C ONTENTS

1. Introduction 3
2. Double Integrations 3
2.1. Problems: Evaluation of Double Integral 5
3. Double Integral in Polar form 7
3.1. Problems on evaluation of Double integrals in Polar
coordinates: 7
4. Change of Variables 8
4.1. Problems: 9
5. Triple Integral: 10
5.1. Problems: 10
1
2

6. Coordinate System: 11
6.1. Problems: 13
 MODULE 2. INTEGRAL CALCULUS

DR.SHWETA NAIK

1. I NTRODUCTION

Double integral are sometimes easier to evaluate when we change the

order of integration or when we change to polar coordinate over the re-
gion whose boundaries are given by the polar equations. General change of
variables involves evaluation of multiple integral by substitution. The aim
of substitution is to replace complicated integrals by one that are easier to
evaluate change of variables simplifies the integrand, the limit of integration
or both. Evaluation of triple integral is simplified by the use of cylindrical
or spherical coordinates in problems of physics, engineering or geometry
involving a cylinder, cone or sphere. Double and triple integrals are useful
in evaluating area, volume, mass, centroid and moments of inertia of plane
and solid regions.

2. D OUBLE I NTEGRATIONS

The definite integral

Z b
f (x) dx
a
is defined as the limit of the sum f (x1 )δx1 + f (x2 )δx2 + . . . + f (xn )δxn ,
where n → ∞ and each of the lengths δx1 , δx2 , . . . , δxn tends to zero. A
double integral is its counterpart in two dimensions.
Let us consider a function f (x, y) of two variable x and y defined in
the finite region R of xy-plane. Divide the region A into elementary areas.
Then ZZ n
X
f (x, y)dA = lim f (xr , yr )δAr
R n→∞
r=1
3
4 DR.SHWETA NAIK

. Double interals would be limited if it were required limit sums to evaluate

them. To evaluate, above integral is expressed as the iterated integral
Z x2 Z y2
f (x, y)dxdy
x1 y1

. Note: Iterated integrals is evaluated as follows:

(i) When y1 and y2 are functions of x and x1 and x2 are constants, f (x, y) is
first integrated with respect to y keeping x fixed between limits y1 , y2 and
then resulting expression is integrated with respect to x within the limits x1 ,
x2 , i.e.,
Z x2 Z y2
I1 = f (x, y)dydx
x1 y1

where integration is carried from the inner to the outer rectangle.

Here AB and CD are the two curves whose equations are y1 = f1 (x) and
y2 = f2 (x). P Q is a vertical strip of width dx.
Then the inner rectangle integral means that the integration is along one
edge of the strip P Q from P to Q, while outer rectangle integral corre-
sponds to the sliding of the edge from AC to BD. Thus the whole region
of integration is the area ABDC.
(ii) When x1 , x2 are functions of y and y1 and y2 are constants, f (x, y)
is first integrated with respect to x keeping y fixed, within limits x1 , x2 and
the resulting expression is integrated with respect to y between the limits
y1 , y2 , i.e.,
Z y2 Z x2
I2 = f (x, y)dxdy
y1 x1

.
 MODULE 2. INTEGRAL CALCULUS 5

Here AB and CD are the curves x1 = f1 (y) and x2 = f2 (y). P Q is the

horizontal strip of width dy. Then inner rectangle indicates that the inte-
gration is alon one edge of this strip from P to Q while the outer rectangle
corresponds to the sliding of this edge from AC to BD. Thus the whole
region is ABDC.
(iii) When both pairs of limits are constants, the region of integration is
the rectangle ABDC. In I1 , we integrate along the vertical strip P Q and

then slide it from AC to BD.

In I2 , we integrate along the horizantal strip P ′ Q′ and then slide it from

AB to CD. Here Obviously I1 = I2 .

