Unit 3 | unit 3 multiple integrals and their applications

M 2

Unit - 3

Multiple integrals and their applications

3.1 Elementary double 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 Rin 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 =

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

Example-4: Evaluate-

Where D is bounded by y = x and .

Sol.

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)

Example-1: 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,

Example-2: 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.

Example-3: Evaluate

Sol. Let the integral,

I =

Put x = sinθ

= π / 24 ans.

3.2 Change of variable (simple transformations)

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-

3.3 Change of order of integration (Cartesian and polar)

Change of order of integration

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

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,

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 x-axis therefore y varies from y = a to y = 2a.

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.

From Equation (1), (2) and (3),

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.

Put a2 – y2 = b2

Example-4: Express as single integral and evaluate.

Soln.:

Given:

The limits of region of integration are

x = –; x = and y = 0, y = 1 and 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 = x2 to y = 3 and x varies from x = –1 to x = 1.

I =

3.4 Applications to find Mass, Area, Volume and Centre of Gravity (Cartesian and polar forms)

Area under and between the curves-

Total shaded area will be as follows of the given figure( by using definite integrals)-

Total shaded area =

Example-1: Determine the area enclosed by the curves-

Sol. We know that the curves are equal at the points of interaction, thus equating the values of y of each curve-

Which gives-

By factorization,

Which means,

x = 2 and x = -3

By determining the intersection points the range the values of x has been found-

-3

-2

-1

And

x	-3	0	2
y = 7 - x	10	7	5

We get the following figure by using above two tables-

Area of shaded region =

= ( 12 – 2 – 8/3 ) – (-18 – 9/2 + 9)

= 125/6 square unit

Example-2: Determine the area bounded by three straight lines y = 4 – x, y = 3x and 3y = x

Sol. We get the following figure by using the equations of three straight lines-

y = 4 – x, y = 3x and 3y = x

Area of shaded region-

Example-3: Find the area enclosed by the two functions-

and g(x) = 6 – x

Sol. We get the following figure by using these two equations

To find the intersection points of two functions f(x) and g(x)-

f(x) = g(x)

On factorizing, we get-

x = 6, -2

Now

Then, area under the curve-

A =

Therefore the area under the curve is 64/3 square unit.

Areas and volumes of revolutions-

The volume of revolution (V) is obtained by rotating area A through one revolution about the x-axis is given by-

Suppose the curve x = f(y) is rotated about y-axis between the limits y = c and y = d, then the volume generated V, is given by-

Example-1: Find the area enclosed by the curves and if the area is rotated about the x-axis then determine the volume of the solid of revolution.

Sol. We know that, at the point of intersection the coordinates of the curve are equal. So that first we will find the point of intersection-

We get,

x = 0 and x = 2

The curve of the given equations will look like as follows-

Then,

The area of the shaded region will be-

A =

So that the area will be 8/3 square unit.

The volume will be

= (volume produced by revolving – (volume produced by revolving

Method of cylindrical shells-

Let f(x) be a continuous and positive function. Define R as the region bounded above by the graph f(x), below by the x-axis, on the left by the line x = a and on the right x = b, then the volume of the solid of revolution formed by revolving R around the y-axis is given by

Example-2: Find the volume of the solid of revolution formed by revolving R around y-axis of the function f(x) = 1/x over the interval [1 , 3].

Sol. The graph of the function f(x) = 1/x will look like-

The volume of the solid of revolution generated by revolving R(violet region) about the y-axis over the interval [1 , 3]

Then the volume of the solid will be-

Example-3: Find the volume of the solid of revolution formed by revolving R around y-axis of the function f(x) = 2x - x² over the interval [0 , 2].

Sol. The graph of the function f(x) = 2x - x² will be-

The volume of the solid is given by-

Key takeaways-

Areas and volumes of revolutions-

The volume of revolution (V) is obtained by rotating area A through one revolution about the x-axis is given by-

2. Method of cylindrical shells-

Centre of gravity

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 x-axis.

Sol.

Here y = 5x - x²when y = 0, x = 0 or x = 5

Therefore the curve cuts the x-axis at 0 and 5 as in the figure-

Now

= 2.5

Therefore the centroid of the area lies at (2.5, 2.5)

3.5 Elementary triple integrals

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: