# Calculate the volume of the solid of revolution of f(x)=e^x

1. Mar 31, 2013

### jasonbay74

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

Calculate the volume of the solid of revolution formed by rotating the region around the y-axis. Apply the shell method.

f(x)=e^x, x=0, y=8

2. Relevant equations

V=∫2∏x((f(x))-g(x))dx

3. The attempt at a solution

This is what I did: (I integrated from 0 to 8)

V=∫ 2∏x(8-e^x)dx
=2∏∫ (8x-xe^x)

I used integration by parts with u=x, du=1dx, v=e^x, and dv e^x(dx)

giving:

2∏[8∫ xdx-(xe^x-∫ e^x(dx)]

When I apply the disk method using x=ln(y) I get 48.13407626.

These two answers should be the same and I think there's an error in my shell method that I can't figure out???
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 31, 2013

### Staff: Mentor

Did you subtract your integral from a cylinder with length (e^8-1) and radius 8?
Anyway, this number is too small.

I don't think the volume should be negative, by the way.

3. Mar 31, 2013

### jasonbay74

This what I did for the disk method (about the y-axis)

V=∫(lny)^2-(0)^2 dy........from 1 to 8

used two integration by parts

1st: u=(lny)^2, du=2lny/y dy, dv=dy, v=y

giving: v=∏[(lny)^2-∫(y)(2lny/y)]

2nd: u=lny, du= 1/y dy, dv=dy, v=y

giving: ∏[y(lny)^2-2ylny+∫y(1/y)dy

answer: ∏[y(lny)^2 -2ylny+2y], evaluated from 1 to 8=∏[8(ln8)^2-16ln8+14]=48.13407626

4. Mar 31, 2013

### Staff: Mentor

You are calculating two different things like this - one volume is between the function and the y-axis, the other volume is between the function and the x-axis.