Unit - 1

Multivariable 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 xy-plane.

Then the double integration over region R can be evaluated by two successive integrations

Evaluation of double integrals-

If A is described as

Then,

]dx

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

= π/2 [ ¾ (1) – 3/7 (1) – ¾ (-1) + 3/7 (-1)] = π/2 [9/14] = 9π/28

Example-3: Evaluate the following double integral,

Sol. Let,

I =

On solving the integral, we get

= 5π/8

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 cos θ

r = 2 cos θ

From the region of integration, r lies from 0 to 2 cos θ and θ varies from 0 to π / 2.

As we know in case of polar coordinates,

Replace x by r cos θ and y by r sin θ, dy dx by r dr dθ,

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 convert into polar coordinates,

By putting

x by r cos θ and y by r sin θ, dy dx by r dr dθ,

Limits of r are0 to 2 cos θ and limits of θ are from 0 to π / 2.

Example-3: Evaluate

Sol. Let the integral,

I =

=

Put x = sin θ

= 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.

Spherical Polar Coordinates

1) For the spheres x2 + y2 + z2 = r2

Put x = r sin θ cos x, y = r sin θ sin , z = r cos θ

Dx dy dz = r2 sin θ dr dθ d

(a) For complete sphere x2 + y2 + z2 = a2, θ = 0 to π, = 0 to 2π, r = 0 to a.

(b) For hemisphere x2 + y2 + z2 = a2 , z > 0, θ = 0 to π/2, = 0 to 2π, r = 0 to a

(c) For positive octant of a sphere x2 + y2 + z2 = a2, θ = 0 to π/2, r = 0 to a

2) For ellipsoid x2/a2 + y2/b2 + z2/c2 = 1

Put x = ar sin θ cos , y = b r sin θ sin , z = c r sin θ

Dx dy dz = abc r2 dr dθ d, θ = 0 to π, = 0 to 2π, r = 0 to 1

Dirichlets Theorem: -

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=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

Ex.3: Evaluate

Solution: -

= 1/18 [e12 – 2e6 – 9e4 + 18e2 – 8]

Ex.4: Evaluate

, taken through out the volume of the sphere x2 + y2 + z2 = 1 in positive octant.

Solution:

Put x = r sin θ sin , z = r cos θ, dx dy dz = r2 sin θ dr dθ d

θ = 0 to π/2, r= 0 to 1 and x2 + y2 + z2 = r2 φ = 0 to π/2

Where I1 = put r = sin t

Dr = cos dt r 0 1

t 0 π/2

= π/4 . ()π/20

= π/4 . π/2

= π2/8

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-ordinate’s 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

= 8dx

=8

=8

=8

Volume = 16

Ex.3: Evaluate

Solution

Put x = r sin θ cos , y = r sin θ sin , z = r cos θ

θ = 0 to π, = 0 to 2π, r = 0 to 1, dx dy dz = r2 sin θ dr dθ d

I =

=

= 4a7π / 35

Ex.4:

Integrate through out the volume bounded x = 0, y=0, z = 0, x/a + y/b+z/c = 1.

Solution: x = au dx = a du

y = bv dy = b dv

z = cw dz = c dw

I =

=

= a3 b2 c2

= a3b2c2

= a3b2c2

= a3b2c2 2!/7!

= a3b2c2/2520

Ex.5:

Evaluate throught the volume of the ellipsoid

Solution: Put

x = ar sin θ cos

y = br sin θ sin

z = cr cos θ

Dxdydz = abc r2 drdθd

θ = 0 to π, = 0 to 2π, r = 0 to 1

I =

= abc

Put r = sin t dr = cos t dt

If r = 0 then t = 0 and r = 1 then t = π/2

Abc

= abc

= abcπ/16

= abcπ/16

= abcπ/8

= abcπ2/4

Example 1: Evaluate

By changing the order of integration.

Solution: The given limits are (inner) y from x to π/2; (Outer) x from 0 to π/2.

We use these to sketch the region of integration.

The given limits have inner variable y. To reverse the order of integration we use horizontal stripes. The limits in this order are

(inner) x from 0 to y; (outer) y from 0 to π/2.

So the integral becomes

We compute the inner, then the outer integrals.

Inner: Outer : - cos =1

Change of Order of Integration:

This is the case where limits are constants. But if they are variables, then by changing the order of integration, the limits of integration also changes.

A rough sketch helps in fixing the new limits of integration.

Example 1: Change the order of integration in

Of hence evaluate the same.

