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The diagram shows a sketch of the loop whose polar equation is

$$r=2(1-\sin\theta),\qquad -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$$

a)Show that the area enclosed by the loop is 16/3.

b)Show that the initial line divides the area enclosed by the loop in the ratio 1:7.

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a) Integrate $\frac12r^2d\theta$, with $\theta$ running over $-\frac{\pi}2$ to $\frac{\pi}2\\$ b) Integrate the same from $\frac{-\pi}2$ to $0$ and see to it, that the answer is $\frac78$ times that you got in a) –  Aneesh Karthik C Jul 11 '12 at 6:23
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Is this homework? If so, please tag it accordingly. –  draks ... Jul 11 '12 at 6:55
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2 Answers

Fleshing out Aneesh's comment: $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{2}r^2d\theta=-2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(1-\sin\theta\right)^2\,\overbrace{d(1-\sin\theta)}^{-\cos\theta}=\left.-\frac{2}{3}\left(1-\sin\theta\right)^3\right|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=$$ $$=-\frac{2}{3}\left[(1-1)^3-(1+1)^3\right]=\frac{16}{3}$$

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If you want the details here it is. The area of the loop has two parts$ I_1 $ and $I_2$ where. $$\int_{-\pi/2}^{0}\frac{1}{2}r^2d\theta \int_{0}^{+\pi/2}\frac{1}{2}r^2d\theta= I_1+I_2 $$ Now $$I1=\int_{-\pi/2}^{0}\frac{1}{2}r^2d\theta=\int_{-\pi/2}^{0}2(1-\sin\theta)^2\cos\theta d\theta$$. Assuming $x=\sin\theta$ and observing $dx=\cos\theta d\theta$ we have $$ I_1=\int_{-1}^{0}2(1-x)^2 dx=14/3$$ Similarly $$I_2=\int_{0}^{+1}2(1-x)^2 dx=2/3$$ Showing that $I_1+I_2=16/3$ and $I2/I1=1:7$

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