Double integral with square root over an asteroid I have to solve the integral $$\int_D \sqrt{x^2+y^2} dx dy$$ where $D=\{(x,y)\in\mathbb{R^2}: x^{2/3}+y^{2/3}\le1\}$.
I am not able to find a parameterization that suits the integrand.
I tried with $$\cases{x=(r\cos t)^3\\y=(r\sin t)^3}$$ in order to reduce the domain to a circle but then the integral becomes $$\int_0^1\int_0^{2\pi} r^3\sqrt{(\cos t)^6+(\sin t)^6}\cdot9r^5\cos^2 t\sin^2 t dt dr$$ and then I cannot proceed to solve the integral.
Should I change the parametrization? 
 A: We may exhibit a parametrization of the integration domain through $x=\rho\cos^3\theta, y=\rho\sin^3\theta$ with $\rho\in[0,1]$ and $\theta\in[0,2\pi)$. By considering the Jacobian of this transformation and symmetry, we get that the original integral equals
$$ 4 \int_{0}^{1}\int_{0}^{\pi/2}3\rho^2\sin^2\theta\cos^2\theta\sqrt{\sin^6\theta+\cos^6\theta}\,d\theta\,d\rho $$
or
$$ 4\int_{0}^{1}x^2\sqrt{1-x^2}\sqrt{x^6+(1-x^2)^3}\,dx\stackrel{x\mapsto\sqrt{x}}{=}2\int_{0}^{1}\sqrt{x(1-x)} \sqrt{1-3x(1-x)}\,dx $$
and by setting $x=\frac{1+t}{2}$ the last integral is converted into
$$ \frac{1}{2}\int_{-1}^{1}\sqrt{1-t^2}\sqrt{1-\frac{3}{4}(1-t^2)}\,dt=\color{blue}{\frac{1}{2}\int_{0}^{1}\sqrt{1-t^2}\sqrt{1+3t^2}\,dt} $$
which can be represented in terms of complete elliptic integrals of the first and second kind:
$$\boxed{ \iint_{D}\sqrt{x^2+y^2}\,dx\,dy = \tfrac{2}{9}\, E\!\left(\tfrac{3}{4}\right)+\tfrac{1}{9}\,K\!\left(\tfrac{3}{4}\right)}$$
Here I am using Mathematica's convention: the argument of $E$ or $K$ is the elliptic modulus and not the elliptic parameter.
Numerically, this is approximately $0.5087364114$. An explicit series representation is 
$$ I=\frac{\pi}{18}\sum_{n\geq 0}\frac{3^n}{64^n}\binom{2n}{n}^2\frac{2n-3}{2n-1}.$$
The inequality $I\leq \frac{1}{\sqrt{3}}$ can be simply derived by applying the Cauchy-Schwarz inequality to the blue integral above.
A: With Polar Coordinates and with the Jacobian Transformation
$$\int_D\sqrt{x^2+y^2}\ dS\implies\iint_{\mathbb R^2} r\cdot rdS\implies\int_0^1\int_0^{2\pi}r^2dtdr$$