Solution:

y = x2 & y = 2 - x

Pt. Of interaction:

x2 + x + 2 = 0

x = 1, 2

y = 1

(1, 1) is coordinate for A

Region of interaction is divided into 2 parts OAM and MAB

For region OAM:

0 ≤ y ≤ 1, 0 ≤ x ≤ y

For region MAB:

1 ≤ y ≤ 2, 0 ≤ x ≤ 2 –y

Example 2: Change the order of integration of evaluate

Solution:

y = x2/4a & y = 2ax

x2 = 4ay & y2 = 4ax

Pt. Of intersection

x4/16a2 = 4ax

x4 – 64 a3x3 = 0

x = 0, 4a

pts, are (0, 0), (4a, 4a)

Determine the image of a region under a given transformation of variables.

Compute the Jacobian of a given transformation.

Evaluate a double integral using a change of variables.

Evaluate a triple integral using a change of variables.

Example 1:

Evaluate .dA

Solution:

Here’s is the region bounded by the lines on the real lines

X+ y=0; x+ y=-1

x- y=0 ; x-y = 8

Let

u=x+ y , 0 u 1

v=x-y, 0 v 8

Now we have two choices

*solve for x and y if possible

*use the inverse transmission

Solving for x and y

[ u = x+y] + [ v = x – y] = u + v = 2x

[ u = x+y ] – [ v = x – y] = u – v = 2y

X=1/2(u)+1/2(v) and y =1/2(u-1/2(v)

Using of inverse method:

and J(x, y) . J(u, v) = 1

J(u, v) = 1/J(u, v)

J(u, v) =

J(u, v) = (- ¼ ) – ( ¼ ) = - ½

½ (e – 1)

J(u, v) = 6(e – 1)

Example 2:

Evaluate .dA

Solution:

Here s is the region bounded by the lines on the real lines

X - 2y=0; x - 2y= 4

3x-y=1 ;3 x-y = 8

Let

u=x – 2y

v= 3x- y

Now we have two choices

*solve for x and y if possible

*use the inverse transmission

Using of inverse method:

and J(x, y) . J(u, v) = 1

J(u, v) = 1/J(u, v)

J(u, v) =

where u/x = 1, u/y = - 2, v/x = 3, v/x = - 1

J(x, y) =

J(u, v) = 1/J(x, y) J(u, v) = 1/5

Example 3:

Given a plane z= 3x+4y+2 that lies above the rectangle [0,5] [1,4]. Find the surface area

Solution:

The area of the surface with equation z=f(x,y),(x,y)∈ D ,where fx and fy are continuous,

Is A(S)= dA

We have z=2+3x+4y.

Then, z/x = 3 and z/y = 4

A(S) = dA

From the dimensions of the rectangle, we get the limits of x as (0,5) and the limits of y as(1,4)

On substituting the known values in the expression for area ,we get

A(S) =

Evaluate the iterated integral.

A(S) =

=

=15 26

Example 4:

As an example, let us consider the following integral in two dimensions:

I=

Solution: Where C is a straight line from the origin to (1,1), as shown the figure, Let s be the arc length measured from the origin. We then have

x =s cos θ = s/2

y=s sin θ = s/2

The endpoint (1,1) corresponds to s=2.Thus , the line integral becomes

Definition: evaluating triple integrals is similar to evaluating nested functions: you work from inside out. Triple integrals look scary, but if you take them step by step, they’re no kore difficult than regular integrals. You start in the centre and work your way out. For example, begin by separating two inner integrals.

Example 1: Evaluate the following:

Solution:

Example 2: Evaluate the following:

Solution: let I =

Example 3: Evaluate the following triple integral

Solution: let

Area of double integration

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,

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 concept 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 + 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,

=

= ½

=

=

=

=

= [ θ + θ/2 + sin 2θ/4 + 2 sin θ – ¼ (2 tan θ/2 + 2/3 tan3 θ/2)]π/20

= [π/2 + π/4 +0+ 2 sin π/2 – ½ tan π/4 – 1/6 tan3 π/4] = [3π/4 + 2 – ½ - 1/6 ] = [3π/4 + 4/3]

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 x-axis,

Example-1: Calculate the volume generated by the revolution of a cardioid,

r = a (1 – cosθ) about its axis

Sol. Here, r = a (1 – cosθ)

Volume =

=

= 2πa3/3

Which is the volume of generated by cardioid.

Example-2: 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,

=

= 2π

= 2π (3y – xy)+4-x2-4-x2

= 2π [34-x2 - x4-x2 + 34-x2 - x4-x2]

= 4π[3 4 – x2 - x4 – x2] dx = 4π [ 3 x/2 4-x2 + 3 4/2 sin-1 x/2 + 1/3(4 – x2)3/2]2-2

= 4π[ 6 π/2+ 6π/2] = 24π2

The volume is 24π².

Example-3: 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,

References:

- E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
- P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.

