Using integration to find Volume with two functions.?

Question

I have the region enclosed by y=x^3 and y = √x rotated about the line x=1.

I'm supposed to find the volume using integration.
The answer in the book is 13pi/30.

However, I keep getting 2pi/5 with respect to y and 5pi/14 with respect to x.

Which, of course, doesn't make sense, because if I was doing this correctly, the answer with respect to either x and y should be the same..right?? uhh..

I don't see what i'm doing wrong.


This is how I started out. A(y)=[ pi(y^2)^2 ] - [ pi(y^(1/3))^2 ]

I didn't think I needed to add or subtract distance because the two functions meet at x=1.

Help please! Ack!

Answer 1

This is a washer method problem. To found the volume first solve for x in terms of y (which you appear to have done). Then you need to realize that you are rotating around the line x=1. So the outside radius will be (1-y^2) and the inside radius will be (1-y^(1/3)). Set up your integral as the integral from 0 to 1 of the outside radius squared minus the integral from 0 to 1 of the inside radius squared. You will get .533 for the first integral and .099 for the second. Lastly, multiply by pi.
To recap:
Solve each equation for x in terms of y.
Find the expression for the outside radius and then the expression for the inside radius.
Set up the integrals. Evaluate. Multiply by pi. You can multiply by pi as you go but it is probably easier to do it at the end. Best of luck.

Answer 2

Using integration to find Volume with two functions.
I have the region enclosed by y=x^3 and y = √x rotated about the line x=1.

Use the method of cylindrical shells.

The two functions meet at (0,0) and (1,1) so our limits of integration are from zero to one.

Volume = ∫2πrhdx = 2π∫[(1 - x)(√x - x³)dx

= 2π∫[√x - x³ - x^(3/2) + x^4]dx

= 2π[(2/3)x^(3/2) - (1/4)x^4 - (2/5)x^(5/2) + (1/5)x^5]

= 2π[2/3 - 1/4 - 2/5 + 1/5] = 2π(13/60) = 13π/30