Subject : Applied Mathematics 2

Topic : Triple integration and Applications of Multiple integrals

Difficulty : Medium

$\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 …

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