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I have problem with this integral $$\int_{-\pi/2}^{\pi/2}\int_{0}^{a\cos(\theta)}\frac{r\cos(\theta)}{\sqrt{(a^2-r^2)}}rdrd\theta$$ $$\ = \frac{1}{2a^2}\int_{-\pi/2}^{\pi/2}\cos(\theta)\arcsin(\cos(\theta))d\theta = a^2$$ I find it in this way, but I do not sure that it is the right answer. Can you help me maybe I make a mistake. I start to do that: $$\int_{0}^{a\cos(\theta)}\frac{r^2}{\sqrt{(a^2-r^2)}}drd\theta$$ $$\int_{0}^{\arcsin(\cos(\theta))}a^2(\sin u)^2du = \frac{a^2}{2}\left (\int_{0}^{\arcsin(\cos(\theta))}du - \int_{0}^{\arcsin(\cos(\theta))}\cos2udu\right) = \frac{a^2}{2}(\arcsin(\cos(\theta)-\sin(\theta)\cos(\theta)))$$

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Hint : First substitute $r=a\cos\phi$

$$\int_{-\pi/2}^{\pi/2}\cos\theta\int_0^{a\cos\theta}{r^2\over\sqrt{a^2-r^2}}dr$$Now take $r=a\sin\phi$$$=\int_{-\pi/2}^{\pi/2}\cos\theta\int_0^{a\cos\theta}{a^2\sin^2\phi\over a\cos\phi}a\cos\phi d\phi \ = {a^2\over 2}\int_{-\pi/2}^{\pi/2}\cos\theta\int_0^{\sin^{-1}(\cos\theta)}(1-\cos2\phi) d \phi$$

Now you complete

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  • $\begingroup$ why can i do such thing ?$$r=a\sin\phi$$ $\endgroup$
    – Viktor
    Dec 16 '16 at 20:17
  • $\begingroup$ @Viktor Don't you know what substitution is? $\endgroup$ Dec 16 '16 at 20:21

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