6.1 Double Integral | unit 6 multiple integrals and their applications

Unit-6

Multiple Integrals and their Applications

6.1 Double Integral

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).

Ex1) 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 =

Ex2) 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 = xa dx 

I =  x (1 – x) dx =  (x– x4/3) dx

I =

I = =

 xy (1 – x –y) dx dy =

Ex3) Evaluate

Sol)

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 =

= 

Ex4) Evaluate e–x2 (1 + y2) x dx dy.

Sol)

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)

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

Ex1) Evaluate  y dx dy over the area bounded by x= 0 y = and x + y = 2 in the first quadrant

Sol)

Ex2 ) Evaluate  over x  1, y 

Sol)

1a1

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 =

Ex3) Evaluate  (+ ) dx dy through the area enclosed by the curves y = 4x, x + y = 3 and y =0, y = 2.

Sol)

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 =

6.2 Evaluation of Double Integral by Changing the Order of integration (Cartesian Form)

Ex1) 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

Ex 2) Evaluate

I =

Sol)

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 =

Ex3) 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. 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).

Ex1) Evaluate

Sol)

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 =

Ex2) Sketch the area of double integration and evaluate

dxdy

Sol)

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 []

I = [/4] =

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

Ex1) Evaluate  r4 cos3 dr d over the interior of the circle r = 2a cos 

Sol)

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 =

Ex2) Evaluate  r sin  dr d over the cardioid r = a (1 – cos ) above the initial line.

Sol)

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 ).

11a 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 =

Ex1) Find the area between the curves and its asymptote.

Sol)

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

Ex2) Show that the area of curve is

Sol)

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

Ex3) Find the Area included between the two cardioids

Sol)

Area =

Ex4) Find the Area common to the two circle

Sol)

By converting the given circle into polar form we get

Area =

NOTES:

6.3 Triple Integration

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

Sol)

Let

I =

(Assuming m = )

= dxdy

= dx

I =

Ex2) Evaluate Where V is annulus between the spheres

and ()

sol)

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

= 8

=8 log

= 8 log

I= 8 log I = 4 log

Ex3) Evaluate

Sol)

Ex.4:Evaluate

Sol)

NOTES:

The volume of solid is given by

Volume =

In Spherical polar system

In cylindrical polar system

Ex1) Find Volume of the tetrahedron bounded by the co-ordinates planes and the plane

Sol)

Volume = ………. (1)

Put ,

From equation (1) we have

V =

=24

=24 (u+v+w=1) By Dirichlet’s theorem.

=24

= = = 4

Volume =4

Ex2) Find volume common to the cylinders, .

Sol)

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)

= 8dx

Volume = 16

Ex3) Evaluate

1. Sol)

Ex4)

Ex5) Evaluate

Sol)

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=

Ex1)

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

Sol)

Ex2) 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.

Sol)

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

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.

Ex1) 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=

Ex2) Find The Mean Value of Over the positive octant of the

Ellipsoid

Sol)

M.V.=

Since

Put

M.V.=

NOTES:

6.4 Center of gravity

DEFINITION

Centre of mass:-The centre of mass of a body is the point through which the resultant mass acts.

Centre of gravity:-The centre of gravity (C.G.) of a body is the point through which the resultant weight acts. Since weight is proportional to mass ,the centre of a mass is the same point as the centre of a gravity.

C.G. of an arc of a curve;

The C.G. of arc of a curve is given by

;

Where is density.

; if is constant.

Remark:

1. If then

2. If then

3. If then

4. If then

5. If then

C.G. of a plane area or Lamina:

Let be the co-ordinates of C.G. of the lamina then;

;

Or in double integral form, Cartesian system

If the curve is given in Polar coordinates,

then

Ex1) A lamina bounded by the parabolas and has a variable density

Given by . Prove that

Sol)

The points of intersection of two parabolas are (0,0), (1,2), density (given)

----------------------------------------(i)

Diagram

dxdy=

------------------------------------------(ii)

=1 ----------------------------------------------(iii)

From (i),(ii) and (iii),

Ex2) Find the Center of gravity of one of the loop of r =

Sol)

The curve r= is four leaved rose lies within the circle r=a,

Consider a loop lies between as shown in the fig.

= …….. (1)

Where N=

= 1

= ………..(2)

= …………………..(3)

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

we have = = , also

Ex2) find the centroid of the loop of the curve.

Sol)

C.G lies on x-axis Therefore,

	0
t	0

C.G. is

6.5 Moment of Inertia:

The moment of inertia (M.I.) is given by Where dm = dx dy

1 ) The M.I. of an area

In Cartesian form:

Polar form:

2) The M.I. of an Arc

, Where dm=ds

Ex1) Prove that the Moment of inertia of the area included in between the parabolas and about X-axis is , where M is the mass of the area included between the curves.