Unit - 1

Multivariable 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 xy-plane.

Then the double integration over region R can be evaluated by two successive integrations

Evaluation of double integrals-

If A is described as

Then,

]dx

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

= π/2 [ ¾ (1) – 3/7 (1) – ¾ (-1) + 3/7 (-1)] = π/2 [9/14] = 9π/28

Example-3: Evaluate the following double integral,

Sol. Let,

I =

On solving the integral, we get

= 5π/8

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 cos θ

r = 2 cos θ

From the region of integration, r lies from 0 to 2 cos θ and θ varies from 0 to π / 2.

As we know in case of polar coordinates,

Replace x by r cos θ and y by r sin θ, dy dx by r dr dθ,

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 convert into polar coordinates,

By putting

x by r cos θ and y by r sin θ, dy dx by r dr dθ,

Limits of r are0 to 2 cos θ and limits of θ are from 0 to π / 2.

Example-3: Evaluate

Sol. Let the integral,

I =

=

Put x = sin θ

= 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.

Spherical Polar Coordinates

1) For the spheres x2 + y2 + z2 = r2

Put x = r sin θ cos x, y = r sin θ sin , z = r cos θ

Dx dy dz = r2 sin θ dr dθ d

(a) For complete sphere x2 + y2 + z2 = a2, θ = 0 to π, = 0 to 2π, r = 0 to a.

(b) For hemisphere x2 + y2 + z2 = a2 , z > 0, θ = 0 to π/2, = 0 to 2π, r = 0 to a

(c) For positive octant of a sphere x2 + y2 + z2 = a2, θ = 0 to π/2, r = 0 to a

2) For ellipsoid x2/a2 + y2/b2 + z2/c2 = 1

Put x = ar sin θ cos , y = b r sin θ sin , z = c r sin θ

Dx dy dz = abc r2 dr dθ d, θ = 0 to π, = 0 to 2π, r = 0 to 1

Dirichlets Theorem: -

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=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

Ex.3: Evaluate

Solution: -

= 1/18 [e12 – 2e6 – 9e4 + 18e2 – 8]

Ex.4: Evaluate

, taken through out the volume of the sphere x2 + y2 + z2 = 1 in positive octant.

Solution:

Put x = r sin θ sin , z = r cos θ, dx dy dz = r2 sin θ dr dθ d

θ = 0 to π/2, r= 0 to 1 and x2 + y2 + z2 = r2 φ = 0 to π/2

Where I1 = put r = sin t

Dr = cos dt r 0 1

t 0 π/2

= π/4 . ()π/20

= π/4 . π/2

= π2/8

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-ordinate’s 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

= 8dx

=8

=8

=8

Volume = 16

Ex.3: Evaluate

Solution

Put x = r sin θ cos , y = r sin θ sin , z = r cos θ

θ = 0 to π, = 0 to 2π, r = 0 to 1, dx dy dz = r2 sin θ dr dθ d

I =

=

= 4a7π / 35

Ex.4:

Integrate through out the volume bounded x = 0, y=0, z = 0, x/a + y/b+z/c = 1.

Solution: x = au dx = a du

y = bv dy = b dv

z = cw dz = c dw

I =

=

= a3 b2 c2

= a3b2c2

= a3b2c2

= a3b2c2 2!/7!

= a3b2c2/2520

Ex.5:

Evaluate throught the volume of the ellipsoid

Solution: Put

x = ar sin θ cos

y = br sin θ sin

z = cr cos θ

Dxdydz = abc r2 drdθd

θ = 0 to π, = 0 to 2π, r = 0 to 1

I =

= abc

Put r = sin t dr = cos t dt

If r = 0 then t = 0 and r = 1 then t = π/2

Abc

= abc

= abcπ/16

= abcπ/16

= abcπ/8

= abcπ2/4

Example 1: Evaluate

By changing the order of integration.

Solution: The given limits are (inner) y from x to π/2; (Outer) x from 0 to π/2.

We use these to sketch the region of integration.

The given limits have inner variable y. To reverse the order of integration we use horizontal stripes. The limits in this order are

(inner) x from 0 to y; (outer) y from 0 to π/2.

So the integral becomes

We compute the inner, then the outer integrals.

Inner: Outer : - cos =1

Change of Order of Integration:

This is the case where limits are constants. But if they are variables, then by changing the order of integration, the limits of integration also changes.

A rough sketch helps in fixing the new limits of integration.

Example 1: Change the order of integration in

Of hence evaluate the same.

