Find the volume of the solid that is generated by rotating the region formed by the graphs of y = x2, y = 2, and x = 0 about the y-axis.
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You can use shells or washers. Shells: $\displaystyle 2{\pi}\int_{0}^{\sqrt{2}}x(2-x^{2})dx$ Washers: $\displaystyle {\pi}\int_{0}^{2}ydy$
Last edited by galactus; Nov 24th 2008 at 05:39 AM.
sorry to be so ignorant at this, but using shells could you show me the rest of the answer?
$\displaystyle 2{\pi}\int_{0}^{\sqrt{2}}x(2-x^{2})dx$ $\displaystyle 2 \pi \int_{0}^{\sqrt{2}} 2x-x^3dx$ Call the integral $\displaystyle F(x)=x^2-\frac{x^3}{3}$ Thus the volume is $\displaystyle 2\pi \left( F(\sqrt{2})-F(0) \right)$ Make sense?
so if I did it right then, V = 6.64257 ?
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