Question Number 118210 by bemath last updated on 16/Oct/20
Answered by john santu last updated on 16/Oct/20
![Reduces the equation of the cylinder to r^2 =2r cos θ or r = 2cos θ, in θ ∈ [0,π ] The volume equals ∫∫(8x^2 +8y^2 )dA =∫_0 ^π ∫_0 ^(2cos θ) (8r^2 ).(rdr dθ) =∫_0 ^π (2r^4 )∣_0 ^(2cos θ) dθ = ∫_0 ^π 32 cos^4 θ dθ = 32∫_0 ^π ((1/2)(1+cos 2θ))^2 dθ = 4∫_0 ^π (3+4cos 2θ+cos (4θ)) dθ = 12π](https://www.tinkutara.com/question/Q118213.png)
$${Reduces}\:{the}\:{equation}\:{of}\:{the}\:{cylinder} \\ $$$${to}\:{r}^{\mathrm{2}} =\mathrm{2}{r}\:\mathrm{cos}\:\theta\:{or}\:{r}\:=\:\mathrm{2cos}\:\theta,\:{in}\:\theta\:\in\:\left[\mathrm{0},\pi\:\right] \\ $$$${The}\:{volume}\:{equals}\: \\ $$$$\int\int\left(\mathrm{8}{x}^{\mathrm{2}} +\mathrm{8}{y}^{\mathrm{2}} \right){dA}\:=\underset{\mathrm{0}} {\overset{\pi} {\int}}\int_{\mathrm{0}} ^{\mathrm{2cos}\:\theta} \left(\mathrm{8}{r}^{\mathrm{2}} \right).\left({rdr}\:{d}\theta\right) \\ $$$$\:=\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\left(\mathrm{2}{r}^{\mathrm{4}} \right)\mid_{\mathrm{0}} ^{\mathrm{2cos}\:\theta} \:{d}\theta\:=\:\int_{\mathrm{0}} ^{\pi} \mathrm{32}\:\mathrm{cos}\:^{\mathrm{4}} \theta\:{d}\theta \\ $$$$\:=\:\mathrm{32}\int_{\mathrm{0}} ^{\pi} \left(\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta\right)\right)^{\mathrm{2}} {d}\theta \\ $$$$\:=\:\mathrm{4}\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\left(\mathrm{3}+\mathrm{4cos}\:\mathrm{2}\theta+\mathrm{cos}\:\left(\mathrm{4}\theta\right)\right)\:{d}\theta \\ $$$$\:=\:\mathrm{12}\pi \\ $$