Volume of Rotated Shapes: Finding the Volume using the Disk Method

[pinkerpikachu]

Homework Statement

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

Homework Equations

PI\int R²dx or perhaps PI\int R² - r²dx

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/3thanks :)

 

[Mark44]

Homework Statement

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

Homework Equations

PI\int R²dx or perhaps PI\int R² - r²dx

It's the second one.

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

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?

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 :)