Unit -5

Content

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,

]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

Example-3: Evaluate the following double integral,

Sol. Let ,

I =

On solving the integral, we get

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 xy (1 – x –y) dx dy.

Soln. :

Given : I = xy (1 – x –y) dx dy.

Here the limits of inner integration are functions of y therefore first integrate w.r.t y.

I = xdx

Put 1 – x = a (constant for inner integral)

I = xdx

Put y = at dy = a dt

y | 0 | a |

t | 0 | 1 |

I = xdx

I = xdx

I = xa dx

I = x (1 – x) dx = (x– x4/3) dx

I =

=

I = =

xy (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

Fig. 6.2

(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.

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)

Fig. 6.4

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 =

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 ,

Q

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 – cos θ) about its axis

Sol. Here, r = a ( 1 – cos θ)

Volume =

=

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,

=

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,

Ex. 1 : Change the order of integration for the integral and evaluate the same with reversed order of integration.

Soln. :

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

Ex. 2 : 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 =

Ex.3 : 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 =

NOTES:

Evaluation of Double Integral by Changing Cartesian to Polar co-ordinates (when limits are given).

Ex. 1 : Evaluate

Soln. :

The region of integration bounded by

y = 0, y = and x = 0, x = 1

y = x2 + y2 = x

The region bounded by these is as shown in Fig. 6.8.

Convert the integration in polar co-ordinates by using x = r cos , y = r sin and dx dy = r dr d

x2 + y2 = x becomes r = cos

y = 0 becomes r sin = 0 = 0

x = 0 becomes r cos = 0 =

And x = 1 becomes r = sec Fig. 6.8

Take a radial strip SR with angular thickness , Along strip constant and r varies from r = 0 to r = cos . Turning strip throughout region therefore varies from = 0 to =

I = r dr d

= 4 cos sin d r dr

= 4 cos sin d [–]

= – 2 cos sin [+1] d

= – 2 [cos sin – cos sin ] d

= –2 + 2 cos sin d

= – + 2

= + 1 =

Ex. 2: Sketch the area of double integration and evaluate

dxdy

Soln. :

Let I = dxdy

The region of integration is bounded by the curves

x = y, x = and y = 0, y = Fig. 6.9

i.e. x = y, x2 + y2 = a2 and y = 0, y =

The region bounded by these is as shown in Fig. 6.9.

The point of intersection of x = y and x2 + y2 = a2 is x =

Convert given integration in polar co-ordinates by using polar transformation x = r cos , y = r sin and dx dy = r dr d

x = y gives r cos = r sin tan = 1 =

x2 + y2 = a2 r2 = a2 r = a

y = 0 gives r sin = 0 = 0.

y = gives r sin = r = cosec

Take a radial strip SR, along SR constant and r varies from r = 0 to r = a. Turning this strip throughout region therefore varies from = 0 to =

I = log r2 r dr d = 2 d r log r dr

I =

= 2 d

= 2 d

= 2 d

= 2 []

I = [/4] =

Evaluation of Double Integral when region of Integration is provided (Polar form)

Ex. 1 : Evaluate r4 cos3 dr d over the interior of the circle r = 2a cos

Soln. :

The region of the integration is as shown in Fig. 6.10.Take a radial strip SR, along strip constant and r varies from r = 0 to r = 2a cos. Now turning this strip throughout region therefore varies from = to =

I = r4 cos3 dr d

= cos3 d

= cos3 cos5 d

= cos8 d Fig. 6.10

=

= 2

I =

r4 cos3 dr d =

Ex. 2 : Evaluate r sin dr d over the cardioid r = a (1 – cos ) above the initial line.

Soln. :

The cardioid r = a (1 – cos ) is as shown in Fig. 6.11. The region of the integration is above the initial line.

Take a radial strip SR, along strip constant and r Varies from r = 0 to r = a (1 – cos ).

New turning the strip throughout region therefore varies from = 0 to = .

I = r sin dr d

= sin d

Fig.6.11

= sin [a2 (1 – cos )2]

=

I = (sin – 2 sin cos + sin ) d

= 2 (sin – sin2 + sin ) d

= +

I = a2= a2

I =

NOTES:

Area enclosed by plane curves expressed in Cartesian coordinates:

Y (x,y) P Q (x+dx,y+dy) y=f(x)

V U

G H dxdy

A R S B

x=a y=0 x=b X

Consider the area enclosed by the curves

And and the ordinates

Area =

Consider the area enclosed by the curves

And and the ordinates

Area =

Area enclosed by plane curves expressed in Polar coordinates:

Consider the area enclosed by the polar curves

And and the line

Area =

Ex.1: Find the area between the curves and its asymptote.

Solution: The curve is symmetrical about x axis not passing through origin. Also is the asymptote to the curve and intersect x axis at for curve doesn’t exists. And for and. Because of symmetry

Put

x | 0 | 2a |

t | 0 | 1 |

Ex: 2: Show that the area of curve is

Solution: The curve is symmetrical about Y - axis passing through origin and there exists a cusp at origin. It intersects Y-axis at (0 , 0) , (0 , 2a) .

Also .

Putting

.