2.1. Problems: Evaluation of Double Integral.

RR
1 Evaluate R f (x, y)dA for f (x, y) = 100 − 6x2 y and
R : 0 ≤ x ≤ 2, −1 ≤ y ≤ 1.
RR R1 R2
sol: R f (x, y)dA= −1 0 (100 − 6x2 y)dxdy
R1
= −1 (200 − 16y)dy=400.
R1R1 y
2 Evaluate 0 0 1+xy dxdy.
R2R1
3 Evaluate 0 −1 (x − y)dydx.
R ln2 R ln5 2x+y
4 Evaluate 0 e dydx.
R 2π R π1
5 Evaluate π 0 (sinx + cosy)dxdy.
6 DR.SHWETA NAIK

6 Find the volume of the region bounded above by the elliptical pa-
raboloid z = 10 + x2 + 3y 2 and below by the rectangle
R : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2.
RR 2
7 Calculate R xyexy dA,R : 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.
8 Calculate R f (x, y)dA for f (x, y) = x2 yy2 +1 and
RR

R : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.
R1R1
9 Evaluate 0 0 √ dxdy √ .
1−x2 1−y 2
R 1 R y2
10 Evaluate the integral 0 0 3y 3 exy .
11 Find the volume of the prism whose base is the triangle in the xy-
plane bounded by the x-axis and the lines y = x and x = 1 and
whose top lies in the plane.
RR
Sol: Let V = R f (x, y)dA
For dA = dx.dy
Horizantal Strip, x = y to x = 1 (variable limits), y = 0 to y = 1
(constant limit)

R1 R1
V = y=0 x=y (3 − x − y)dxdy
R1
= y=0 ( 52 − 4y + 23 y 2 )dy = 1
For dA = dy.dx
Vertical Strip, y = 0 to y = x (variable limits), x = 0 to x = 1
(constant limit)

R1 Rx
V = x=0 y=0
(3 − x − y)dy dx
 MODULE 2. INTEGRAL CALCULUS 7
R1
= x=0
(3x − 23 x2 )dx = 1
.
RR
12 Evaluate R
(x + y)dydx. R is region bounded by x = 0, x = 2,
y = x, y = x + 2.
RR
13 Evaluate R xydxdy. Where R is the quadrant of the circle
x2 + y 2 = a2 where x ≥ 0 and y ≥ 0.
RR
14 Evaluate R x3 ydxdy. Where R is the region enclosed by the el-
2
x2
lipse + yb2 = 1 in the first quadrant.
a2
(x + y)2 dxdy over the area bounded by the ellipse
RR
15 Evaluate
x2 y2
a2
+ = 1.
b2
RR 2
16 IR = R
x dxdy. Where R is the region in the first quadrant
bounded by the hyperbola xy = 16 and the lines y = x, y = 0
and x = 8.
17 Find the area of the region R bounded by y = x and y = x2 in the
first quadrant.
18 Find the area of the region R enclosed by the parabola y = x2 and
the line y = x + 2.

3. D OUBLE I NTEGRAL IN P OLAR FORM

Double integrals are sometimes easier to evaluate if we change to polar

coordinates. This section shows how to accomplish the change and how to
evaluate double integrals over regions whose boundaries are given by polar
equations.

3.1. Problems on evaluation of Double integrals in Polar coordinates:

R π Ra
1 Evaluate 0
2
a(1−cosθ
r2 drdθ.
Sol:
8 DR.SHWETA NAIK

Rπ Ra
0
2
dθ a(1−cosθ r2 dr
3 R π
= a3 02 [1 − (1 − cosθ)3 ]dθ
3 R π a3
= a3 02 [cosθ3 − 3cos2 θ + 3cosθ]dθ = 36
[44 − 9π]

2 Find the area enclosed by the lemniscate r2 = 4cos2θ.

RR 2 2
3 Evaluate R ex +y dydx, where R is the semicircular region bounded
√
by the x-axis and the curve y = 1 − x2 .
4 Using polar integration, find the area of the region R in the xy-plane
enclosed by the circle x2 + y 2 = 4, above the line y = 1, and below
√
the line y = 3x.
5 Change the cartesian integral into an equivalent polar integral and
R 1 R √1−x2
evaluate −1 0 dydx.
5 Change the cartesian integral into an equivalent polar integral and
R 1 R √1−x2 2 2
evaluate 0 0 (x + y 2 ) dydx.

