<s>[INST] Change of variables for double integral Problem: Using the change of variables $$x=\sqrt2u-\sqrt\frac{2}{3}v,y=\sqrt2u+\sqrt\frac{2}{3}v$$
Calculate the double integral $$\iint_Rx^4-2x^3y+3x^2y^2-2xy^3+y^4dA$$ where $R$ is the region bound by $x^2-xy+y^2=2$.
My work so far: The Jacobian is fairly trivial: $\frac{4}{\sqrt3}$
The region $R$ becomes $$(\sqrt2u-\sqrt\frac{2}{3}v)^2-(\sqrt2u-\sqrt\frac{2}{3}v)(\sqrt2u+\sqrt\frac{2}{3}v)+(\sqrt2u+\sqrt\frac{2}{3}v)^2=2$$$$\rightarrow2u^2+2v^2=2$$$$\rightarrow u^2+v^2=1$$
which is the unit disk. And after a lengthy calculation, I believe $$x^4-2x^3y+3x^2y^2-2xy^3+y^4$$$$\rightarrow \frac{8}{3}(-3u^4+6u^2v^2+v^2)$$
So we have
$$\iint_S\frac{8}{3}(-3u^4+6u^2v^2+v^2)\frac{4}{\sqrt3} \ du \ dv$$
where $S$ is the unit disk.
Now what to do? I tried proceeding using polar coordinates, but didn't find it an easy integral to compute. Am I just very bad at polar coordinates, or is there another way? Or have I miscalculated somewhere?
 [/INST] A comment disputes some of your coefficients, but note that $$\iint_S u^k v^l dudv=\frac{1}{k+l+2}\int_0^{2\pi}\cos^k\theta\sin^l\theta d\theta,$$so you only need to know $$\int_0^{2\pi}\cos^4\theta d\theta=\frac14\int_0^{2\pi}(1+\cos2\theta)^2d\theta=\frac14\int_0^{2\pi}(1+2\cos2\theta+\cos^22\theta)d\theta=\frac{3\pi}{4}$$(and similarly $\int_0^{2\pi}\cos^4\theta d\theta=\frac{3\pi}{4}$) and$$\int_0^{2\pi}\cos^2\theta \sin^2\theta d\theta=\frac14\int_0^{2\pi}\sin^22\theta d\theta=\frac{\pi}{4}.$$For each of these calculations, I use the fact that $\cos^2\phi,\,\sin^2\phi$ both average to $\frac12$.
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