Module3
Integral Calculus
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 Rin xyplane.
Then the double integration over region R can be evaluated by two successive integration
Evaluation of double integrals
If A is described as
Then,
]dx
Let do some examples to understand more about double integration
Example: Evaluate , where dA is the small area in xyplane.
Sol. Let , I =
=
=
=
= 84 sq. Unit.
Which is the required area,
Example2: 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
Example3: Evaluate the following double integral,
Sol. Let,
I =
On solving the integral, we get
Double integration in polar coordination 
In polar coordinates, we need to evaluate
Over the region bounded by θ1 and θ2.
And the curves r1(θ) and r2(θ)
Example1: Evaluate the following by changing to polar coordinates,
Sol. In this problem, the limits for y are 0 to and the limits for are 0 to 2.
Suppose,
y =
Squaring both sides,
y² = 2x  x²
x² + y² = 2x
But in polar coordinates,
We have,
r² = 2r cosθ
r = 2 cosθ
From the region of integration, r lies from 0 to 2 cosθ and θ varies from 0 to π / 2.
As we know in case of polar coordinates,
Replace x by r cosθ and y by r sinθ, dy dx by r drdθ,
We get,
Example2: Evaluate the following integral by converting into polar coordinates.
Sol. Here limits of y,
y =
y² = 2x  x²
x² + y² = 2x
x² + y²  2x = 0 ………………(1)
Eq. (1) represent a circle whose radius is 1 and centre is ( 1, 0)
Lower limit of y is zero.
Region of integration in upper half circle,
First we will covert into polar coordinates,
By putting
x by r cos θ and y by r sinθ , dy dx by r drdθ,
Limits of r are 0 to 2 cosθ and limits of θ are from 0 to π / 2.
Example3: Evaluate
Sol. Let the integral,
I =
=
Put x = sinθ
= π / 24 ans.
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.
Example1: Evaluate
Solution:
Let
I =
=
(Assuming m = )
= dxdy
=
=
= dx
= dx
=
=
I =
Example2: Evaluate Where V is annulus between the spheres and ()
Solution: It is convenient to transform the triple integral into spherical polar coordinate by putting
, ,
, dxdydz = sindrdd,
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
Example3: Evaluate
Solution:
Ex.4: Evaluate
Solution:
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.
Example1: 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: Change the order of integration in
Sol.
The limits are given as
It means
Also
Now we have here
Example2: 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 IIelto xaxis therefore y varies from y = a to y = 2a.
= dyxy dx .....…(2)
2nd Region:
Along strip, y constant and x varies from x = 0 to x= . Slide strip IIel to xaxis therefore xvaries from y = 0 to y = a.
= dyxy dx …(3)
From Equation (1), (2) and (3),
= dyxy dx + dyxy dx
= + y dy
=dy + (ay) dy = y (4a2 – 4ay + y2) dy + ay2dy
= (4a2 y – 4ay2 + y3) dy + y2dy
=
= +
= a4
Example4: Express as single integral and evaluate dy dx + dy dx.
Sol:
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
To consider the complete region take a vertical strip SR along the strip y varies from y to y = 3 and x varies from x = –1 to x = 1.
I =
Key takeaways
2.
Area of double integration
Area in Cartesian coordinates
Example1: 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.
Example2: Find the area between the parabola y ² = 4ax and another parabola x² = 4ay.
Sol. Let,
y² = 4ax ………………..(1)
And
x² = 4ay…………………..(2)
Then if we solve these equations , we 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
Example3: 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.
Example4: Find the are lying inside a cardioid r = 1 + cosθ and outside the parabola r(1 + cosθ) = 1.
Sol. Let,
r = 1 + cosθ ……………………..(1)
r(1 + cosθ) = 1……………………..(2)
Solving these equations, we get
(1 + cosθ )(1 + cosθ ) = 1
(1 + cosθ )² = 1
1 + cosθ = 1
Cosθ = 0
θ = ±π / 2
So that, limits of r are,
1 + cosθ and 1 / 1 + cosθ
The area can be founded as below,
Double integrals as volumes
Suppose we have a curve y = f(x) is revolved about an axis , then a solid is generated , now we need to find out the volume of the solid generated ,
The formula for volume of the solid generated about xaxis,
Example1: Calculate the volume generated by the revolution of a cardioid,
r = a ( 1 – cosθ) about its axis
Sol. Here,
r = a ( 1 – cosθ)
Volume =
=
Which is the volume of generated by cardioid.
Example2: Find the volume generated by revolving a circle x ² + y² = 4 about the line x= 3.
Sol. We know that,
Volume =
Here , PQ = 3 – x,
=
The volume is 24π².
Example3: Calculate the volume under the surface z = 3 + x²  2y over the region R defined as 0≤ x ≤ 1 and x ≤ y ≤ x
Sol. The is a double integral of z = 3 + x²  2y over the region R.
Volume will be,
CENTER OF MASS AND GRAVITY
Centre of mass
We have
Mass = volume
Example: Find the mass of a plate formed by the coordinate planes and the plane
The variable density
Sol.
Here we the mass will be
The limits of y are from y = 0 to y = b(1 – x/a) and limits of x are from 0 to a.
Therefore the required mass will be
Which is the required answer.
Centre of gravity
The centre of gravity of a lamina is the point where it balances perfectly which means the lamina’s centre of mass.
If and are the coordinates of the centroid C of area A, then
And
Example: Determine the coordinates of the centroid of the area lying between the curve y = 5x  x² and the xaxis.
Sol.
Here y = 5x  x²when y = 0, x = 0 or x = 5
Therefore the curve cuts the xaxis at 0 and 5 as in the figure
Now
= 2.5
Therefore the centroid of the area lies at (2.5, 2.5)
Key takeaways
 Centre of mass
We have
Mass = volume
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
Solution:
= 2 π/3
BETA FUNCTION MORE PROBLEMS
Relation between Beta and Gamma functions :
 
Example(3): Prove that Solution : Let Put or ,Example(4): Evaluate Solution :Let Put or ,,When ,;,Also
Example(5): Show that Solution : = (0<p<1)(by above result) 
Evaluation of integrals
1.
by using beta and gamma functions
Evaluation of the first interal
Evaluation of the second integral
Example: Prove that
Where a>0.
Sol.
We know that
Put m – 1 = 1 or m = 2
We get
Hence proved
Key takeaways
And
2.
3.
4.
References:
 C.Ray Wylie, Louis C.Barrett : "Advanced Engineering Mathematics", 6th Edition, 2. McGrawHill Book Co., New York, 1995.
 James Stewart : “Calculus Early Transcendentals”, Cengage Learning India Private Ltd., 2017.
 B.V.Ramana: "Higher Engineering Mathematics" 11th Edition, Tata McGrawHill, 2010.
 Srimanta Pal & Subobh C Bhunia: “Engineering Mathematics”, Oxford University Press, 3rd Reprint, 2016.
 Gupta C.B., Singh S.R. And Mukesh Kumar: “Engineering Mathematics for Semester I & II”, McGraw Hill Education (India) Pvt.Ltd., 2015.