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DEPARTMENT OF MATHEMATICS, IIT - GUWAHATI
Even Semester of the Academic Year 2019-
MA 101 Mathematics I
Problem Sheet 4: Multiple integrals and applications, Green’s Theorem, Stokes Theorem
and Divergence Theorem.
Instructors: Dr. J. C. Kalita and Dr. S. Bandopadhyay
- Evaluate the integrals:
(a)
3
0
1
0
x + ydx dy,
(b)
R
xy
2
x 2
dA, R = {(x, y)| 0 ≤ x ≤ 1 , − 3 ≤ y ≤ 3 }
(c)
R
x sin(x + y)dA, R =
[
π
]
×
[
π
]
Solution:
(a)
3
0
1
0
x + ydx dy =
3
0
[
(x + y)
3 2
] 1
0
dy
0
[
(1 + y)
3 (^2) − y
3 2
]
dy
[
(1 + y)
5 (^2) − y
5 2
] 3
0
=
4 15
(b) Ans = 9 ln(2).
(c) Ans =
1 2
π 12
- Find the volume of the solid lying under the plane z = 2x + 5y + 1 and above the rectangle
{(x, y)| − 1 ≤ x ≤ 1 , 1 ≤ y ≤ 4 }.
Solution: The required volume is given by:
V =
R
z dA =
4
1
1
− 1
(2x + 5y + 1)dx dy = 81.
0 4
5
10
1
z
3
15
y
20
x
2 0 -0. (^1) -
z=2x+5y+
- Find the volume of the solid lying under the elliptic paraboloid
x 2
y 2
square R = [− 1 , 1] × [− 2 , 2].
Solution: The required volume is given by:
V =
R
z dA =
− 2
− 1
x
2
y
2
)dx dy =
- Evaluate the iterated integrals:
2
y
0
x
0 0.5 (^1)
1
z
1
x
2 /4+y
2 /9+z=
(a)
1
0
e y
y
xdx dy
(b)
∫ π 2
0
∫ (^) cos θ
0
e
sin θ dr dθ.
Solution:
(a)
0
∫ (^) ey
y
xdx dy
0
(e
y )
3 (^2) − y
3 2
dy
e
3 y (^2) −
y
5 2
0
e
3 (^2) −
(b)
∫ π 2
0
cos θ
0
e
sin θ dr dθ
∫ π 2
0
[
re
sin θ
]cos θ
0
dθ
∫ π 2
0
e
sin θ cos θdθ
[
e
sin θ
] 1
0
= e − 1.
- Evaluate the double integrals:
(a)
D
2 y
x 2
dA, R = {(x, y)| 0 ≤ x ≤ 1 , 0 ≤ y ≤
x}
(b)
D
x cos ydA, D is bounded by y = 0, y = x
2 , x = 1
(c)
D
y
3 dA, D is the triangular region with vertices (0,2), (1,1) and (3,2).
Solution:
(a) The required integral is equal to: ∫ (^1)
0
√ x
0
2 y
x 2
dy dx
1
0
x 2
[
y
2
]
√ x
0
dx
[
ln(x
2
] 1
0
ln(2).
Solution: (a) The required volume is:
R
z dA =
1
0
x
x^2
(x
2
2 )dy dx
1
0
y
y^2
(x
2
2 )dx dy
0
[
x
2 y +
y
3
]
√ x
x 2
dx
1
0
(x
5 (^2) +
x
3 2
− x
4 −
x
6
)dx
18 105
(b) The required volume is:
R
z dA =
1
∫ (^7) − 3 y
1
xydx dy
1
∫ 7 −x 3
1
xydy dx
1
y
[
x
2
] 7 − 3 y
1
dy
1
y
(7 − 3 y)
2 − 1
dy
31 8
(c) The required volume is:
R
z dA =
2
0
∫ 2 −x 2
0
9 − x 2 dy dx
1
0
2 − 2 y
0
9 − x 2 dx dy
0
9 − x 2
2 − x
dx
0
9 − x 2
2 − x
dx
0
9 − x 2 dx −
0
x
9 − x 2 dx
9 2 sin
− 1 (
2 3
1 6
3 (^2) − 27).
