# Thread: Washer Method

1. ## Washer Method

I understand how to do it and i've gotten every problem right so far but i can't see where my error is in this problem. Can someone point it out? I use a variation of the formula to make it easier on myself. The equations are y=x^2 and y =4x-x^2. They intersect at (0,0) and (2,4). It says to revolve the region to find the volume of the solid about the line y=6. That means that the total volume is pi * integral 6^2 from 0 to 2 with respect to x. Then minus the area of the open area... which would be pi*integral (6-(4x-x^2))^2 from 0 to 2 with respect to x and also subtract pi*integral (6-x^2)^2 from 0 to 2 with respect to x. Where is my error. The correct answer is 64pi/3.

2. Here's your set up.

$\displaystyle {\pi}\int_{0}^{2}\left[(6-x^{2})^{2}-(6-(4x-x^{2}))^{2}\right]dx$

$\displaystyle {\pi}\int_{0}^{2}\left[8x^{3}-40x^{2}+48x\right]dx$

Just for fun, here's the shells method also:

$\displaystyle 2{\pi}\int_{0}^{4}(y-6)(-\sqrt{4-y}+2-\sqrt{y})dy$

3. thanks i haven't learned the shell method yet. Is it easier?

4. i see yours. Inner minus outer. I just don't understand why my variation doesnt work

5. If your method is the same as I showed, then you probably just have a arithmetic error. That's mostly the problem.

Yes, sometimes the shells method is easier, sometimes not. Depends.