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Unit-3Multiple Integration Q1) Evaluate

A1) 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 =

As we know that by beta function,

Which gives,

Now put the value of c, we get

Q2) Evaluate the following by changing to polar coordinates,

A2)

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,

Q3) Evaluate

A3)

Let the integral,

I =

Put x = sinθ

= π / 24 ans.

Q4) Evaluate

A4)

Let

I =

(Assuming m = )

= dxdy

= dx

I =

Q5) Evaluate

A5)

Q6) Change the order of integration in the double integral-

A6)

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.

Q7) Evaluate-

By changing into polar coordinates.A7)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-

Q8) Find the area between the parabola y ² = 4ax and another parabola x² = 4ay.A8)

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 =

Q9) Find the are lying inside a cardioid r = 1 + cosθ and outside the parabola r(1 + cosθ) = 1.A9)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,

Q10) Find the volume generated by revolving a circle x ² + y² = 4 about the line x= 3.A10)We know that,

Volume =

Here , PQ = 3 – x,

The volume is 24π².

Q11) Find the mass of a plate formed by the coordinate planes and the plane-

The variable density

A11)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.

Q12) Evaluate

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

A12)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.

Q13) 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.A13)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

The equation of the line OB is y = x

Now by stoke’s theorem,

Q14) By using divergence theorem to evaluate-

A14)Where-

And S is the surface of the sphere .

Sol.

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

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