(d) The required volume is:
R
z dA =
0
∫ (^1) −y
0
(1 − x − y)dx dy
0
∫ (^1) −x
0
(1 − x − y)dy dx
1 6
- Get an upper bound and lower bound of each of the integrals given below by using the result that if
m, M are such that m ≤ f (x, y) ≤ M for all (x, y) ∈ D then m × A(D) ≤
D
f (x, y)dA ≤ M × A(D),
where A(D) is the area of D.
(a)
D
x 3
- y 3 dA, D = [0, 1] × [0, 1]
(b)
D
e
x 2 +y 2 dA, D being the disk with center at origin and radius 0.5.
Solution: (a) Note that if there exists m and M such that m ≤ f (x, y) ≤ M for all (x, y) ∈ D
then
m × A(D) ≤
D
f (x, y)dA ≤ M × A(D), where A(D) is the area of D.
Since 0 ≤ f (x, y) =
x 3
2 for all (x, y) ∈ D,
D
x 3
2 × 1.
(b) As in part (a),
since 0 ≤ x
2
2 ≤ 0 .5 for all (x, y) ∈ D,
1 ≤ f (x, y) = e x 2 +y 2 ≤ e
- 5 for all (x, y) ∈ D
⇒ 1 × A(D) ≤
D
e
x 2 +y 2 dA ≤ e
- 5 × A(D), where A(D) is the area of the disc
⇒ 1 ×
π
D
e
x 2 +y 2 dA ≤ e
- 5 ×
π
- Using polar coordinates, find:
a
0
∫ π 2
0
a 2 − r 2 rdθdr
4 3
πa 3 .
(d) The volume of the closed region enclosed by the cylinder x
2
2 = 4 and the ellipsoid 4x
2
2
2 = 64.
Solution: The required volume is given by:
R
64 − 4 x 2 − 4 y 2 dA,
where R = {(x, y) : x
2
2 ≤ 4 }.
0
√ 4 −x^2
0
64 − 4 x 2 − 4 y 2 dydx (which is not easy to calculate).
After changing to polar coordinates the integral is given by:
∫ π 2
0
0
64 − 4 r^2 rdrdθ = − 4 π
[
(16 − r
2 )
3 2
] 2
0
8 π
(e)
0
√ 1 −x^2
0
e
x 2 +y 2 dydx.
Solution: The required integral after changing to polar coordinates is given by: ∫ π 2
0
1
0
e
r 2 rdrdθ
π 4
(e − 1).
(f)
2
0
4 −y^2
−
4 −y^2
x
2 y
2 dxdy.
Solution: The required integral after changing to polar coordinates is given by:
∫ π 2
0
0
r
2 cos
2 θ sin
2 θrdrdθ
4 π 3
- Find the mass, center of mass and the moments of inertia Ix, Iy and Io of the lamina D bounded
by the parabola x = y
2 and the line y = x − 2; the density is ρ(x, y) = 3.
Solution: By definition, the mass and the coordinates of the center of mass are given by:
M =
D
3 dA, ¯x =
D 3 xdA
M
, ¯y =
D 3 ydA
M
M =
2
− 1
y+
y^2
3 dxdy =
27 2
¯x =
D 3 xdA
M
2 − 1
y+ y^2
3 x dxdy
M
¯x =
¯y =
D
3 ydA
M
− 1
∫ (^) y+
y^2 3 y dxdy
M
¯y =
Ix =
− 1
∫ (^) y+
y^2
3 y
2 dxdy =
, Iy =
− 1
∫ (^) y+
y^2
3 x
2 dxdy =
and
I 0 = Ix + Iy =
- Find the area of the surface:
(a) The part of the plane 3x + 2y + z = 6 that lies in the first octant.
