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Question Number 101551 by mhmd last updated on 03/Jul/20

Answered by bobhans last updated on 03/Jul/20

$$\left.\left(\mathrm{Q2}\right)\:\mathrm{The}\:\mathrm{area}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\underset{\mathrm{0}} {\overset{{x}} {\int}}\:{dy}\:{dx}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left({y}\right)\right]_{\mathrm{0}} ^{{x}} \:\:{dx} \\$$$$=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:{x}\:{dx}\:=\:\:\left[\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:\right]_{\mathrm{0}} ^{\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:\heartsuit \\$$

Commented by mhmd last updated on 03/Jul/20

$${sir}\:{i}\:{want}\:{in}\:{polar} \\$$

Answered by mr W last updated on 04/Jul/20

$${Q}\mathrm{2}\left({b}\right) \\$$$$\mathrm{1}−{r}\:\mathrm{cos}\:\theta={r}\:\mathrm{sin}\:\theta \\$$$$\Rightarrow{r}=\frac{\mathrm{1}}{\mathrm{sin}\:\theta+\mathrm{cos}\:\theta} \\$$$${A}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}} \rho{d}\rho{d}\theta \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)^{\mathrm{2}} }{d}\theta \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:\mathrm{2}\theta}{d}\theta \\$$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:\mathrm{2}\theta}{d}\left(\mathrm{2}\theta\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:{t}}{dt} \\$$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tan}\:\frac{{t}}{\mathrm{2}}}\right]_{\pi} ^{\mathrm{0}} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{0}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}} \\$$