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.
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washer method ... $\displaystyle R = 2 + \sqrt{x}$ $\displaystyle r = 2 - \sqrt{x}$ $\displaystyle R^2 - r^2 = (4 + 4\sqrt{x} + x) - (4 - 4\sqrt{x} + x) = 8\sqrt{x}$ $\displaystyle V = \pi \int_0^4 8\sqrt{x} \, dx$
For shells, see if you get the same solution. $\displaystyle 2{\pi}\int_{0}^{4}(y-2)y^{2}dy$
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
Last edited by elitespart; Nov 19th 2008 at 04:23 PM.
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