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7.9kviews
Find the area outside the circle $r = 2acos\theta$ and inside the cardioids $r=a(1+cos\theta)$.

Subject : Applied Mathematics 2

Topic : Triple integration and Applications of Multiple integrals

Difficulty : Medium

2 Answers
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1.5kviews

$\text{Area} = \int \int r \hspace{0.1cm}dr \hspace{0.1cm}d \theta\\ \hspace{0.8cm}= 2 \int^{\frac{\pi}{2}}_0 \int^{a(1+cos \theta)}_{r = 2acos \theta} r \hspace{0.1cm}dr\hspace{0.1cm} d \theta\\ \hspace{0.8cm}= 2 \int^{\frac{\pi}{2}}_0 \big(\frac{r^2}{2}\big)^{a(1+cos \theta)}_{r = 2acos \theta}\hspace{0.1cm} d \theta\\ \hspace{0.8cm}= 2 \int^{\frac{\pi}{2}}_0 [a^2(1+cos \theta)^2-4a^2 cos^2 \theta]\\ \hspace{0.8cm}= a^2 \int^{\frac{\pi}{2}}_0(1+2cos \theta+cos^2 \theta - 4 cos^2 \theta)d \theta\\ \hspace{0.8cm}= a^2 \int^{\frac{\pi}{2}}_0 (1 + 2cos \theta -3cos^2 \theta)d \theta\\ \hspace{0.8cm}= a^2 \int^{\frac{\pi}{2}}_0 \big(1+2cos \theta - \frac{3}{2}(1+cos 2 \theta)\big)d \theta\\ \hspace{0.8cm}= a^2 \Big[\theta +2 sin \theta - \frac{3}{2} \theta- \frac{3}{2} \frac{sin 2 \theta}{2}\Big]_0^{\frac{\pi}{2}}\\ \hspace{0.8cm} = a^2 \Big[\frac{\pi}{2}+2 - \frac{3}{2} \frac{\pi}{2}\Big] = a^2\Big[\frac{\pi}{2} - \frac{3 \pi}{4}+2\Big]\\ \hspace{0.8cm}=a^2\Big(2 - \frac{\pi}{4}\Big) $

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My answer is πa2 which is calculated by considering seperate curve and then subtracting

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