UNIT 4
Multiple Integrals
Double integral –
Before studying about multiple integrals, first let’s go through the definition of definition of definite integrals for function of single variable.
As we know, the integral
Where is belongs to the limit a ≤ x ≤ b
This integral can be written as follows-
Now suppose we have a function f(x , y) of two variables x and y in two dimensional finite region R in xy-plane.
Then the double integration over region R can be evaluated by two successive integration
Evaluation of double integrals-
If A is described as 
Then,
Let do some examples to understand more about double integration-
Example-1: Evaluate
, where dA is the small area in xy-plane.
Sol. Let, I = 
= 
= 
= 
= 84 sq. Unit.
Which is the required area.
Example-2: Evaluate 
Sol. Let us suppose the integral is I,
I = 
Put c = 1 – x in I, we get
I = 
Suppose, y = ct
Then dy = c
Now we get,
I = 
I = 
I = 
I = 
I = 
As we know that by beta function,
Which gives,
Now put the value of c, we get
Example-3: Evaluate the following double integral,
Sol. Let ,
I = 
On solving the integral, we get
Double Integral over Rectangular and general regions
Consider a function f (x, y) defined in the finite region R of the x-y plane. Divide R into n elementary areas A1, A2,…,An. Let (xr, yr) be any point within the rth elementary are Ar
Fig. 6.1
f (x, y) dA = f (xr, yr) A
Evaluation of Double Integral when limits of Integration are given(Cartesian Form).
Ex. 1 : Evaluate 
ey/x dy dx.
Soln. :
Given : I = 
ey/x dy dx
Here limits of inner integral are functions of y therefore integrate w.r.t y,
I = 
dx
= 
= 
I = 
= =
ey/x dy dx =
Ex. 2 : Evaluate x
y
(1 – x –y)
dx dy.
Soln. :
Given : I = x
y
(1 – x –y)
dx dy.
Here the limits of inner integration are functions of y therefore first integrate w.r.t y.
I = x
dx 
Put 1 – x = a (constant for inner integral)
I = x
dx 
Put y = at dy = a dt
y | 0 | a |
t | 0 | 1 |
I = x
dx 
I = x
dx 
I = x
a dx
I = x
(1 – x) dx = (x
– x4/3) dx
I = 
=
I = =
x
y
(1 – x –y)
dx dy =
Ex. 3: Evaluate 
Soln. :
Let, I = 
Here limits for both x and y are constants, the integral can be evaluated first w.r.t any of the variables x or y.
I = dy 
I = 
= 
= 
= 
=
= 
=
Ex. 4: Evaluate e–x2 (1 + y2) x dx dy.
Soln. :
Let I = e–x2 (1 + y2) x dy = dy e–x2 (1 + y2) x dy
= dy e– x2 (1 + y2) 
dx
= dy [∵ f (x) ef(x) dx = ef(x) ]
= (–1) dy (∵ e– = 0)
= = =
e–x2 (1 + y2) xdx dy =
NOTES:
Type II: Evaluation of Double Integral when region of Integration is provided (Cartesian form)
Ex.1: Evaluate y dx dy over the area bounded by x= 0 y = 
and x + y = 2 in the first quadrant
Soln. :
The area bounded by y = x2 (parabola) and x + y = 2 is as shown in Fig.6.2
The point of intersection of y = x2 and x + y = 2.
x + x2 = 2 x2 + x – 2 = 0
x = 1, – 2
At x = 1, y = 1 and at x = –2, y = 4
(1, 1) is the point of intersection in Ist quadrant. Take a vertical strip SR, Along SR x constant and y varies from S to R i.e. y = x2 to y = 2 – x.
Now slide strip SR, keeping IIel to y-axis, therefore y constant and x varies from x = 0 to x = 1.
I = 
= 
= 
= (4 – 4x + 
– 
) dx
= =
I = 16/15
Ex. 2 : Evaluate 
over x 1, y 
Soln. :
Let I = 
over x 1, y 
The region bounded by x 1 and y 
Is as shown in Fig. 6.3.
Take a vertical strip along strip x constant and y varies from y = 
To y = . Now slide strip throughout region keeping parallel to y-axis. Therefore y constant and x varies from x = 1 to x = .
I = 
= 
= 
[ ∵
dx = tan–1 (x/a)]
= 
=
= – = (0 – 1)
I =
Ex. 3 : Evaluate (
+ 
) dx dy through the area enclosed by the curves y = 4x, x + y = 3 and y =0, y = 2.
