ) to a sin θ and θ varies from 0 to
π/2 (can also be verified from above figure). Let A be the required area, then
Z π/2 Z a sin θ
A= rdrdθ
0 a(1−cos θ)
π/2 2 a sin θ
Z 
r
= dθ
0 2 a(1−cos θ)
π/2
a2
Z
= [2 cos θ − 1 − cos 2θ]dθ
2 0
π/2
a2

sin 2θ
= 2 sin θ − θ −
2 2 0
2
a (4 − π)
=
4
x2 y2 z2
8. Find by double integration, the volume of the ellipsoid a2
+ b2
+ c2
= 1.
Solution:
x2 y2 z2
The ellipsoid a2
+ b2
+ c2
= 1 is shown by a given figure below:

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 It can be notice that if we project the ellipsoid to the xy -plane (z q = 0) it gives
x2 y2 2 2
the ellipse a2 + b2 = 1. The two heights at each (x, y) points are z = ±c 1 − xa2 − yb2 .
q
2 2
Thus, the difference, 2c 1 − xa2 − yb2 , is the length of thin rectangle above that point
√
and is the function to be integrated. For each x the ellipse goes from y = −b
a
a2 − x2 to
√
y = ab a2 − x2 and x goes from −a to a. Thus, the volume of the ellipse is given by

b
√ r
a a2 −x2
x2 y 2
Z Z
a
V = √
2c 1 − − 2 dxdy
−a −b
a
a2 −x2 a2 b
b
√ r
aZ a2 −x2
x2 y 2
Z
a
=8c 1− − 2 dxdy.
0 0 a2 b

Since the integral Z A √ π

A2 − x2 dx = A2
0 4
the change of varible x = y/b brings
s  2
Z bA
y π
A2 − dy = A2 b .
0 b 4

Therefore, using the above result we can write

b
√ r
a a2 −x2 a
x2 y 2 bπ a2 − x2
Z Z Z
a
V = 8c 1 − 2 − 2 dxdy = 8c dx
0 0 a b 0 4 a2
Z a 2
a − x2
= 2πbc dx
0 a2
a
x3

= 2πbc x − 2
3a 0
a
= 2πbc(a − )
3
4πabc
= .
3

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