Solution:

y = x2 & y = 2 - x

Pt. Of interaction:

x2 + x + 2 = 0

x = 1, 2

y = 1

(1, 1) is coordinate for A

Region of interaction is divided into 2 parts OAM and MAB

For region OAM:

0 ≤ y ≤ 1, 0 ≤ x ≤ y

For region MAB:

1 ≤ y ≤ 2, 0 ≤ x ≤ 2 –y

Example 2: Change the order of integration of evaluate

Solution:

y = x2/4a & y = 2ax

x2 = 4ay & y2 = 4ax

Pt. Of intersection

x4/16a2 = 4ax

x4 – 64 a3x3 = 0

x = 0, 4a

pts, are (0, 0), (4a, 4a)

Determine the image of a region under a given transformation of variables.

Compute the Jacobian of a given transformation.

Evaluate a double integral using a change of variables.

Evaluate a triple integral using a change of variables.

Example 1:

Evaluate .dA

Solution:

Here’s is the region bounded by the lines on the real lines

X+ y=0; x+ y=-1

x- y=0 ; x-y = 8

Let

u=x+ y , 0 u 1

v=x-y, 0 v 8

Now we have two choices

*solve for x and y if possible

*use the inverse transmission

Solving for x and y

[ u = x+y] + [ v = x – y] = u + v = 2x

[ u = x+y ] – [ v = x – y] = u – v = 2y

X=1/2(u)+1/2(v) and y =1/2(u-1/2(v)

Using of inverse method:

and J(x, y) . J(u, v) = 1

J(u, v) = 1/J(u, v)

J(u, v) =

J(u, v) = (- ¼ ) – ( ¼ ) = - ½

½ (e – 1)

J(u, v) = 6(e – 1)

Example 2:

Evaluate .dA

Solution:

Here s is the region bounded by the lines on the real lines

X - 2y=0; x - 2y= 4

3x-y=1 ;3 x-y = 8

Let

u=x – 2y

v= 3x- y

Now we have two choices

*solve for x and y if possible

*use the inverse transmission

Using of inverse method:

and J(x, y) . J(u, v) = 1

J(u, v) = 1/J(u, v)

J(u, v) =

where u/x = 1, u/y = - 2, v/x = 3, v/x = - 1

J(x, y) =

J(u, v) = 1/J(x, y) J(u, v) = 1/5

Example 3:

Given a plane z= 3x+4y+2 that lies above the rectangle [0,5] [1,4]. Find the surface area

Solution:

The area of the surface with equation z=f(x,y),(x,y)∈ D ,where fx and fy are continuous,

Is A(S)= dA

We have z=2+3x+4y.

Then, z/x = 3 and z/y = 4

A(S) = dA

From the dimensions of the rectangle, we get the limits of x as (0,5) and the limits of y as(1,4)

On substituting the known values in the expression for area ,we get

A(S) =

Evaluate the iterated integral.

A(S) =

=

=15 26

Example 4:

As an example, let us consider the following integral in two dimensions:

I=

Solution: Where C is a straight line from the origin to (1,1), as shown the figure, Let s be the arc length measured from the origin. We then have

x =s cos θ = s/2

y=s sin θ = s/2

The endpoint (1,1) corresponds to s=2.Thus , the line integral becomes

Definition: evaluating triple integrals is similar to evaluating nested functions: you work from inside out. Triple integrals look scary, but if you take them step by step, they’re no kore difficult than regular integrals. You start in the centre and work your way out. For example, begin by separating two inner integrals.

Example 1: Evaluate the following:

Solution:

Example 2: Evaluate the following:

Solution: let I =

Example 3: Evaluate the following triple integral

Solution: let

Area of double integration

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,

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 concept 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 + 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,

=

= ½

=

=

=

=

= [ θ + θ/2 + sin 2θ/4 + 2 sin θ – ¼ (2 tan θ/2 + 2/3 tan3 θ/2)]π/20

= [π/2 + π/4 +0+ 2 sin π/2 – ½ tan π/4 – 1/6 tan3 π/4] = [3π/4 + 2 – ½ - 1/6 ] = [3π/4 + 4/3]

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 x-axis,

Example-1: Calculate the volume generated by the revolution of a cardioid,

r = a (1 – cosθ) about its axis

Sol. Here, r = a (1 – cosθ)

Volume =

=

= 2πa3/3

Which is the volume of generated by cardioid.

Example-2: 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,

=

= 2π

= 2π (3y – xy)+4-x2-4-x2

= 2π [34-x2 - x4-x2 + 34-x2 - x4-x2]

= 4π[3 4 – x2 - x4 – x2] dx = 4π [ 3 x/2 4-x2 + 3 4/2 sin-1 x/2 + 1/3(4 – x2)3/2]2-2

= 4π[ 6 π/2+ 6π/2] = 24π2

The volume is 24π².

