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Question Number 118210 by bemath last updated on 16/Oct/20

Answered by john santu last updated on 16/Oct/20

$$Reducestheequationofthecylinder$$$$to{r}^2=2r\cos \theta orr=2\cos \theta ,in\theta \in [0,\pi ]$$$$Thevolumeequals$$$$\int \int (8x^2+8y^2)dA=\underset{0}{\stackrel{\pi }{\int }}{\int }_0^{2\cos \theta }\left(8r^2\right).\left(rdrd\theta \right)$$$$=\underset{0}{\stackrel{\pi }{\int }}\left(2r^4\right){\mid }_0^{2\cos \theta }d\theta ={\int }_0^{\pi }32\cos {}^4\theta d\theta$$$$=32{\int }_0^{\pi }{\left(\frac{1}{2}(1+\cos 2\theta )\right)}^{2}d\theta$$$$=4\underset{0}{\stackrel{\pi }{\int }}(3+4\cos 2\theta +\cos \left(4\theta \right))d\theta$$$$=12\pi$$

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