# Volume by Cylindrical shells

1. Feb 14, 2012

### Biosyn

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

Rotate around the y-axis the region above the graph of y=x3 that is bounded by the lines x=1 and y=8

2. Relevant equations

dV= (2pix)(y)(dx)

3. The attempt at a solution

dV = (2pix)(y)dx
dV = (2pix)(x^3) dx
= 2pix^4

I integrated from y = 1 to y=8 and I get the wrong answer.

V = 2∏x=1x=2x^4 dx
V = 2∏[1/5x^5]21

Can someone please explain to me, how I would set up this equation? Something to do with the first function subtracting the second function which I think is x=1

Here is a graph: http://www3.wolframalpha.com/Calculate/MSP/MSP15491a07c8557bi4590b00001ibd52eb6gcc3177?MSPStoreType=image/gif&s=21&w=366&h=298&cdf=RangeControl [Broken]

Last edited by a moderator: May 5, 2017
2. Feb 14, 2012

### SammyS

Staff Emeritus
The problem is with the following:
dV= (2πx)(y)(dx)
The height of the cylindrical shell is 8-y, not y.

So dV = (2πx)(8-y)(dx)

Last edited by a moderator: May 5, 2017
3. Feb 14, 2012

### Biosyn

[STRIKE]So dV = (2∏x)(8-x^3)dx
= 16∏x-2∏x^4 dx

V= 2∏∫(8x-x^4) dx ?

[/STRIKE]

Thank you for helping me!