<s>[INST] 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
 [/INST] 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 $\pi$.
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