4. C HANGE OF VARIABLES

Double integration can be solved easily by change of independent vari-

RR
ables. Let the double integrals be R f (x, y)dxdy. It is to be changed by
new variables u and v.
Note:

Definition 4.1. The Jacobian determinant or Jacobian of the coordinate

transformation x = g(u, v), y = h(u, v) is
∂x ∂x
∂u ∂v ∂x ∂y ∂y ∂x
J(u, v) = ∂y ∂y
= −
∂u ∂v ∂u ∂v
∂u ∂v
 MODULE 2. INTEGRAL CALCULUS 9

.
The Jacobian can also be denoted by
∂(x, y)
J(u, v) =
∂(u, v)

Theorem 4.1. Suppose that f (x, y) is continuous over the region R. Let G
be the preimage of R under the transformation x = g(u, v), y = h(u, v),
assumed to be one-to-one on the interior of G, then
RR RR ∂(x, y)
R
f (x, y)dxdy = G
f (g(u, v), h(u, v)) | ∂(u, v)
| dudv.

4.1. Problems:
1 Evaluate
ZZ
(x + y)2 dxdy
R

, where R is the parallelogram in the xy-plane with vertices (1, 0),

(3, 1), (2, 2), (0, 1) using the transformation u = x + y
and v = x − 2y.
Sol:The region of integration is a parallelogram ABCD wher A(1, 0),
B(3, 1), C(2, 2) and D(0, 1) in the xy-plane. The new region of in-
tegration is a rectangle A’B’C’D’ in uv-plane.

xy-plane A(x, y) = (1, 0) B(x, y) = (3, 1) C(x, y) = (2, 2) D(x, y) = (0, 1)

uv-plane A (u, v) = (1, 1) B (u, v) = (4, 1) C (u, v) = (4, −2) D′ (u, v) = (1, −2)
′ ′ ′

∂(x, y)
J= = − 13 , dxdy = |J|dudv = 31 dudv
∂(u, v)
RR 2 R 1 R 4 21
R
(x + y) dxdy = −2 1
u 3 dudv = 21.
10 DR.SHWETA NAIK

2 Using the transformation x + y = u, y = uv. Show that

Z 1 Z 1−x
y 1
e (x+y) dydx = (e − 1)
0 0 2
.
3 Evaluate ZZ
(x2 + y 2 )dxdy
R
, where R is the Region in the first quadrant bounded by x2 −y 2 = a,
x2 − y 2 = b, 2xy = c, 2xy = d 0 < a < b, 0 < c < d.
4 Evaluate ZZ
(x + y)2 dxdy
R
where R is region bounded by the parallelogram x + y = 0,
x + y = 2, 3x − 2y = 0, 3x − 2y = 3.

5. T RIPLE I NTEGRAL :

Consider a function f (x, y, z) defined at every point of the 3-dimensional

finite region V . Divide V into n elementary volumes δV1 , δV2 , . . . , δVn . Let
(xr , yr , zr ) be any point within the rth sub-division δVr . Consider the sum
P∞
r=1 f (xr , yr , zr )δVr .

The limit of this sum, if it exists as n → ∞ and δVr → 0 is called the

triple integral of f (x, y, z) over the region V and is denoted by
ZZZ
f (x, y, z)dV

. For purposes of evaluation, it can also be expressed as the repeated integral

Z x2 Z y2 Z z2
f (x, y, z) dx dy dz
x1 y1 z1

5.1. Problems:
1 Evaluate Z 1 Z 2−x Z 2−x−y
dz dy dx
0 0 0
.
 MODULE 2. INTEGRAL CALCULUS 11

2 Evaluate
Z eZ e2 Z e3
1
dx dy dz
1 1 1 xyz
.
3 Evaluate
Z 2 Z 3 Z 2
xy 2 zdz dy dx
0 1 1

.
4 Evaluate
Z 1 Z √
1−x2 Z √1−x2 −y2
dx dy dz
0 0 0

6. C OORDINATE S YSTEM :

1. Cartesian Coordinates: Take a point O in the space draw three

mutually perpendicular lines through O. O is known as origin and
these three lines are known as x-axis, y-axis, z-axis. There are three

coordinate planes.

xy plane: The plane passing through z-axis and y-axis is known

as xy-plane.
yz plane: The plane passing through y-axis and z-axis is known
as yz-plane.
zx plane: The plane passing through z-axis and x-axis is known
as zx-plane.
12 DR.SHWETA NAIK

Consider a point P in the space draw perpendicular P Q to xy-plane.