Solution: If z = f (x, y) then the required surface area is given by: ∫ ∫
R
1 + f 2 x
2
0
∫ 6 − 3 x 2
0
2
2 dydx
3
0
∫ 6 − 2 y 3
0
2
2 dxdy
(b) The part of the hyperbolic paraboloid z = y
2 − x
2 that lies between the cylinders x
2
2 = 1
and x
2
2 = 4.
Solution: If z = f (x, y) then the required area is given by: ∫ ∫
R
1 + f 2 x
After changing (1) to polar coordinates the above integral is equal to
1
∫ π 2
0
1 + 4r 2 rdθdr
π 6
( The expression for (1) in rectangular coordinates is given by:
1
√ 4 −x^2
√ 1 −x^2
1 + (− 2 x) 2
which is difficult to integrate ).
(d) Cut from the paraboloid z = r
2 by the cylinder r = 1.
Solution: If z = f (x, y) then the required area is given by: ∫ ∫
R
1 + f 2 x +^ f^
2 y dA,
where R is the unit disc centered at the origin.
The above integral I is equal to
0
√ 1 −x^2
0
1 + f 2 x
1
0
√ 1 −x^2
0
1 + (2x)
2
2 dydx.
By changing to polar coordinates we get:
I = 4
1
0
∫ π 2
0
1 + 4r 2 rdθdr
0
π
1 + 4r^2 rdr =
π
- Evaluate the triple integrals.
(a)
E
2 xdV , where E =
(x, y, z)| 0 ≤ y ≤ 2 , 0 ≤ x ≤
4 − y 2 , 0 ≤ z ≤ y
Solution: The required integral is given by: ∫ 2
0
y
0
4 −y^2
0
2 xdxdzdy
0
∫ (^) y
0
[
x
2
]
4 −y^2
0 dzdy
0
∫ (^) y
0
(4 − y
2 )dzdy
2
0
y(4 − y
2 )dy
Aliter: The required integral is also given by: ∫ 2
0
4 −y^2
0
y
0
2 xdzdxdy
2
0
4 −y^2
0
2 x [z]
y 0 dxdy
2
0
4 −y^2
0
2 xydxdy
0
y
[
x
2
]
4 −y^2
0
dy
0
y(4 − y
2 )dy
(b)
E
xdV , where E is bounded by the paraboloid x = 4y
2
2 and the plane x = 4.
Solution: The required integral is given by:
0
√ x 2
0
√ x− 4 y^2 2
0
xdzdydx
4
0
√ x 2
0
x − 4 y 2 xdydx
4
0
√ x 2
0
x − 4 y 2 xdydx
Take u such that
xu = 2y then
xdu = 2dy then the above integral reduces to:
2 ×
4
0
1
0
x − xu 2 x
xdudx
0
0
1 − u^2 x
2 dudx
4
0
x
2
[
sin
− 1 (u)
] 1
0
dx
π
[
x
3
] 4
0
π 4
2
16 π
Aliter: The required integral is also given by:
I =
D
4 z^2 +4y^2
xdx)dA, where D is the unit disc centered at the origin in the Y, Z plane.
Hence I =
D
2 − 4 z
2 − 4 y
2 )
2 )dA.
