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

Answered by bobhans last updated on 03/Jul/20

(Q2) The area = ∫_0 ^1  ∫_0 ^x  dy dx = ∫_0 ^1  (y)]_0 ^x   dx  = ∫_0 ^1  x dx =  [ (1/2)x^2  ]_0 ^1 = (1/2) ♥

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

Commented by mhmd last updated on 03/Jul/20

sir i want in polar

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

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

Q2(b)  1−r cos θ=r sin θ  ⇒r=(1/(sin θ+cos θ))  A=∫_0 ^(π/2) ∫_0 ^(1/(sin θ+cos θ)) ρdρdθ  =(1/2)∫_0 ^(π/2) (1/((sin θ+cos θ)^2 ))dθ  =(1/2)∫_0 ^(π/2) (1/(1+sin 2θ))dθ  =(1/4)∫_0 ^(π/2) (1/(1+sin 2θ))d(2θ)  =(1/4)∫_0 ^π (1/(1+sin t))dt  =(1/4)[(2/(1+tan (t/2)))]_π ^0   =(1/2)(1−0)  =(1/2)

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

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