P Q is known as z-coordinate; from Q draw perpendicular lines QR
and QS to x-axis and y-axis respectively. QS is x-coordinates, QR is
y-coordinate and small element of the solid in cartesian coordinates
is dx dy dz.
2. Spherical Coordinates: In the adjoining figure join OP. OP is de-
noted by r, θ is called the angle between OP i.e, r and z-axis. ϕ is
the angle between x-axis and OQ. Then the coordinates (r, θ, ϕ) of
a point P are known as spherical coordinates. Z = P Q = OT =
OP cosθ = rcosθ
OQ = P T = OP sinθ = rsinθ.
In right angled triangled triangle QRO, x = OR = OQcosϕ =
rsinθcosϕ Again in right angled triangle QSO, y = OQsinϕ =
rsinθsinϕ.

So, x = r sinθ cosϕ; y = r sinθ sinϕ; z = rcosθ. √

x2 +y 2
2 2
Also r = x + y + z ; ϕ =2 2
tan−1 xy ; θ −1
= tan z
If J =
∂(x, y z)
∂(r, θ ϕ)
then dx dy dz = r2 sinθdr dθ dϕ.
3. Cylindrical Coordinates: If P Q = z, OQ = ρ and OQ makes
angle ϕ with the x-axis then (z, ρ, ϕ) are known as cylindrical coor-
dinates.
 MODULE 2. INTEGRAL CALCULUS 13

x = ρ cosϕ; y = ρ sinϕ; z = z.
p
Also ρ = x2 + y 2 ; ϕ = tan−1 xy ; z = z
∂(x, y z)
If J = ∂(ρ, ϕ,z)
then dx dy dz = ρ dρ dϕ dz.

6.1. Problems:
R 2π R 1 R √1−r2
1 0 0 −√1−r2 rdzdrdθ
Ans: 4π .
R π R 23 R tanθ
2 0
4
0 0
(z − r2 )rdzdrdθ
√
Ans:1 − π4 + 4ln( 22 ).
RRR
3 B
(zrsinθ)rdrdθdz where the cylindrical box B is B = {(r, θ, z)|0 ≤
r ≤ 2, 0 ≤ θ ≤ π2 , 0 ≤ z ≤ 4}.
Ans: 64
3
.
4 By transforming into cylindrical coordinates evaluate the integral
RRR 2 2 2
(x +y +z ) dx dy dz taken over the region 0 ≤ z ≤ x2 +y 2 ≤
1.
Sol: Introducing cylindrical coordinates x = rcosθ, y = rsinθ,
R 1 RR
z = z, the given integral becomes 0 R (x2 + y 2 + z 2 ) dx dy dz =
R 1 R 2π R 1 2 2
0 0 0
(r cos θ + r2 sin2 θ + z 2 ) dr dθ dz .
Where R: circular region bounded by the circle of radius one and
centre at the origin: x2 + y 2 = 1. So that r varies fro 0 to 1 and θ
from 0 to 2π.

1
1 z2
Z
5π
= 2π ( + )dz =
0 4 2 6
14 DR.SHWETA NAIK
R 2π R π R 1
5 Evaluate the triple integral θ=0
2
ϕ=0 ρ=0
ρ2 sinϕdρdϕdθ.
Ans: 2π
3
.
R π R π R 3√2
6 Evaluate the triple integral ϕ=0
4 2
θ=0 ρ=0
ρ4 sinϕdρdθdϕ.
RπR ϕ R
2 2
7 Evaluate the triple integral ϕ
6
5ϕ 0
ρ sinϕdρdϕdθ.
2 6
8 Use Spherical coordinates to evaluate the integral
ZZZ
dx dy dz
p
x2 + y 2 + z 2
taken over the region V in the first octant bounded by the cones
π
θ= 4
and θ = arc tanz and the sphere x2 + y 2 + z 2 = 6.
Sol: Introducing spherical coordinate

x = ρsinθcosϕ
y = ρsinθsinϕ
x = ρcosθ
p
we have x2 + y 2 + z 2 = ρ
√
Jacobian = ρ2 sinθ and radius of the sphere is 6
Thus the given integral in spherical coordinates is
RRR dx dy dz R π R arctan2 R √6 1 2
√ = 0
2
π
0 ρ
ρ sinθdρdθdϕ
x2 +y 2 +z 2 4
√
since restricted to first octant, ρ varies from 0 to 6, θ:cone angle
π π
varies from θ1 = to θ2 = arctan2 and ϕ from 0 to
4 2
Z π Z arctan2
2
I=3 sinθdθdϕ
π
0 4

3π 1 1
(√ − √
2 2 5
.