By changing to polar coordinates we get:
I = 4
0
∫ π 2
0
1 − r
4
rdθdr =
16 π 3
Aliter: 4
1
0
4
4 y^2
√ x− 4 y^2 2
0
xdzdxdy
1
0
4
4 y^2
x − 4 y 2
xdxdy
1
0
4
4 y^2
(x − 4 y
2 )
3 (^2) + 4y^2
x − 4 y 2
dxdy = 2
1
0
[
(x − 4 y
2 )
5 (^2) +
× 4 y
2 (x − 4 y
2 )
3 2
] 4
4 y^2
dy
1
0
× (
5 (1 − y
2 )
5 (^2) ) +
× 2
3 × (4y
2 (1 − y
2 )
3 (^2) )
dy
1
0
6
(1 − y
2 )
5 (^2) −
6
(1 − y
2 − 1)(1 − y
2 )
3 2
dy
1
0
7
(1 − y
2 )
5 (^2) +
6
(1 − y
2 )
3 2
dy
16 π 3
(the above calculations are lengthy and points out the merit of using polar coordinates)
1
0
2 π
0
2 r
4 (cos 2θ^ + 1)
2
dθdr
[
r
5
] 1
0
[
sin 2θ
θ
] 2 π
0
2 π 5
. Aliter: By using spherical coordinates the required integral is given by:
∫ π 2
cot−^1
sin φ
0
∫ π 2
0
ρ
2 cos
2 θ sin
2 φ
J
x, y, z
ρ, φ, θ
dθdρdφ
∫ π 2
cot−^1
sin φ
0
∫ π 2
0
ρ
2 cos
2 θ sin
2 φρ
2 sin φ dθdρdφ
[
∫ π 2
cot−^1
sin φ
0
ρ
4 sin
3 φ dρdφ
] [
∫ π 2
0
cos
2 θdθ
]
= π
∫ π 2
cot−^1
sin φ
0
ρ
4 sin
3 φdρdφ
= π
∫ π 2
cot − 1 2
sin
5 φ
sin
3 φdφ
π 5
∫ π 2
cot−^1
cosec
2 φdφ =
π
[− cot φ]
π 2 cot−^1
2 π
Aliter:
1
− 1
1 −y^2
−
1 −y^2
2
x^2 +y^2
0
x
2 dzdxdy
0
1 −y^2
0
x^2 + y^2 x
2 dxdy ( which is difficult to compute ).
(c) Evaluate
x 2
- y 2
- z 2 dV , where E is bounded below by the cone φ =
π 6 and above
by the sphere ρ = 2.
Solution: By using spherical coordinates the integral becomes: ∫ 2 π
0
∫ π 6
0
2
0
ρ
J
x, y, z
ρ, φ, θ
dρdφdθ
∫ (^2) π
0
∫ π 6
0
0
ρ(ρ
2 sin φ)dρdφdθ
= 4π(2 −
- Find the volume and centroid of the solid E that lies above the cone z =
x 2
sphere x 2
Solution: By using spherical coordinates the volume of the enclosed solid is given by: ∫ (^2) π
0
∫ π 4
0
0
J
x, y, z
ρ, φ, θ
dρdφdθ, (the angle φ =
π 4
for all points on the boundary of the cone)
2 π
0
∫ π 4
0
1
0
(ρ
2 sin φ)dρdφdθ,
2 π 3
1 √ 2
- Show that
∞
−∞
∞
−∞
∞
−∞
x 2
−(x 2 +y 2 +z 2 ) dxdydz = 2π.
Solution: By using spherical coordinates we get: ∫ 2 π
0
π
0
∞
0
ρe
−ρ 2
J
x, y, z
ρ, φ, θ
dρdφdθ
∫ (^2) π
0
∫ (^) π
0
0
e
−ρ 2 ρ(ρ
2 sin φ)dρdφdθ
= 2π [− cos φ]
π 0
0
e
−ρ 2 ρ
3 dρ
= 2π × 2 ×
[
−e
−ρ 2 ρ
2 − e
−ρ 2
]∞
0 = 2π.
- Use the given transformation to evaluate the integral:
(a)
R
(3x + 4y)dA, where R is the region bounded by the lines y = x, y = x − 2, y = − 2 x and
y = 3 − 2 x; x =
(u + v), y =
(v − 2 u).
Solution: Let I =
R
(3x + 4y)dA.
