0-3-0-3-9-y-2-dydx-

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Question Number 147 by novrya last updated on 25/Jan/15

$$\underset{0}{\stackrel{3}{\int }}\underset{0}{\stackrel{3}{\int }}\sqrt{9-y^2}dydx=\dots .$$

Answered by vkulkarni last updated on 11/Dec/14

$$\int \sqrt{a^2-x^2}dx=\frac{x}{2}{\sin }^{-1}\frac{x}{a}+\frac{a^2}{2}\sqrt{a^2-x^2}+C$$$$\mathrm{This}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{by}\mathrm{putting}x=a\sin \theta$$$${\int }_0^3{\int }_0^3\sqrt{9-y^2}dydx={\int }_0^3{[\frac{y}{2}{\sin }^{-1}\frac{y}{3}+\frac{9}{2}\sqrt{9-y^2}]}_0^{3}dx$$$${\int }_0^3[\frac{3}{2}{\sin }^{-1}\left(1\right)-\frac{27}{2}]dx=3\cdot [\frac{3}{2}{\sin }^{-1}\left(1\right)-\frac{27}{2}]$$

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