Help for solving integral in spherical coordinates with domain that is elipse

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

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,

$$r^6\leq\cfrac{16\sin\theta\cos\theta}{\sqrt{6}}$$

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 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.

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