Volume using the disk method

1. Mar 4, 2010

pinkerpikachu

1. The problem statement, all variables and given/known data
Use the functions:
y=x+1, y=0, x=0, x=2

Find the volume when rotated around y=3
Find the volume when rotated around y=-1

2. Relevant equations
PI$$\int$$ R2dx or perhaps PI$$\int$$ R2 - r2dx

3. The attempt at a solution

i don't think that this is a particularly hard question, but I just can't seem to get the right answer. (perhaps I'm making a silly mistake)

So first I graphed the equations, all in the first quadrant, vertical line at x=2. bounded region looks like a trapezoid.

attempt:
PI$$\int$$ (from 0-2) of [3-(x+1)]2 dx

however the answer is 46pi/3 and I get 8pi/3 <-- very obviously wrong.
is this a washer method problem?

for the next, PI$$\int$$ (from 0-2) of [1+(x+1)]2 dx

i get 32pi/3 and the answer is 50pi/3

thanks :)

2. Mar 5, 2010

Staff: Mentor

It's the second one.
The volume of your typical volume element is $\Delta V = \pi (R^2 - r^2) \Delta x$. It's not $\Delta V = \pi (R - r)^2 \Delta x$. Do you see what you're doing wrong?