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Question Number 190052 by Mastermind last updated on 26/Mar/23

$$\mathrm{Evaluate}\int {\int }_{\mathrm{A}}{(\mathrm{x}+\mathrm{y})}^2\mathrm{dxdy}\mathrm{over}\mathrm{the}$$$$\mathrm{area}\mathrm{bounded}\mathrm{by}\mathrm{the}\mathrm{ellipse}$$$$\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$$$$$$$$$$$$\mathrm{Anybody}?$$

Answered by CElcedricjunior last updated on 26/Mar/23

$$\int {\int }_{\mathit{A}}{(\mathbf{x}+\mathbf{y})}^2\mathbf{dxdy}$$$$\mathbf{A}=\{(\mathit{x};\mathit{y});\frac{{\mathit{x}}^2}{{\mathit{a}}^2}+\frac{{\mathit{y}}^2}{{\mathit{b}}^2}=1\}$$$$\mathit{E}\mathit{n}\mathit{c}\mathit{h}\mathit{a}\mathit{n}\mathit{g}\mathit{e}\mathit{a}\mathit{n}\mathit{t}\mathit{l}\mathit{e}\mathit{s}\mathit{c}\mathit{o}\mathit{o}\mathit{r}\mathit{d}\mathit{o}\mathit{n}\mathit{n}\mathit{e}\mathit{e}\mathit{s}\mathit{d}\mathit{e}\mathit{c}\mathit{a}\mathit{r}\mathit{t}\mathit{e}\mathit{s}\mathit{i}\mathit{e}\mathit{n}\mathit{n}\mathit{e}\stackrel{`}{\mathit{a}}\mathit{p}\mathit{o}\mathit{l}\mathit{a}\mathit{i}\mathit{r}\mathit{e}$$$$\mathit{o}\mathit{n}\mathit{a}:\{\begin{array}{l}\mathit{x}=\mathit{a}\mathit{r}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\\ \mathit{y}=\mathit{b}\mathit{r}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\end{array}1>\mathit{r}>0\mathit{\theta }\in [0;2\mathit{\pi }]$$$$\mathit{e}\mathit{t}\mathit{d}\mathit{x}\mathit{d}\mathit{y}=\mid \left|\begin{array}{c}\mathit{a}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }-\mathit{a}\mathit{r}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\\ \mathit{b}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{b}\mathit{r}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\end{array}\right|\mid \mathit{d}\mathit{r}\mathit{d}\mathit{\theta }$$$$\mathit{d}\mathit{x}\mathit{d}\mathit{y}=\mathit{a}\mathit{b}\mathit{r}\mathit{d}\mathit{r}\mathit{d}\mathit{\theta }$$$$A=\{(\mathit{r};\mathit{\theta })0<\mathit{r}<1;0⩽\mathit{\theta }⩽2\mathit{\pi }\}◼\mathit{M}oivre$$$$\int {\int }_{\mathit{A}}{\mathit{a}\mathit{b}\mathit{r}}^3{(\mathit{a}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }+\mathit{b}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta })}^2\mathit{d}\mathit{r}\mathit{d}\mathit{\theta }=\mathit{k}$$$$\mathit{k}={\int }_0^{2\mathit{\pi }}{\left[\frac{\mathit{a}\mathit{b}}{4}{\mathit{r}}^4{(\mathit{a}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }+\mathit{b}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta })}^2\right]}_0^1\mathit{d}\mathit{\theta }$$$$=\frac{\mathit{a}\mathit{b}}{4}{\int }_0^{2\mathit{\pi }}({\mathit{a}}^2\mathit{c}\mathit{o}\mathit{s}\stackrel{2}{\mathit{\theta }}+2\mathit{a}\mathit{b}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }+{\mathit{b}}^2\mathit{s}\mathit{i}\stackrel{2}{\mathit{n}\theta })\mathit{d}\mathit{\theta }$$$$\frac{\mathit{a}\mathit{b}}{4}{[{\mathit{a}}^2(\frac{1}{2}\mathit{\theta }+\frac{1}{4}\mathit{s}\mathit{i}\mathit{n}2\mathit{\theta })-\frac{\mathit{a}\mathit{b}\mathit{c}\mathit{o}\mathit{s}2\mathit{\theta }}{2}+{\mathit{b}}^2(\frac{1}{2}\mathit{\theta }-\frac{1}{4}\mathit{s}\mathit{i}\mathit{n}2\mathit{\theta })]}_0^{2\mathit{\pi }}$$$$=\frac{\mathit{a}\mathit{b}}{4}\left(\mathit{\pi }({\mathit{a}}^2+\mathit{b})\right)★\mathcal{C}edricjunior$$$$=\frac{\mathit{\pi }}{4}\mathit{a}\mathit{b}({\mathit{a}}^2+{\mathit{b}}^2)$$

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