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Question Number 27500 by abdo imad last updated on 07/Jan/18

$$find\int {\int }_{\mathrm{\Delta }}\sqrt{4-x^2-y^2}dxdywith$$$$\mathrm{\Delta }=\{(x,y)\in {\mathbb{R}}^2/x^2+y^2⩽2x\}$$

Commented by abdo imad last updated on 10/Jan/18

$$letusethechangementx=rcos\theta andy=rsin\theta$$$$x^2+y^2⩽2x\iff r^2⩽2rcos\theta \iff 0<r⩽2cos\theta$$$$I=\int {\int }_{{-}_{}\frac{\pi }{2}<\theta <\frac{\pi }{2}and0<r⩽2cos\theta }\sqrt{4-r^2}rdrd\theta$$$$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left({\int }_0^{2cos\theta }r\sqrt{4-r^2}dr\right)d\theta but$$$${\int }_0^{2cos\theta }r\sqrt{4-r^2}dr={[-\frac{1}{3}{(4-r^2)}^{\frac{3}{2}}]}_0^{2cos\theta }$$$$=-\frac{1}{3}({(4-4{cos}^2\theta )}^{\frac{3}{2}}-4^{\frac{3}{2}}$$$$=-\frac{1}{3}(4^{\frac{3}{2}}{\left({sin}^2\right)}^{\frac{3}{2}}-8)=-\frac{1}{3}(8{sin}^3\theta -8)$$$$I=-\frac{8}{3}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}({sin}^3\theta -1)d\theta =\frac{8\pi }{3}-\frac{8}{3}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{sin}^3\theta d\theta$$$$wefindthevalueofIbylinearisationof{sin}^3\theta ....$$$$$$$$$$$$$$$$$$

Commented by abdo imad last updated on 10/Jan/18

$$thefonctionisimparso{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{sin}^{3}dx=0andI=\frac{8\pi }{3}.$$

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