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Question Number 72466 by Learner-123 last updated on 29/Oct/19

$$Provethat{\int }_0^2{\int }_{-\sqrt{1-{(y-1)}^2}}^0{xy}^{2}dxdy=-\frac{4}{5}$$$$\mathit{a}\mathit{f}\mathit{t}\mathit{e}\mathit{r}\mathrm{changing}\mathrm{the}\mathrm{integral}\mathrm{to}\mathbf{polar}\mathbf{form}.$$

Commented by Abdo msup. last updated on 30/Oct/19

$$I={\int }_0^2\left({\int }_{-\sqrt{1-{(y-1)}^2}}^{0}xdx\right)y^{2}dybut$$$${\int }_{-\sqrt{1-{(y-1)}^2}}^{0}xdx={\left[\frac{x^2}{2}\right]}_{-\sqrt{1-{(y-1)}^2}}^0=-\frac{1-{(y-1)}^2}{2}$$$$=-\frac{1-y^2+2y-1}{2}=\frac{y^2-2y}{2}\Rightarrow$$$$I=\frac{1}{2}{\int }_0^2(y^2-2y)y^{2}dy=\frac{1}{2}{\int }_0^2(y^4-2y^3)dy$$$$=\frac{1}{2}{[\frac{y^5}{5}-\frac{1}{2}{y}^4]}_0^2=\frac{1}{2}\{\frac{32}{5}-8\}=\frac{1}{2}\times \frac{-8}{5}=-\frac{4}{5}$$$$withoutuseofpolarcoordinates.$$$$$$

Commented by Learner-123 last updated on 30/Oct/19

$$thankssir.$$

Answered by mind is power last updated on 29/Oct/19

$$\mathrm{x}=\mathrm{rcos}\left(\theta \right)$$$$\mathrm{y}=\mathrm{rsin}\left(\theta \right)$$$$\mathrm{dxdy}=\mathrm{rdrd}\theta$$$${0⩾\mathrm{x}⩾-\sqrt{1-(\mathrm{y}-1})}^2$$$$0⩽{\mathrm{x}}^2⩽1-{(\mathrm{y}-1)}^2=2\mathrm{y}-{\mathrm{y}}^2\Rightarrow 0⩽{\mathrm{x}}^2+{\mathrm{y}}^2⩽2\mathrm{y}$$$$0⩽{\mathrm{r}}^2⩽2\mathrm{rsin}\left(\theta \right)\Rightarrow 0⩽\mathrm{r}⩽2\sin \left(\theta \right)$$$$0⩽\mathrm{y}⩽2$$$$\Rightarrow 0⩽\mathrm{rsin}\left(\theta \right)⩽2\Rightarrow \sin \left(\theta \right)⩾0\Rightarrow \theta \in [0,\pi ]$$$$\mathrm{but}\mathrm{x}⩽0\Rightarrow \cos \left(\theta \right)⩽0\Rightarrow \theta \in [\frac{\pi }{2},2]$$$$\mathrm{our}\mathrm{integral}\iff {\int }_{\frac{\pi }{2}}^{\pi }{\int }_0^{2\sin \left(\theta \right)}.\mathrm{rcos}\left(\theta \right).{\mathrm{r}}^2{\sin }^2\left(\theta \right).\mathrm{rdrd}\theta$$$$={\int }_{\frac{\pi }{2}}^{\pi }{\int }_0^{2\sin \left(\theta \right)}.{\mathrm{r}}^4\cos \left(\theta \right){\sin }^2\left(\theta \right)\mathrm{drd}\theta$$$$$$$$$$

Commented by Learner-123 last updated on 29/Oct/19

$$OnemoreQues.::$$$${\int }_{-a}^a{\int }_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}dydx.$$

Commented by Learner-123 last updated on 29/Oct/19

$$thankyousir!$$

Commented by mind is power last updated on 29/Oct/19

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{Welcom}$$

Commented by mind is power last updated on 29/Oct/19

$$\mathrm{x}=\mathrm{rcos}\left(\mathrm{b}\right)$$$$\mathrm{y}=\mathrm{rsin}\left(\mathrm{b}\right)$$$$\mathrm{we}\mathrm{have}-\mathrm{a}<\mathrm{x}<\mathrm{a}$$$$-\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}<\mathrm{y}<\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}\Rightarrow 0<{\mathrm{x}}^2+{\mathrm{y}}^2⩽{\mathrm{a}}^2$$$$\Rightarrow 0<\mathrm{R}<\mathrm{a}$$$$\mathrm{b}\in [0,2\pi ]$$$${\int }_0^{2\pi }{\int }_0^{\mathrm{a}}\mathrm{rdrdb}$$$$=2\pi .\left(\frac{{\mathrm{a}}^2}{2}\right)=\pi {\mathrm{a}}^2$$

Commented by Learner-123 last updated on 29/Oct/19

$$thanksagain,sir!$$

Commented by mind is power last updated on 29/Oct/19

$$\mathrm{most}\mathrm{Welcom}$$

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