Hello i want to ask for help with understanding how to integrate a integral with a elipse. I have the following domain $D=(\frac{x^2}{2} + \frac{y^2}{2})^4 \le \frac{xy}{\sqrt6}$ and the integral $$\iint_D \sqrt {xy}$$ I want to move into spherical coordinates but i can't simplify the domain to find the limits if someone can help me with some guide how to approach to this problem. I am not skilled in maths so i apologize if the question is not relevant. Thank you in advance


1 Answer 1


You mean polar coordinates right? Because spherical coordinates are for 3-dimensional space. Anyways, here we go. Substitute $x=r\cos\theta,y=r\sin\theta$ to get

$$\left(\cfrac{r^2\cos^2\theta+r^2\sin^2\theta}{2}\right)^4\leq \cfrac{r^2\sin\theta\cos\theta}{\sqrt{6}}$$

so using the Pythagorean identity,


Now hit the integral, remembering what $D$ is:

$$\begin{align}&\iint\limits_D\sqrt{xy}\,\mathrm dA\\ =& \int\int_0^{\sqrt[6]{\frac{16\sin\theta\cos\theta}{\sqrt{6}}}}r\sqrt{r^2\sin\theta\cos\theta}\,\mathrm dr\mathrm\, d\theta\\ =&\int\frac{1}{3}\sqrt{\frac{16\sin\theta\cos\theta}{\sqrt{6}}}\sqrt{\sin\theta\cos\theta}\, d\theta\\ =&\int\frac{1}{3}\sqrt{\frac{16}{\sqrt{6}}}\sin\theta\cos\theta\, d\theta\\ \end{align}$$

Now what are the bounds for this? If you think carefully, $\sqrt{xy}$ exists so $xy$ has to be positive. Thus the bounds are simply $0\to\frac{\pi}{2}$ and $\pi\to\frac{3}{2}\pi$.

Thus we obtain $$\cfrac{1}{3}\sqrt{\cfrac{16}{\sqrt{6}}}=\cfrac{4}{3\sqrt[4]{6}}$$ as the final answer because the antiderivative of $\sin\cos$ is $\sin^2$.

and a proof

and an image because why not? This was just a sanity check by me to make sure I had done everything correctly and desmos reassures me I had.

  • $\begingroup$ Thank you very much for showing me in such great detail. I will examine it now. $\endgroup$ May 24, 2019 at 7:03

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