int_0^a x^2 (a^2 - x^2)^{3/2} dx = | vedclass

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(A) Let $I = \int_0^a x^2 (a^2 - x^2)^{3/2} dx$. Substitute $x = a \sin 	heta$,then $dx = a \cos 	heta d	heta$. When $x = 0$,$	heta = 0$. When $x

Vedclass · Accepted Answer

$\frac{\pi a^6}{32}$

Vedclass · Answer

(A) Let $I = \int_0^a x^2 (a^2 - x^2)^{3/2} dx$. Substitute $x = a \sin 	heta$,then $dx = a \cos 	heta d	heta$. When $x = 0$,$	heta = 0$. When $x = a$,$	heta = \frac{\pi}{2}$. $I = \int_0^{\pi/2} (a^2 \sin^2 	heta) (a^2 - a^2 \sin^2 	heta)^{3/2} (a \cos 	heta) d	heta$ $I = \int_0^{\pi/2} (a^2 \sin^2 	heta) (a^2 \cos^2 	heta)^{3/2} (a \cos 	heta) d	heta$ $I = \int_0^{\pi/2} (a^2 \sin^2 	heta) (a^3 \cos^3 	heta) (a \cos 	heta) d	heta$ $I = a^6 \int_0^{\pi/2} \sin^2 	heta \cos^4 	heta d	heta$ Using the formula $\int_0^{\pi/2} \sin^m 	heta \cos^n 	heta d	heta = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$: $I = a^6 \frac{\Gamma(\frac{3}{2}) \Gamma(\frac{5}{2})}{2 \Gamma(\frac{8}{2})} = a^6 \frac{(\frac{1}{2} \sqrt{\pi}) (\frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi})}{2 \cdot 3!} = a^6 \frac{\frac{3}{8} \pi}{2 \cdot 6} = a^6 \frac{3\pi}{96} = \frac{\pi a^6}{32}$.

$\int_0^a x^2 (a^2 - x^2)^{3/2} dx = $

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