# Integral Calc: Volume of Solid of Revolution

1. Jan 23, 2014

### cathy

1. The problem statement, all variables and given/known data

Find the volume of the first quadrant region bounded by x=y-y3, x=1 and y=1 that is revolved about the y-axis.

2. The attempt at a solution

v=∏ ∫ from 0 to 1 of (y-y^3)^2 dy
and doing this, I got the answer to be 8∏/105.

Did I set up that integral correctly? I am confused as to if I did this right or not. Please advise. Thank you in advance.

2. Jan 23, 2014

### dirk_mec1

What is y3?

3. Jan 23, 2014

### cathy

the equation is x=y-y^3

4. Jan 23, 2014

### HallsofIvy

Staff Emeritus
No, that is not set up correctly. $y- y^3$ is the distance from the y-axis to $x= y- y^3$ but you said the region to be rotated was bounded by $x= y- y^3$ and x=1.

What you have will give the volume from rotating the region bounded by x= 0 and $x= y- y^3$, the region inside the graph while you want the region outside that graph and inside x= 1.

You could use the "washer method" but equivalent, and simpler in my opinion, is to find the volume inside the rotated graph, using the integral as you have it set up, and then subtract that from the volume of the cylinder of radius 1 and height 1 leaving only the volume outside the graph.

5. Jan 23, 2014

### cathy

Okay, I understand what I was doing wrong. I was looking for area inside, rather than outside and within bounds. Thank you.

But, the teacher gave us the answer as V=2∏∫ from 0 to 1 of (y)(1-y+y^3) dy.

And I have a question about this. She said that r=y and h= 1-x, but if you were to draw this as shells, the "stick" would run from the line y=1 to the line x=0. The "stick" never actually touches the y-y^3 part, so how is that included? We take a vertical stick, correct? I'm confused. :(

Last edited: Jan 23, 2014
6. Jan 23, 2014

### cathy

And just to clarify, the "r" is talking about the radius of the shell formed, correct?
And since it is a vertical cylinder, why isn't the answer in terms of x?

Last edited: Jan 23, 2014