Soln. :
Let I = (
+ 
) dx dy
The area enclosed by the curves y = 4x, x + y =3, y = 0 and y = 2 is as shown in Fig. 6.4.
(find the point of intersection of x + y = 3 and y = 4x)
Take a horizontal strip SR, along SR y constant and x varies from x = to x = 3 – y. Now slide strip keeping IIel to x axis therefore x constant and y varies from y = 0 to y = 2.
I = dy (
+ 
) dx
= 
= 
+
dy
I = 
= 
= 
= + – 6 + 18
I =
Triple integrals
Definition: Let f(x,y,z) be a function which is continuous at every point of the finite region (Volume V) of three dimensional space. Divide the region V into n sub regions of respective volumes
. Let (
) be a point in the 
sub region then the sum:
= 
Is called triple integration of f(x, y, z) over the region V provided limit on R.H.S of above Equation exists.
Spherical Polar Coordinates
Where the integral is extended to all positive values of the variables subjected to the condition 
Ex.1: Evaluate
Solution : Let
I = 
= 
(Assuming m = 
)
= 
dxdy
= 
=
= 
dx
= 
dx
= 
=
I =
Ex.2: Evaluate 
Where V is annulus between the spheres 
And 
(
)
Solution: It is convenient to transform the triple integral into spherical polar co-ordinate by putting
, 
, 
, dxdydz=
sin
drd
d
,
and 
For the positive octant, r varies from r =b to r =a , 
varies from 
And 
varies from 
I= 
= 8
=8
=8
=8
=8 log
= 8 log
I= 8 log
I = 4 log
Ex.3: Evaluate 
Solution:-
Ex.4:Evaluate 
Solution:- 
NOTES:
The volume of solid is given by
Volume =
In Spherical polar system
In cylindrical polar system
Ex.1: Find Volume of the tetrahedron bounded by the co-ordinates planes and the plane
Solution: Volume =
………. (1)
Put 
, 
From equation (1) we have
V = 
=24
=24
(u+v+w=1) By Dirichlet’s theorem.
=24 
=
=
= 4
Volume =4
Ex.2: Find volume common to the cylinders
, 
.
Solution: For given cylinders,
, 
.
Z varies from
Z=-
to z = 
Y varies from
y= -
to y = 
x varies from x= -a to x = a
By symmetry,
Required volume= 8 (volume in the first octant)
=8 
=8
= 8
dx
=8
=8
=8
Volume = 16
Ex.3: Evaluate
1. 
Solution:-
Ex.4:
Ex.5: Evaluate 
Solution:- Put
The limits of integration change when we change the order of integration.
We draw the rough diagram of the region of integration to find the new limits.
Example-1: Change the order of integration in the double integral-
Sol.
Limits are given-
x = 0, x = 2a
And
And
The area of integration is the shaded portion of OAB. On changing the order of integration first we will integrate with respect to x, the area of integration has three portions BCE, ODE and ACD,
Now-
Which is the required answer.
Example-2: Change the order of integration for the integral 
and evaluate the same with reversed order of integration.
Sol:
Given, I = 
…(1)
In the given integration, limits are
y = 
, y = 2a – x and x = 0, x = a
The region bounded by x2 = ay, x + y = 2a Fig.6.5
And x = 0, x = a is as shown in Fig. 6.5
Here we have to change order of Integration. Given the strip is vertical.
Now take horizontal strip SR.
To take total region, Divide region into two parts by taking line y = a.
1 st Region:
Along strip, y constant and x varies from x = 0 to x = 2a – y. Slide strip IIel to x-axis therefore y varies from y = a to y = 2a.
I1 = dy xy dx …(2)
2nd Region:
Along strip, y constant and x varies from x = 0 to x= . Slide strip IIel to x-axis therefore x-varies from y = 0 to y = a.
I2 = dy xy dx …(3)
From Equation (1), (2) and (3),
= dy xy dx + dy xy dx
= 
+ y dy 
=
dy + (ay) dy = y (4a2 – 4ay + y2) dy + ay2 dy
= (4a2 y – 4ay2 + y3) dy + y2 dy
= 
= 
+ 
= a4
Example-3: Evaluate
I = 
Soln. :
Given : I = 
…(1)
In the given integration, limits are
x = 0, x = a, y = 0, y = 
The bounded region is as shown in Fig. **.
