Where is belongs to the limit a ≤ x ≤ b

If A is described as 

Then,

]dx

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

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 convert 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.

=

is called triple integration of f (x, y, z) over the region V provided limit on R.H.S of above Equation exists.

Example-1: Evaluate

Solution:

Let

I     =    

=    

                                            (Assuming m =  )

= dxdy

=

=

= dx

=  dx

= 

=

I =

Example-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

Example-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

1. 
2. The volume of solid is given by

Volume =

3.     In Spherical polar system 

4.     In cylindrical polar system

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 = 

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,

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  =

           =

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,