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

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,

Let the integral,

I =

=

Put x = sinθ

= π / 24   ans.

Let

I     =    

=    

                                            (Assuming m =  )

= dxdy

=

=

= dx

=  dx

= 

=

I =

Limits are given-

x = 0, x = 2a

And

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.

We have

The limits of x and y are 0 to

Hence the region of integration is in the first quadrant,

The region is covered by the radius strip from r = 0 to r = ∞ and it starts from θ = 0 to θ = π/2

Hence the integral becomes-

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 conceot of double integral,

Area = 

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,

We know that,

Volume =

Here , PQ = 3 – x,

                                   =

The volume is 24π².

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.

by using Green’s theorem, where C is a triangle formed by

First we will draw the figure-

Here the vertices of triangle OED are (0,0), (

Now by using Green’s theorem-

Here P = y – sinx, and Q  =cosx

So that-

and

Now-

=

Which is the required answer.

If and C is the boundary of the triangle with vertices at (0, 0, 0), (1, 0, 0) and (1, 1, 0), then evaluate by using Stoke’s theorem.

here we see that z-coordinates of each vertex of the triangle is zero, so that the triangle lies in the xy-plane and

Now,

Curl

Curl

The equation of the line OB is y = x

Now by stoke’s theorem,

Where-

And S is the surface of the sphere .

Sol.

On putting x = , y = , z = , we get-