Module-3

Integral Calculus

Question-1: 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

Question-2: 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,

Question-3: Evaluate

Solution:

Let

I =

=

(Assuming m = )

= dxdy

=

=

= dx

= dx

=

=

I =

Question-4: Change the order of integration in the double integral-

Sol.

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.

Question-5: 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 IIelto x-axis therefore y varies from y = a to y = 2a.

=dyxy 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.

=dyxy dx…(3)

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

= dyxy dx + dyxy dx

=+ y dy

=dy + (ay) dy = y (4a2 – 4ay + y2) dy + ay2dy

= (4a2 y – 4ay2 + y3) dy + y2dy

=

=+

= a4

Question-6: 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 conceot of double integral,

Area =

Question-7: 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.

Question-8: 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.

Question-9: Find the mass of a plate formed by the coordinate planes and the plane-

The variable density

Sol.

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.

Question-10: Determine the coordinates of the centroid of the area lying between the curve y = 5x - x² and the x-axis.

Sol.

Here y = 5x - x²when y = 0, x = 0 or x = 5

Therefore the curve cuts the x-axis at 0 and 5 as in the figure-

Now

= 2.5

Therefore the centroid of the area lies at (2.5, 2.5)

Question-11: Evaluate dx

Solution dx = dx

= γ(5/2)

= γ(3/2+ 1)

= 3/2 γ(3/2 )

= 3/2. ½ γ(½ )

= 3/2. ½ π

= ¾ π

Question-12: Evaluate dx.

Solution: Let dx.

x | 0 | |

t | 0 |

Put or ; 4x dx = dt

dx

Question-13: Evaluate

Solution:

= 2 π/3

Question-14: Evaluate

Solution :Let

Put or ,,

When,;,

Also

o | ||

1 | 0 |

Question-15: Prove that-

Where a>0.

Sol.

We know that-

Put m – 1 = 1 or m = 2

We get-

Hence proved