If int_0^{2a} x^2 sqrt{2ax-x^2} dx = ka^4, th | vedclass

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(C) Let $I = \int_0^{2a} x^2 \sqrt{2ax-x^2} dx$. Substitute $x = 2a \sin^2 	heta$, then $dx = 4a \sin 	heta \cos 	heta d	heta$. The limits change:

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$5$

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(C) Let $I = \int_0^{2a} x^2 \sqrt{2ax-x^2} dx$. Substitute $x = 2a \sin^2 	heta$, then $dx = 4a \sin 	heta \cos 	heta d	heta$. The limits change: when $x=0, 	heta=0$; when $x=2a, 	heta=\frac{\pi}{2}$. Also, $\sqrt{2ax-x^2} = \sqrt{2a(2a \sin^2 	heta) - (2a \sin^2 	heta)^2} = \sqrt{4a^2 \sin^2 	heta \cos^2 	heta} = 2a \sin 	heta \cos 	heta$. Substituting these into the integral: $I = \int_0^{\pi/2} (2a \sin^2 	heta)^2 (2a \sin 	heta \cos 	heta) (4a \sin 	heta \cos 	heta) d	heta$ $I = \int_0^{\pi/2} (4a^2 \sin^4 	heta) (8a^2 \sin^2 	heta \cos^2 	heta) d	heta$ $I = 32a^4 \int_0^{\pi/2} \sin^6 	heta \cos^2 	heta d	heta$. Using Wallis' formula $\int_0^{\pi/2} \sin^m 	heta \cos^n 	heta d	heta = \frac{(m-1)!!(n-1)!!}{(m+n)!!} \cdot \frac{\pi}{2}$ (for even $m, n$): $I = 32a^4 \cdot \frac{(5 \cdot 3 \cdot 1) \cdot (1)}{(8 \cdot 6 \cdot 4 \cdot 2)} \cdot \frac{\pi}{2} = 32a^4 \cdot \frac{15}{384} \cdot \frac{\pi}{2} = \frac{5\pi a^4}{8}$. Given $I = ka^4$, we have $k = \frac{5\pi}{8}$, so $k : \pi = 5 : 8$.

If $\int_0^{2a} x^2 \sqrt{2ax-x^2} dx = ka^4$, then $k : \pi =$ (in $:8$)

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