# Finding the double integral over a polar region

I am not sure if I simplified wrong, and if I didn't, how would I go about integrating this?

In the problem statement, the region is $$D = \Big\{(r,\theta) \mid 2\le r \le 3, \frac{\pi}{3} \le \theta \le \frac{2\pi}{3}\Big\},$$ and the function is $$f(x,y) = x^4+2x^2y^2+y^4.$$ So converting this to polar coordinates and setting up the double integral over the region I achieve $$\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \left(\int_{2}^{3}\left[r^4\cos^4(\theta)+2\left(r^2\cos^2(\theta)r^2\sin^2(\theta)\right)+r^4\sin^4(\theta)\right]r \;\mathrm{d}r \right) \mathrm{d}\theta$$

After all my simplifying through grouping and recognizing trig identities I can use I achieve $$\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \left(\int_{2}^{3} r^5\left(\cos^4(\theta)+\sin^4(\theta)\right)+2r^3 \; \mathrm{d}r \right) \mathrm{d}\theta = \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \left[\frac{665}{6}\left(\cos^4(\theta)+\sin^4(\theta)\right) + \frac{65}{2} \right]\mathrm{d}\theta.$$

I can easily integrate $$\frac{65}{2}$$ by itself, but how would I integrate the former with $$\cos^4(\theta)+\sin^4(\theta)$$? Obviously, I can pull the constant out and then split them up into two different integrals (and multiply the constant back after integrating both) but to integrate $$\cos^4(\theta)$$ and $$\sin^4(\theta)$$ doesn't seem like it would be pretty to do.

Did I create the right setup and simplify correctly?

Observe that \begin{align*} \cos^4x+\sin^4 x &=\left(\cos^2x+\sin^2x\right)^2-2(\sin x\cos x)^2 \\ &=1-2 \left(\frac{1}{2}\sin(2x) \right)^2\\ &=1-\frac{1}{2}\sin^2(2x)\\ &=1-\frac{1}{2}\left(\frac{1-\cos(4x)}{2}\right)\\ &=\frac{1}{4}\cos(4x)+\frac{3}{4}. \end{align*} This should then be easy to integrate.
Observe that $$f(x,y)=(x^2+y^2)^2$$ so your integral is
$$\int_{\pi/3}^{2\pi/3} \int_2^3 f(r\cos\theta,r\sin\theta) rdrd\theta$$
$$=\int_{\pi/3}^{2\pi/3} \int_2^3 (r^2\cos^2\theta+r^2\sin^2\theta)^2 rdrd\theta$$
$$=\int_{\pi/3}^{2\pi/3} \int_2^3 r^4.rdrd\theta$$
$$=\int_{\pi/3}^{2\pi/3} d\theta\times \int_2^3 r^5 dr$$
$$=\frac{ (3^6-2^6) }{18}\pi$$