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!

  • 2
    $\begingroup$ Let $x^2=r\cos t$ and $y^2=r\sin t$. $\endgroup$ – 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: $$ \mathfrak{I}=\int_{-\infty}^{+\infty}\frac{u^2}{1+u^4}\,du =\color{blue}{\frac{\pi}{\sqrt{2}}}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.