Double integral with Polar coordinates - hard example Calculate using polar coordinates:
$$\iint_{D}^{} (x^2+y^2)^\frac{1}5 \ dx \ dy $$
where D is the region inside the circle with radius 1.
Working:
D: $ \ x^2+y^2=1 \\ $ so
$ 0 \leq r \leq 1 \ \ , $ $0 \leq \theta \leq \pi \ $
and $x=rcos\theta \ ,$ $y=rsin\theta $ so
$$\int_{0}^{1} \int_{0}^{\pi} r^2(cos^2\theta + sin^2\theta)^\frac{1}5 \ d\theta \ dr $$ 
But here is where I get stuck! Isn't $ (cos^2\theta + sin^2\theta) = 1$? So its just
$$\int_{0}^{1} \int_{0}^{\pi} r^2 \ d\theta \ dr $$ 
I am really confused as what to do! Any help is appreciated
 A: According to your calculations it should go on as:
$$\iint_{D}^{} (x^2+y^2)^\frac{1}5 \ dx \ dy $$ 
$$=\int_{0}^{1} \int_{0}^{2\pi} (r^2\cos^2\theta + r^2\sin^2\theta)^\frac{1}{5} \cdot r d\theta \ dr $$
$$=\int_{0}^{1} \int_{0}^{2\pi} r^\frac{2}{5} \cdot  r d\theta \ dr$$
$$=\int_{0}^{1} r^\frac{7}{5} dr\int_{0}^{2\pi}  d\theta $$
$$=\frac{5}{12} \cdot 2\pi$$
$$=\frac{5\pi}{6}$$
NOTE: The limits for $\theta$ must be $0$ to $2\pi$ and not $0$ to $\pi$.
A: When converting to polar coordinates we make the substitution:
$$x = r\cos(\theta)$$
$$y = r\sin(\theta)$$
$$dA = rdrd\theta$$
With this the integral becomes:
$$\iint_D (x^2 + y^2)^\frac{1}{5} dA = \int_0^{2\pi} \int_0^1 (r^2)^\frac{1}{5} rdrd\theta$$
$$=\int_0^{2\pi} \int_0^1 r^\frac{7}{5} drd\theta$$
$$=\int_0^{2\pi} d\theta \int_0^1 r^\frac{7}{5} dr$$
$$=2\pi \cdot \Big[ \frac{5}{12}r^\frac{12}{5} \Big|_0^1 \Big]$$
$$= \frac{5\pi}{6}$$
where splitting up the double integral into two separate integrals is allowed since we are integrating over a "rectangular region".