Example-3: 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,

References:

- E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
- P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.

Unit - 1

Multivariable 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 xy-plane.

Then the double integration over region R can be evaluated by two successive integrations

Evaluation of double integrals-

If A is described as

Then,

]dx

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

= π/2 [ ¾ (1) – 3/7 (1) – ¾ (-1) + 3/7 (-1)] = π/2 [9/14] = 9π/28

Example-3: Evaluate the following double integral,

Sol. Let,

I =

On solving the integral, we get

= 5π/8

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 cos θ

r = 2 cos θ

From the region of integration, r lies from 0 to 2 cos θ and θ varies from 0 to π / 2.

As we know in case of polar coordinates,

Replace x by r cos θ and y by r sin θ, dy dx by r dr dθ,

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 convert into polar coordinates,

By putting

x by r cos θ and y by r sin θ, dy dx by r dr dθ,

Limits of r are0 to 2 cos θ and limits of θ are from 0 to π / 2.

Example-3: Evaluate

Sol. Let the integral,

I =

=

Put x = sin θ

= 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.

Spherical Polar Coordinates

1) For the spheres x2 + y2 + z2 = r2

Put x = r sin θ cos x, y = r sin θ sin , z = r cos θ

Dx dy dz = r2 sin θ dr dθ d

(a) For complete sphere x2 + y2 + z2 = a2, θ = 0 to π, = 0 to 2π, r = 0 to a.

(b) For hemisphere x2 + y2 + z2 = a2 , z > 0, θ = 0 to π/2, = 0 to 2π, r = 0 to a

(c) For positive octant of a sphere x2 + y2 + z2 = a2, θ = 0 to π/2, r = 0 to a

2) For ellipsoid x2/a2 + y2/b2 + z2/c2 = 1

Put x = ar sin θ cos , y = b r sin θ sin , z = c r sin θ

Dx dy dz = abc r2 dr dθ d, θ = 0 to π, = 0 to 2π, r = 0 to 1

Dirichlets Theorem: -

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=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

Ex.3: Evaluate

Solution: -

= 1/18 [e12 – 2e6 – 9e4 + 18e2 – 8]

Ex.4: Evaluate

, taken through out the volume of the sphere x2 + y2 + z2 = 1 in positive octant.

Solution:

Put x = r sin θ sin , z = r cos θ, dx dy dz = r2 sin θ dr dθ d

θ = 0 to π/2, r= 0 to 1 and x2 + y2 + z2 = r2 φ = 0 to π/2

Where I1 = put r = sin t

Dr = cos dt r 0 1

t 0 π/2

= π/4 . ()π/20

= π/4 . π/2

= π2/8

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-ordinate’s 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

= 8dx

=8

=8

=8

Volume = 16

Ex.3: Evaluate

Solution

Put x = r sin θ cos , y = r sin θ sin , z = r cos θ

θ = 0 to π, = 0 to 2π, r = 0 to 1, dx dy dz = r2 sin θ dr dθ d

I =

=

= 4a7π / 35

Ex.4:

Integrate through out the volume bounded x = 0, y=0, z = 0, x/a + y/b+z/c = 1.

Solution: x = au dx = a du

y = bv dy = b dv

z = cw dz = c dw

I =

=

= a3 b2 c2

= a3b2c2

= a3b2c2

= a3b2c2 2!/7!

= a3b2c2/2520

Ex.5:

Evaluate throught the volume of the ellipsoid

Solution: Put

x = ar sin θ cos

y = br sin θ sin

z = cr cos θ

Dxdydz = abc r2 drdθd

θ = 0 to π, = 0 to 2π, r = 0 to 1

I =

= abc

Put r = sin t dr = cos t dt

If r = 0 then t = 0 and r = 1 then t = π/2

Abc

= abc

= abcπ/16

= abcπ/16

= abcπ/8

= abcπ2/4

Example 1: Evaluate

By changing the order of integration.

Solution: The given limits are (inner) y from x to π/2; (Outer) x from 0 to π/2.

We use these to sketch the region of integration.

The given limits have inner variable y. To reverse the order of integration we use horizontal stripes. The limits in this order are

(inner) x from 0 to y; (outer) y from 0 to π/2.

So the integral becomes

We compute the inner, then the outer integrals.

Inner: Outer : - cos =1

Change of Order of Integration:

This is the case where limits are constants. But if they are variables, then by changing the order of integration, the limits of integration also changes.

A rough sketch helps in fixing the new limits of integration.

Example 1: Change the order of integration in

Of hence evaluate the same.

