# Thread: Washer method

1. ## Washer method

Find the volume generated by revolving the area bounded by x=y^2 and x=4 about the line y=2.

Is it the integral from 0 to 4 of pi(4-(2-sqrt x)^2)?

Sorry I don't know how to use latex yet.

2. washer method ...

$\displaystyle R = 2 + \sqrt{x}$

$\displaystyle r = 2 - \sqrt{x}$

$\displaystyle R^2 - r^2 = (4 + 4\sqrt{x} + x) - (4 - 4\sqrt{x} + x) = 8\sqrt{x}$

$\displaystyle V = \pi \int_0^4 8\sqrt{x} \, dx$

3. For shells, see if you get the same solution.

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

4. Originally Posted by skeeter
washer method ...

$\displaystyle R = 2 + \sqrt{x}$

$\displaystyle r = 2 - \sqrt{x}$

$\displaystyle R^2 - r^2 = (4 + 4\sqrt{x} + x) - (4 - 4\sqrt{x} + x) = 8\sqrt{x}$

$\displaystyle V = \pi \int_0^4 8\sqrt{x} \, dx$
yeah i forgot that the graph is not bounded by y=0 and -sqrt(x) needs to be considered as well. Thanks for your help. appreciate it.

And yeah I get the same answer with the shell method - 128pi/3