# Calculate the double integral over a region $\int_{D}(x^2+y^2)dxdy$

Calculate the double integral $$\int_{D}(x^2+y^2)\,dx\,dy$$ over the region $D = \{x^4+y^4 \leqslant 1\}$.

I tried to solve this by switching to polar coordinates and got $$\int\frac{1}{\sin^4\phi + \cos^4\phi}\,d\phi$$

As you can see, this is a fairly complex integral.

Perhaps there is another ways to solve it. I would be grateful for any advice!

• Let $x^2=r\cos t$ and $y^2=r\sin t$. – Nosrati Oct 17 '17 at 19:31

By setting $x=\sqrt{\rho\cos\theta},y=\sqrt{\rho\sin\theta}$ we have
$$\mathfrak{I}=\iint_{D}(x^2+y^2)\,dx\,dy = 4\int_{0}^{1}\int_{0}^{\pi/2}\frac{\rho(\cos\theta+\sin\theta)}{4\sqrt{\sin\theta\cos\theta}}\,d\theta\,d\rho$$ or: $$\mathfrak{I}=\frac{1}{2}\int_{0}^{\pi/2}\sqrt{\tan\theta}+\sqrt{\cot\theta}\,d\theta=\int_{0}^{\pi/2}\sqrt{\tan\theta}\,d\theta=\int_{0}^{+\infty}\frac{\sqrt{u}}{1+u^2}\,du$$ or: $$\mathfrak{I}=\int_{-\infty}^{+\infty}\frac{u^2}{1+u^4}\,du =\color{blue}{\frac{\pi}{\sqrt{2}}}.$$