Solution:

y = x2 & y = 2 - x

Pt. Of interaction:

x2 + x + 2 = 0

x = 1, 2

y = 1

(1, 1) is coordinate for A

Region of interaction is divided into 2 parts OAM and MAB

For region OAM:

0 ≤ y ≤ 1, 0 ≤ x ≤ y

For region MAB:

1 ≤ y ≤ 2, 0 ≤ x ≤ 2 –y

Example 2: Change the order of integration of evaluate

Solution:

y = x2/4a & y = 2ax

x2 = 4ay & y2 = 4ax

Pt. Of intersection

x4/16a2 = 4ax

x4 – 64 a3x3 = 0

x = 0, 4a

pts, are (0, 0), (4a, 4a)

Determine the image of a region under a given transformation of variables.

Compute the Jacobian of a given transformation.

Evaluate a double integral using a change of variables.

Evaluate a triple integral using a change of variables.

Example 1:

Evaluate .dA

Solution:

Here’s is the region bounded by the lines on the real lines

X+ y=0; x+ y=-1

x- y=0 ; x-y = 8

Let

u=x+ y , 0 u 1

v=x-y, 0 v 8

Now we have two choices

*solve for x and y if possible

*use the inverse transmission

Solving for x and y

[ u = x+y] + [ v = x – y] = u + v = 2x

[ u = x+y ] – [ v = x – y] = u – v = 2y

X=1/2(u)+1/2(v) and y =1/2(u-1/2(v)

Using of inverse method:

and J(x, y) . J(u, v) = 1

J(u, v) = 1/J(u, v)

J(u, v) =

J(u, v) = (- ¼ ) – ( ¼ ) = - ½

½ (e – 1)

J(u, v) = 6(e – 1)

Example 2:

Evaluate .dA

Solution:

Here s is the region bounded by the lines on the real lines

X - 2y=0; x - 2y= 4

3x-y=1 ;3 x-y = 8

Let

u=x – 2y

v= 3x- y

Now we have two choices

*solve for x and y if possible

*use the inverse transmission

Using of inverse method:

and J(x, y) . J(u, v) = 1

J(u, v) = 1/J(u, v)

J(u, v) =

where u/x = 1, u/y = - 2, v/x = 3, v/x = - 1

J(x, y) =

J(u, v) = 1/J(x, y) J(u, v) = 1/5

Example 3:

Given a plane z= 3x+4y+2 that lies above the rectangle [0,5] [1,4]. Find the surface area

Solution:

The area of the surface with equation z=f(x,y),(x,y)∈ D ,where fx and fy are continuous,

Is A(S)= dA

We have z=2+3x+4y.

Then, z/x = 3 and z/y = 4

A(S) = dA

From the dimensions of the rectangle, we get the limits of x as (0,5) and the limits of y as(1,4)

On substituting the known values in the expression for area ,we get

A(S) =

Evaluate the iterated integral.

A(S) =

=

=15 26

Example 4:

As an example, let us consider the following integral in two dimensions:

I=

Solution: Where C is a straight line from the origin to (1,1), as shown the figure, Let s be the arc length measured from the origin. We then have

x =s cos θ = s/2

y=s sin θ = s/2

The endpoint (1,1) corresponds to s=2.Thus , the line integral becomes

Definition: evaluating triple integrals is similar to evaluating nested functions: you work from inside out. Triple integrals look scary, but if you take them step by step, they’re no kore difficult than regular integrals. You start in the centre and work your way out. For example, begin by separating two inner integrals.

Example 1: Evaluate the following:

Solution:

Example 2: Evaluate the following:

Solution: let I =

Example 3: Evaluate the following triple integral

Solution: let

Area of double integration

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,

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 concept 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 + 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,

=

= ½

=

=

=

=

= [ θ + θ/2 + sin 2θ/4 + 2 sin θ – ¼ (2 tan θ/2 + 2/3 tan3 θ/2)]π/20

= [π/2 + π/4 +0+ 2 sin π/2 – ½ tan π/4 – 1/6 tan3 π/4] = [3π/4 + 2 – ½ - 1/6 ] = [3π/4 + 4/3]

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 x-axis,

Example-1: Calculate the volume generated by the revolution of a cardioid,

r = a (1 – cosθ) about its axis

Sol. Here, r = a (1 – cosθ)

Volume =

=

= 2πa3/3

Which is the volume of generated by cardioid.

Example-2: 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,

=

= 2π

= 2π (3y – xy)+4-x2-4-x2

= 2π [34-x2 - x4-x2 + 34-x2 - x4-x2]

= 4π[3 4 – x2 - x4 – x2] dx = 4π [ 3 x/2 4-x2 + 3 4/2 sin-1 x/2 + 1/3(4 – x2)3/2]2-2

= 4π[ 6 π/2+ 6π/2] = 24π2

The volume is 24π².

