Question: Find the area inside the circle $r = a sin \theta$ and outside the cardioide $r = a (1 + cos \theta)$
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Subject : Applied Mathematics 2

Topic : Differentiation under Integral sign, Numerical Integration and Rectification

Difficulty: High

mumbai university • 56 views
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modified 9 weeks ago  • written 9 weeks ago by gravatar for Ankit Pandey Ankit Pandey60
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Solution:

$\begin{aligned} \text { Intersection of cardioide and circle is, } \\ r=a(1+\cos \theta) \text { and } r=\operatorname{asin} \theta \end{aligned}$

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$\begin{aligned} \operatorname{asin} \theta &=a(1+\cos \theta)=\gt\theta=90^{\circ} \\ a(1+\cos \theta) & \leq r \leq \operatorname{asin} \theta \\ \frac{\pi}{2} & \leq \theta \leq \pi \end{aligned}$

Area of region bounded by given circle and cardioide,

$\begin{aligned} \mathrm{I} &=\int_{\frac{\pi}{2}}^{\pi} \int_{a \sin \theta}^{a(1+\cos \theta)} r d r d \theta \\ &=\int_{\frac{\pi}{2}}^{\pi} \frac{a^{2}}{2}\left(\sin ^{2} \theta-1-2 \cos \theta-\cos ^{2} \theta\right) \mathrm{d} \theta \end{aligned}$

$\begin{aligned} &=\int_{\frac{\pi}{2}}^{\pi} \frac{a^{2}}{2}(-1-2 \cos \theta-\cos 2 \theta) \mathrm{d} \theta \\ &=\frac{a^{2}}{2}\left[-\theta-2 \sin \theta-\frac{\sin 2 \theta}{2}\right]^{\pi}_\frac{\pi}{2} \end{aligned}$

$I=\frac{a^{2}}{2}\left[(-\pi-0-0)-\left(-\frac{\pi}{2}-2-0\right)\right]$

Required area is

$=I=\frac{a^{2}}{2}\left(2-\frac{\pi}{2}\right)$

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written 9 weeks ago by gravatar for Ankit Pandey Ankit Pandey60
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