I want to calculate $I=\iint_R y^2dA$ where $R$ is the surface bounded by the curve $r=a(1+\cos\theta)$ for $\theta$ ranging between $0$ and $2\pi$.
Attempt: To calculate this as a double integral in polar coordintes we have:
$$I=\iint_R r^2\sin^2\theta=\int\limits_{0}^{2\pi}d\theta\int\limits_{0}^{a(1+\cos\theta)} r^2\sin^2\theta\cdot r\,dr\\=\int\limits_{0}^{2\pi}\sin^2\theta\,dθ\int\limits_{0}^{a(1+\cos\theta)}r^3\,dr=\int\limits_{0}^{2\pi}\sin^2\theta\frac{a^4(1+\cos\theta)^4}{4}\,d\theta$$
But I am stuck from here on. Is there a better way for solving this from scratch? Or can my attempt be completed? Any help, hint or solution would be highly appreciated!
Well, if I understood you properly then we have something like this $$\int \int_R y^2 ~dA$$
where $R$ is the region bounded by $ r= a(1+\cos\theta)$ and $\theta$ goes from 0 to $2\pi$. $$ \int_{\theta = 0}^{2\pi} \int_{r=0}^{a(1+\cos\theta)} r^3 \sin^2 \theta dr d\theta$$
$$\int_{\theta=0}^{2\pi} \sin^2\theta \frac{a^4(1+\cos\theta)^4}{4} d\theta$$ You can integrate the above expression to get the value of $\frac{21}{32}\pi a^4$.
Now, your question about how to carry out the integral is easy to answer, just take $\frac{a^4}{4}$ out of the integral and then expand the $\sin^2\theta (1+\cos\theta)^4$. After expansion the integrals will become some of very common ones (that is they will involve only some powers of $\sin$ and $\cos$).
Hope it helps!