Example-3: 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,

References:

- E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
- P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.

Unit - 1

Multivariable Integral Calculus

Unit - 1

Multivariable Integral Calculus

Unit - 1

Multivariable Integral Calculus

Unit - 1

Multivariable Integral Calculus

Double integral –

As we know, the integral

Where is belongs to the limit a ≤ x ≤ b

This integral can be written as follows-

Then the double integration over region R can be evaluated by two successive integrations

Evaluation of double integrals-

If A is described as

Then,

]dx

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

= π/2 [ ¾ (1) – 3/7 (1) – ¾ (-1) + 3/7 (-1)] = π/2 [9/14] = 9π/28

Example-3: Evaluate the following double integral,

Sol. Let,

I =

On solving the integral, we get

= 5π/8

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 cos θ

r = 2 cos θ

From the region of integration, r lies from 0 to 2 cos θ and θ varies from 0 to π / 2.

As we know in case of polar coordinates,

Replace x by r cos θ and y by r sin θ, dy dx by r dr dθ,

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 convert into polar coordinates,

By putting

x by r cos θ and y by r sin θ, dy dx by r dr dθ,

Limits of r are0 to 2 cos θ and limits of θ are from 0 to π / 2.

Example-3: Evaluate

Sol. Let the integral,

I =

=

Put x = sin θ

= π / 24 ans.

Triple integrals

Spherical Polar Coordinates

1) For the spheres x2 + y2 + z2 = r2

Put x = r sin θ cos x, y = r sin θ sin , z = r cos θ

Dx dy dz = r2 sin θ dr dθ d

(a) For complete sphere x2 + y2 + z2 = a2, θ = 0 to π, = 0 to 2π, r = 0 to a.

(b) For hemisphere x2 + y2 + z2 = a2 , z > 0, θ = 0 to π/2, = 0 to 2π, r = 0 to a

(c) For positive octant of a sphere x2 + y2 + z2 = a2, θ = 0 to π/2, r = 0 to a

2) For ellipsoid x2/a2 + y2/b2 + z2/c2 = 1

Put x = ar sin θ cos , y = b r sin θ sin , z = c r sin θ

Dx dy dz = abc r2 dr dθ d, θ = 0 to π, = 0 to 2π, r = 0 to 1

Dirichlets Theorem: -

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 ()

, ,

, 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

Ex.3: Evaluate

Solution: -

= 1/18 [e12 – 2e6 – 9e4 + 18e2 – 8]

Ex.4: Evaluate

, taken through out the volume of the sphere x2 + y2 + z2 = 1 in positive octant.

Solution:

Put x = r sin θ sin , z = r cos θ, dx dy dz = r2 sin θ dr dθ d

θ = 0 to π/2, r= 0 to 1 and x2 + y2 + z2 = r2 φ = 0 to π/2

Where I1 = put r = sin t

Dr = cos dt r 0 1

t 0 π/2

= π/4 . ()π/20

= π/4 . π/2

= π2/8

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-ordinate’s 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

= 8dx

=8

=8

=8

Volume = 16

Ex.3: Evaluate

Solution

Put x = r sin θ cos , y = r sin θ sin , z = r cos θ

θ = 0 to π, = 0 to 2π, r = 0 to 1, dx dy dz = r2 sin θ dr dθ d

I =

=

= 4a7π / 35

Ex.4:

Integrate through out the volume bounded x = 0, y=0, z = 0, x/a + y/b+z/c = 1.

Solution: x = au dx = a du

y = bv dy = b dv

z = cw dz = c dw

I =

=

= a3 b2 c2

= a3b2c2

= a3b2c2

= a3b2c2 2!/7!

= a3b2c2/2520

Ex.5:

Evaluate throught the volume of the ellipsoid

Solution: Put

x = ar sin θ cos

y = br sin θ sin

z = cr cos θ

Dxdydz = abc r2 drdθd

θ = 0 to π, = 0 to 2π, r = 0 to 1

I =

= abc

Put r = sin t dr = cos t dt

If r = 0 then t = 0 and r = 1 then t = π/2

Abc

= abc

= abcπ/16

= abcπ/16

= abcπ/8

= abcπ2/4

Example 1: Evaluate

By changing the order of integration.

Solution: The given limits are (inner) y from x to π/2; (Outer) x from 0 to π/2.

We use these to sketch the region of integration.

(inner) x from 0 to y; (outer) y from 0 to π/2.

So the integral becomes

We compute the inner, then the outer integrals.

Inner: Outer : - cos =1

Change of Order of Integration:

A rough sketch helps in fixing the new limits of integration.

Example 1: Change the order of integration in

Of hence evaluate the same.