Ex.3: Find the Area included between the two cardioids

Solution: - Area =

=

=

=

Ex.4: Find the Area common to the two circle

Solution: - By converting the given circle into polar form we get

Area =

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 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 hapf circle,

First we will covert into polar coordinates,

By putting

x by r cosθ and y by r sinθ , dy dx by r dr dθ,

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.

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, 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 + 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,

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=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:-

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

= 8dx

=8

=8

=8

Volume = 16

Ex.3: Evaluate

1. Solution:-

Ex.4:

Ex.5: Evaluate

Solution:- Put

NOTES:

MASS OF A LAMINA :- If the surface density ρ of a plane lamina is a function of the position of a point of the lamina, then the mass of an elementary area dA is ρ dA and the total mass of the lamina is

In Cartesian coordinates, if ρ = f(x, y) the mass of lamina, M=

In polar coordinates, if ρ = F(r, Ө) the mass of lamina, M=

Ex.1: A lamina is bounded by the curves and . If the density at any point is then find mass of lamina.

Solution:

Ex.2:If the density at any point of a non uniform circular lamina of radius’ a’ varies as its distance from a fixed point on the circumference of the circle then find the mass of lamina.

Solution:

Take the fixed point on the circumference of the circle as origin and diameter through it as x axis. The polar equation of circle

And density.

Mean Value:

The mean value of the ordinate y of a function over the range to is the limit of mean value of the equidistant ordinates as

Mean Square values of function over the range to is defined as

Mean Square values of function

Mean Square values of function

Root Mean Square Value: (R.M.S. Value):

If y is a periodic function of x of period p, the root mean square value of y is the square root of the mean value of over the range to , c is constant.

Ex. 1: Find mean value and R.M.S. Value of the ordinate of cycloid

, over the range to .

Sol Let P(x,y) be any point on the cycloid . Its ordinate is y.

=

=

=

R.M.S.Value=

Ex. 2: Find The Mean Value of Over the positive octant of the

Ellipsoid

Sol: M.V.=

Since

Put

M.V.=

Cartesian coordinate into spherical coordinates-

The relationship between the Cartesian and spherical coordinates is given by,

x = r sinθ cos∅

y = r sinθ sin∅

z = x = r cosθ

Dx dy dz = r² x = r sinθ dr dθ d∅

Example-: Evaluate the following integral throughout the volume of the sphere,

Sol. We are given,

Change the coordinates into spherical coordinates,

Put,

x = r sinθ cos∅

y = r sinθ sin∅

z = x = r cosθ

Dx dy dz = r² x = r sinθ dr dθ d∅

Limits are as below,

Limit of r - 0 to a

Limit of θ = 0 to π/2

Limit of ∅ = 0 to π/2

[ the given shere lies in 8th quadrant]

Example: Evaluate the following triple integral over the firsr octant of the sphere x² + y² + z² = a²

Sol. We are given,

Change the coordinates to spherical,

x = r sinθ cos∅

y = r sinθ sin∅

z = x = r cosθ

Dx dy dz = r² x = r sinθ dr dθ d∅

Limits are as below,

Limit of r - 0 to a

Limit of θ = 0 to π/2

Limit of ∅ = 0 to π/2

Now we can find the integral as below,

Cartesian coordinates into cylindrical coordinates-

We will change the Cartesian coordinates to cylindrical coordinate as follows-

x = r cos θ

y = r sin θ

z = z

Dx dy dz = r dr dz

Example: Evaluate the triple integral , where the limits of r , , z are 0≤ r ≤ 2 , 0 ≤ θ ≤ π / 2 and 0 ≤ z ≤ 4.

Sol. We can use the method of triple integration as follows,

On solving the above integral, it gives

Therefore the value of the integral is 64/3

If u and v be continuous and differentiable functions of two other independent variables x and y such as

, then we define the determine

as Jacobian of u, v with respect to x, y

Similarly ,

JJ’ = 1

Actually Jacobins are functional determines

Ex.

- Calculate
- If
- If

ST

4. find

5. If and , find

6.

7. If

8. If , ,

JJ1 = 1

If ,

JJ1=1

Jacobian of composite function (chain rule)

Then

Ex.

- If

Where

2. If and

Find

3. If

Find

Jacobian of Implicit function

Let u1, u2 be implicit functions of x1, x2 connected by f1, f2 such there

,

Then

Similarly,

Ex.

If

If

Find

Partial derivative of implicit functions

Consider four variables u, v, x, y related by implicit function.

,

Then

Ex.

If and

Find

If and

Find

Find

If

Find

We can solve the double integration by changing the independent variable.

Suppose there is an integral

It is to be changed by the new variables u and v.

The relationship between x and y can be given as-

x = ∅( u , v)

y = g(u , v)

Then the double integration can be converted into,

Where,

Dx dy = |J| du dv = = du dv

Example; Using the transformation x + y = u and y = uv, find the value of the following double integral , taken over the area of the triangle bounded by the lines x = 0 and y = 0 , x + y = 1.

Sol. It is given that,

We have,

x + y = u or x = u – y or x = u – uv

=

x = 0 ,

u ( 1 – v) = 0

u = 0 , v = 0

y = 0

Uv = 0

u = 0 , v = 0

x+y = 1

u = 1

Therefore the limit of u is from 0 to1 and the limit of v are from 0 to 1.

On putting x = u – uv , y = uv , dx dy = u du dv , we get