How do you find the volume of the solid obtained by revolving the curve given by $x = 3 {cos}^{3} (t)$ $x=3cos^3(t)$ , $y = 5 {sin}^{3} (t)$ $y=5sin^3(t)$ about the $x$ $x$ -axis?

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How do you find the volume of the solid obtained by revolving the curve given by $x = 3 {cos}^{3} (t)$ $x=3cos^3(t)$ , $y = 5 {sin}^{3} (t)$ $y=5sin^3(t)$ about the $x$ $x$ -axis?

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Answer 1 · 2014-09-20T12:48:25

$y^{2} = 25 {sin}^{6} t = 25 {(1 - {cos}^{2} t)}^{3} = 25 {[1 - {(\frac{x}{3})}^{\frac{2}{3}}]}^{3}$ $y^2=25sin^6t=25(1-cos^2t)^3=25[1-(x/3)^{2/3}]^3$

By Disk Method,

$V = π \int_{- 3}^{3} y^{2} d x = 25 π \int_{- 3}^{3} {[1 - {(\frac{x}{3})}^{\frac{2}{3}}]}^{3} d x$ $V=pi int_{-3}^3y^2 dx=25piint_{-3}^3[1-(x/3)^{2/3}]^3dx$

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How do you find the volume of the solid obtained by revolving the curve given by $x = 3 {cos}^{3} (t)$ $x=3cos^3(t)$ , $y = 5 {sin}^{3} (t)$ $y=5sin^3(t)$ about the $x$ $x$ -axis?

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How do you find the volume of the solid obtained by revolving the curve given by $x = 3 {cos}^{3} (t)$ $x=3cos^3(t)$ , $y = 5 {sin}^{3} (t)$ $y=5sin^3(t)$ about the $x$ $x$ -axis?