Then I =
3
0
∫ (^) x
− 2 x
(3x + 4y)dydx +
2 3
∫ (^) x
x− 2
(3x + 4y)dydx +
3
1
∫ (^3) − 2 x
x− 2
(3x + 4y)dydx,
calculations for which is lengthy.
By the given change of variables we get:
since x =
(u + v), y =
(v − 2 u)
- Evaluate the integral by making an appropriate change of variables:
(a)
R
cos
y − x
y + x
dA, where R is the trapezoidal region with vertices (1, 0), (2, 0), (0, 2) and
Solution: Let I =
R
cos
y − x
y + x
dA.
Then I =
1
0
2 −y
1 −y
cos
y − x
y + x
dxdy +
2
1
2 −y
0
cos
y − x
y + x
dxdy
(which is clearly difficult to compute).
By change of variables take u = x + y and v = y − x, then x =
u−v 2 and y =
u+v 2
The lines x = 0, y = 0, x + y = 1, x + y = 2 gives the lines v = u, v = −u, u = 1, u = 2
respectively in the changed coordinate system.
The Jacobian is given by:
J(
x,y u,v
1 2
1 2 1 2
1 2
1 2
Hence I =
2
1
u
−u
cos
v
u
J(
x, y
u, v
dvdu
1
[
u sin
v
u
)]
u
−u
du
3 2 sin(1).
(b)
R
(x 2
dxdy, where R is the first quadrant region bounded by the circles x 2
x
2
2 = 6x and the circles x
2
2 = 2y, x
2
2 = 8y.
Solution: Let I =
R
(x 2
dxdy.
Note that the above integral in rectangular coordinates is not easy to compute:
Take u =
2 x x^2 +y^2
and v =
2 y x^2 +y^2
, then y =
2 v u^2 +v^2
and x =
2 u u^2 +v^2
The circles x 2
- y 2 = 2x, x 2
- y 2 = 6x, x 2
- y 2 = 2y, x 2
- y 2 = 8y, gives the lines u = 1, u = 1 3
, v = 1, v =
1 4
, respectively.
The Jacobian is given by:
J(
x, y
u, v
2 (u^2 +v^2 )
4 u 2
(u^2 +v^2 )^2
4 uv (u^2 +v^2 )^2
−
4 uv (u^2 +v^2 )^2
2 (u^2 +v^2 )
4 v 2
(u^2 +v^2 )^2
(u 2
Hence I =
1 4
1 3
(u
2
2 )
2
J(
x, y
u, v
dudv
1 4
1 3
(u
2
2 )
(u 2
dudv
1 8
2 x + y = 2 8y
2 x + y =
2 6x
2 x + y =
2 2x
2 x + y =
2 2y
x
y
0 1/3 2/3^1
1
u=
v=1/
v=
1/
1/
3/
u=1/
R
S
u
v
0
- Let R be the solid ellipsoid with constant density δ and boundary surface
x 2
a 2
y 2
b 2
z 2
c 2
Use appropriate transformations to show that the mass M of R is
πδabc.
Solution: The mass M of the solid ellipsoid is given by: ∫ ∫ ∫
R
δdV = 8δ
∫ (^) a^2
0
∫ b a
√ a^2 −x^2
0
c
x 2
a 2
y 2
b 2
dydx, which is not easy to calculate:
Consider x = aρ sin φ cos θ, y = bρ sin φ sin θ, z = cρ cos φ.
The Jacobian is given by:
J(
x,y,z ρ,φ,θ
a sin φ cos θ aρ cos φ cos θ −aρ sin φ sin θ
b sin φ sin θ bρ cos φ sin θ bρ sin φ cos θ
c cos φ −cρ sin φ 0
= abcρ
2 sin φ.
Hence the above integral is given by:
∫ π 2
0
∫ π 2
0
1
0
δ
J
x, y, z
ρ, θ, φ
dρdθdφ,
= 8δ
∫ π 2
0
∫ π 2
0
0
abcρ
2 sin φdρdθdφ,
πδabc.