Solution:

y = x2 & y = 2 - x

Pt. Of interaction:

x2 + x + 2 = 0

x = 1, 2

y = 1

(1, 1) is coordinate for A

Region of interaction is divided into 2 parts OAM and MAB

For region OAM:

0 ≤ y ≤ 1, 0 ≤ x ≤ y

For region MAB:

1 ≤ y ≤ 2, 0 ≤ x ≤ 2 –y

Example 2: Change the order of integration of evaluate

Solution:

y = x2/4a & y = 2ax

x2 = 4ay & y2 = 4ax

Pt. Of intersection

x4/16a2 = 4ax

x4 – 64 a3x3 = 0

x = 0, 4a

pts, are (0, 0), (4a, 4a)

Determine the image of a region under a given transformation of variables.

Compute the Jacobian of a given transformation.

Evaluate a double integral using a change of variables.

Evaluate a triple integral using a change of variables.

Example 1:

Evaluate .dA

Solution:

Here’s is the region bounded by the lines on the real lines

X+ y=0; x+ y=-1

x- y=0 ; x-y = 8

Let

u=x+ y , 0 u 1

v=x-y, 0 v 8

Now we have two choices

*solve for x and y if possible

*use the inverse transmission

Solving for x and y

[ u = x+y] + [ v = x – y] = u + v = 2x

[ u = x+y ] – [ v = x – y] = u – v = 2y

X=1/2(u)+1/2(v) and y =1/2(u-1/2(v)

Using of inverse method:

and J(x, y) . J(u, v) = 1

J(u, v) = 1/J(u, v)

J(u, v) =

J(u, v) = (- ¼ ) – ( ¼ ) = - ½

½ (e – 1)

J(u, v) = 6(e – 1)

Example 2:

Evaluate .dA

Solution:

Here s is the region bounded by the lines on the real lines

X - 2y=0; x - 2y= 4

3x-y=1 ;3 x-y = 8

Let

u=x – 2y

v= 3x- y

Now we have two choices

*solve for x and y if possible

*use the inverse transmission

Using of inverse method:

and J(x, y) . J(u, v) = 1

J(u, v) = 1/J(u, v)

J(u, v) =

where u/x = 1, u/y = - 2, v/x = 3, v/x = - 1

J(x, y) =

J(u, v) = 1/J(x, y) J(u, v) = 1/5

Example 3:

Given a plane z= 3x+4y+2 that lies above the rectangle [0,5] [1,4]. Find the surface area

Solution:

The area of the surface with equation z=f(x,y),(x,y)∈ D ,where fx and fy are continuous,

Is A(S)= dA

We have z=2+3x+4y.

Then, z/x = 3 and z/y = 4

A(S) = dA

From the dimensions of the rectangle, we get the limits of x as (0,5) and the limits of y as(1,4)

On substituting the known values in the expression for area ,we get

A(S) =

Evaluate the iterated integral.

A(S) =

=

=15 26

Example 4:

As an example, let us consider the following integral in two dimensions:

I=

x =s cos θ = s/2

y=s sin θ = s/2

The endpoint (1,1) corresponds to s=2.Thus , the line integral becomes

Example 1: Evaluate the following:

Solution:

Example 2: Evaluate the following:

Solution: let I =

Example 3: Evaluate the following triple integral

Solution: let

Area of double integration

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,

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 concept of double integral,

Area =

=

=

Area in polar coordinates-

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.

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,

=

= ½

=

=

=

=

= [ θ + θ/2 + sin 2θ/4 + 2 sin θ – ¼ (2 tan θ/2 + 2/3 tan3 θ/2)]π/20

= [π/2 + π/4 +0+ 2 sin π/2 – ½ tan π/4 – 1/6 tan3 π/4] = [3π/4 + 2 – ½ - 1/6 ] = [3π/4 + 4/3]

Double integrals as volumes

The formula for volume of the solid generated about x-axis,

Example-1: Calculate the volume generated by the revolution of a cardioid,

r = a (1 – cosθ) about its axis

Sol. Here, r = a (1 – cosθ)

Volume =

=

= 2πa3/3

Which is the volume of generated by cardioid.

Example-2: 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,

=

= 2π

= 2π (3y – xy)+4-x2-4-x2

= 2π [34-x2 - x4-x2 + 34-x2 - x4-x2]

= 4π[3 4 – x2 - x4 – x2] dx = 4π [ 3 x/2 4-x2 + 3 4/2 sin-1 x/2 + 1/3(4 – x2)3/2]2-2

= 4π[ 6 π/2+ 6π/2] = 24π2

The volume is 24π².

Sol. The is a double integral of z = 3 + x² - 2y over the region R

Volume will be,

References:

- E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.