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Question Number 198282 by MathedUp last updated on 16/Oct/23

Answered by witcher3 last updated on 25/Oct/23

$$\{\begin{array}{l}2\cos \left(\mathrm{t}\right)\\ 2\sin \left(\mathrm{t}\right)\end{array},\mathrm{t}\in [0,2\pi ]\mathrm{circle}\mathrm{radius}=2\mathrm{origine}(0,0)$$$${\int }_{\mathrm{C}}(-\frac{\mathrm{xy}}{5}\mathrm{dx}+2\mathrm{ydy})=\int {\int }_{\mathrm{D}}(\partial \frac{2\mathrm{y}}{\partial \mathrm{x}}-\frac{\partial }{\partial \mathrm{y}}(-\frac{\mathrm{xy}}{5}))\mathrm{dA}$$$$=\int {\int }_{\mathrm{D}}\left(\frac{\mathrm{x}}{5}\right)\mathrm{dA}$$$$\mathrm{D}\mathrm{disc}\mathrm{of}\mathrm{radius}2,\mathrm{center}(0,0)$$$$\mathrm{x}=\mathrm{rcos}\left(\mathrm{a}\right),\mathrm{dA}=\mathrm{rdrda}$$$$={\int }_0^{2\pi }{\int }_0^2\left(\frac{1}{5}\mathrm{rcos}\left(\mathrm{a}\right)\right)\mathrm{rdrda}$$$$={\int }_0^2\frac{{\mathrm{r}}^2}{5}.{\int }_0^{2\pi }\cos \left(\mathrm{a}\right)\mathrm{da}=0$$$$$$

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