In the given, strip is vertical. Now take horizontal strip SR. Along strip y constant and x varies from x = 0 to
x = 
. Slide strip IIel to X-axis therefore y varies from y = 0 to y = a.
I = dy 
= 
Put a2 – y2 = b2
I = 
= 
= 
= 
dy = 
dy
= 
=
= 
= 
[∵ a = a loge]
I = dy 
= 
Example-4: Express as single integral and evaluate dy dx + dy dx.
Soln. :
Given : I = dy dx + dy dx
I = I1 + I2
The limits of region of integration I1 are
x = – ; x = and y = 0, y = 1 and I2 are x = – 1,
x = 1 and y = 1, y = 3.
The region of integration are as shown in Fig. 6.7
To consider the complete region take a vertical strip SR along the strip y varies from y = x2 to
y = 3 and x varies from x = –1 to x = 1. Fig. 6.7
I = 
We can solve the double integration by changing independent variables.
Let the double integral is-
It’s to be changed by the new variables-u and v
The relationship between x, y and u, v are-
Then the double integration is converted into-
Where-
Example-: Evaluate-
Where-
R is the region bounded by a parallelogram- x + y = 0, x + y = 2, 3x – 2y = 0, 3x – 2y = 3.
Sol.
By changing the variables x, y to the new variables u and v, by the substitution x + y = u, 3x – 2y = v, the given parallelolgram R reduces to a rectangle 
The required Jacobian is-
Since u = x + y and u = x + y = 2, here u varies from 0 to 2 while v varies from 0 to 3.
Since-
3x – 2y = v = 0, 3x – 2y = v = 3
Therefore the integral will be in new variables-
Which is the required answer.
Example: Evaluate 
the transformation is x = v and y = uv.
Sol.
Here the region of integration R is the triangle which is bounded by y = 0, x = 1 and y = x
Here put,
X = u and y = uv, we get-
Here x varies from 0 to 1 while y varies from 0 to x.
Since u = x so u varies from 0 to 1
Here, similarly, since 
, so that v varies from 0 to 1. Thus-
Area in Cartesian coordinates-
Example-1: Find the area enclosed by two curves using double integration.
y = 2 – x and y² = 2 (2 – x)
Sol. Let,
y = 2 – x ………………..(1)
And y² = 2 (2 – x) ………………..(2)
On solving eq. (1) and (2)
We get the intersection points (2,0) and (0,2) ,
We know that,
Area = 
Here we will find the area as below,
Area = 
Which gives,
= (- 4 + 4 /2 ) + 8 / 3 = 2 / 3.
Example-2: Find the area between the parabola y ² = 4ax and another parabola x² = 4ay.
Sol. Let,
y ² = 4ax ………………..(1)
And
x² = 4ay…………………..(2)
Then if we solve these equations , w e get the values of points where these two curves intersect
x varies from y²/4a to 
and y varies from o to 4a,
Now using the conceot of double integral,
Area = 
Area in polar coordinates-
Example-3: Find the area lying inside the cardioid r = a(1+cosθ) and outside the circle r = a, by using double integration.
Sol. We have,
r = a(1+cosθ) …………………….(1)
And
r = a ……………………………….(2)
On solving these equations by eliminating r , we get
a(1+cosθ) = a
(1+cosθ) = 1
Cosθ = 0
Here a θ varies from – π/2 to π/2
Limit of r will be a and 1+cosθ)
Which is the required area.
Example-4: Find the are lying inside a cardioid r = 1 + cos θ and ouside the parabola r(1 + cos θ) = 1.
Sol. Let,
r = 1 + cos θ ……………………..(1)
r(1 + cos θ) = 1……………………..(2)
Solving these equestions , we get
(1 + cos θ )( 1 + cos θ ) = 1
(1 + cos θ )² = 1
1 + cos θ = 1
Cos θ = 0
θ = ±π / 2
So that, limits of r are,
1 + cos θ and 1 / 1 + cos θ
The area can be founded as below,
Volume by triple integral-
The volume of solid is given by
Volume =
In Spherical polar system
In cylindrical polar system
Ex.1: Find Volume of the tetrahedron bounded by the co-ordinates planes and the plane
Solution: Volume =
………. (1)
Put 
, 
From equation (1) we have
V = 
=24
=24
(u+v+w=1) By Dirichlet’s theorem.
=24 
=
=
= 4
Volume =4
Ex.2: Find volume common to the cylinders
, 
.
Solution: For given cylinders,
, 
.
