# Washer method

• Nov 19th 2008, 11:13 AM
elitespart
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.
• Nov 19th 2008, 01:52 PM
skeeter
washer method ...

$R = 2 + \sqrt{x}$

$r = 2 - \sqrt{x}$

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

$V = \pi \int_0^4 8\sqrt{x} \, dx$
• Nov 19th 2008, 02:32 PM
galactus
For shells, see if you get the same solution.

$2{\pi}\int_{0}^{4}(y-2)y^{2}dy$
• Nov 19th 2008, 04:13 PM
elitespart
Quote:

Originally Posted by skeeter
washer method ...

$R = 2 + \sqrt{x}$

$r = 2 - \sqrt{x}$

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

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

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