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 | 1 | initial version | posted 2014-07-23 16:58:11 -0600  |
I agree. But I think it should be $$\int_{0}^{2 \pi}\int_{0}^{2}\int_{r^{2}}^{8-r^{2}}r^{2} rdzdrd \theta$$ $$\int_{0}^{2 \pi}\int_{0}^{2}\int_{r^{2}}^{8-r^{2}}r^{3} dzdrd \theta$$ $$=\int_{0}^{2 \pi}\int_{0}^{2}8r^{3}-2r^{5}drd\theta$$ $$\int_{0}^{2 \pi}\frac{32}{3}d\theta$$ $$=\frac{64 \pi}{3}$$ Let me know if you get a different answer.
| | 2 | No.2 Revision |
I agree. But I think it there should be
be an r in there to account for the larger volume at greater radii like we said in class.
$$\int_{0}^{2 \pi}\int_{0}^{2}\int_{r^{2}}^{8-r^{2}}r^{2} rdzdrd \theta$$
$$\int_{0}^{2 $$=\int_{0}^{2 \pi}\int_{0}^{2}\int_{r^{2}}^{8-r^{2}}r^{3} dzdrd \theta$$
$$=\int_{0}^{2 \pi}\int_{0}^{2}8r^{3}-2r^{5}drd\theta$$
$$\int_{0}^{2 $$=\int_{0}^{2 \pi}\frac{32}{3}d\theta$$
$$=\frac{64 \pi}{3}$$
Let me know if you get a different answer.