Z varies from
Z=-
to z = 
Y varies from
y= -
to y = 
x varies from x= -a to x = a
By symmetry,
Required volume= 8 (volume in the first octant)
=8 
=8
= 8
dx
=8
=8
=8
Volume = 16
Ex.3: Evaluate
1. 
Solution:-
Ex.4:
Ex.5: Evaluate 
Solution:- Put
The beta and gamma functions are defined as-
And
These integrals are also known as first and second Eulerian integrals.
Note- Beta function is symmetrical with respect to m and n.
Some important results-
Ex.1: Evaluate 
dx
Solution
dx = 
dx
= γ(5/2)
= γ (3/2+ 1)
= 3/2 γ(3/2 )
= 3/2. ½ γ (½ )
= 3/2. ½ π
= ¾ π
Ex. 2: Find γ(-½)
Solution: (-½) + 1 = ½
γ(-1/2) = γ(-½ + 1) / (-½)
= - 2 γ(1/2 )
= - 2 π
Ex. 3. Show that 
Solution:
= 
= 
= 
) .......................
= 
= 
Ex. 4: Evaluate 
dx.
Solution: Let 
dx
X | 0 |  |
t | 0 |  |
Put 
or 
;dx =2t dt .
dt
dt
Ex. 5: Evaluate 
dx.
Solution: Let 
dx.
x | 0 |  |
t | 0 |  |
Put 
or 
; 4x dx = dt
dx
Evaluation of beta function 𝛃 (m, n)-
Here we have-
Or
Again integrate by parts, we get-
Repeating the process above, integrating by parts we get-
Or
Evaluation of gamma function-
Integrating by parts, we take 
as first function-
We get-
Replace n by n+1,
Relation between beta and gamma function-
We know that-
………… (1)
…………………..(2)
Multiply equation (1) by 
, we get-
Integrate both sides with respect to x within limits x = 0 to x = 
, we get-
But
By putting λ = 1 + y and n = m + n
We get by using this result in (2)-
So that-
Definition : Beta function
|
Properties of Beta function : |
2.  |
3.  |
4.  |
Example(1): Evaluate I = 
Solution:
= 2 π/3
Example (2): Evaluate: I = 02 x2 / (2 – x ) . Dx
Solution:
Letting x = 2y, we get
I = (8/2) 01 y 2 (1 – y ) -1/2dy
= (8/2). B (3, 1/2 )
= 642 /15
BETA FUNCTION MORE PROBLEMS
Relation between Beta and Gamma functions :
  | ||||||
Example(1): Evaluate: I = 0a x4 (a2 – x2 ) . Dx Solution: Letting x2 = a2 y , we get I = (a6 / 2) 01 y 3/2 (1 – y )1/2dy = (a6 / 2) . B(5/2 , 3/2 ) = a6 /3 2 Example(2): Evaluate: I = 02 x (8 – x3 ) . Dx Solution: Let x3 = 8y I = (8/3) 01 y-1/3 (1 – y ) 1/3 . Dy
= (8/3) B(2/3 , 4/3 ) = 16 π / ( 9 3 ) Example(3): Prove that  Solution : Let   Put            Example(4): Evaluate  Solution :Let  Put      When    
Also             
Example(5): Show that  Solution :  =             |
Let l, m and n are all three positive variables then triple integral
Is known as Dirichlet’s integral.
Note- Dirichlet’s theorem for n variables-
Example: Evaluate the integral-
Where x, y and z are positive with conditions-
Sol. Let us put-
So that the required integral will be-
Where u + v + w ≤ 1
By Dirichlet’s integral-
Example: The plane 
meets the axes in A, B and C. Find the volume of the tetrahedron OABC by using Dirichlit’s integral.
Sol.
The volume of the tetrahedron is given by-
For all positive values of x, y and z subjected to the conditions 
Putting-
And
We get-
Where u ≥ 0, v ≥ 0 and w ≥ 0 subjected to the conditions u + v + w ≤ 1
Now by using Dirichlet’s integral-
= abc/6
Extension of Dirchlet’s theorem ( Liouville’s Extension)-
Error functions-
Example: Evaluate 
, the integral extending over all positive and zero values of x, y, w subjected to x + y + z < 1.
Sol.
Here-
0 ≤ x + y + x < 1
So that-
By extension of Dririchlet’s theorem-
Example: Show that-
The conditions are given x + y = 0, y = 0, z = 0, x + y + z = 1.
Sol.
By Lioville’s theorem when 0 < x + y + z < 1
Prove
