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Question Number 17206 by Arnab Maiti last updated on 02/Jul/17

$$\mathrm{What}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{vallu}\mathrm{of}{\int }_{-\mathrm{a}}^{\mathrm{a}}{\mathrm{x}}^2\mathrm{y}\mathrm{dx}?$$$$\mathrm{Where}{\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2\mathrm{and}\mathrm{y}⩾0$$

Answered by mrW1 last updated on 02/Jul/17

$$\mathrm{let}\mathrm{x}=\mathrm{a}\cos \theta$$$$\mathrm{and}\mathrm{y}=\mathrm{a}\sin \theta$$$$$$$$\mathrm{dx}=-\mathrm{asin}\theta \mathrm{d}\theta$$$$\mathrm{x}\in [-\mathrm{a},\mathrm{a}]\mathrm{and}\mathrm{y}⩾0\equiv \theta \in [\pi ,0]$$$$$$$${\int }_{-\mathrm{a}}^{\mathrm{a}}{\mathrm{x}}^2\mathrm{ydx}={\int }_{\pi }^0{\mathrm{a}}^2{\cos }^2\theta \mathrm{a}\sin \theta (-\mathrm{asin}\theta )\mathrm{d}\theta$$$$=-{\mathrm{a}}^4{\int }_{\pi }^0{\cos }^2\theta {\sin }^2\theta \mathrm{d}\theta$$$$={\mathrm{a}}^4{\int }_0^{\pi }{\cos }^2\theta {\sin }^2\theta \mathrm{d}\theta$$$$=\frac{{\mathrm{a}}^4}{4}{\int }_0^{\pi }{\left(2\cos \theta \sin \theta \right)}^2\mathrm{d}\theta$$$$=\frac{{\mathrm{a}}^4}{4}{\int }_0^{\pi }{\sin }^{2}2\theta \mathrm{d}\theta$$$$=\frac{{\mathrm{a}}^4}{8}{\int }_0^{\pi }[1-\cos 4\theta ]\mathrm{d}\theta$$$$=\frac{{\mathrm{a}}^4}{8}{[\theta -\frac{1}{4}\sin 4\theta ]}_0^{\pi }$$$$=\frac{{\mathrm{a}}^4}{8}\times \pi$$$$=\frac{\pi {\mathrm{a}}^4}{8}$$

Commented by Arnab Maiti last updated on 02/Jul/17

$$\mathcal{E}\times \mathfrak{c}{\mathrm{ille}}^{\mathrm{nt}}!!$$

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