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?

  • 1
    $\begingroup$ WA does not agree with your integrand function. $\endgroup$
    – farruhota
    May 9, 2019 at 15:27

2 Answers 2


We can save a lot of pain and get the correct answer by completing the square:

\begin{align} x^4-2x^3 y+3x^2y^2-2xy^3+y^4 &= (x-y)^4 + 2x^3 y -3x^2 y^2 +2x y^3 \end{align}

Pull a common factor $xy$ out of the leftover terms and complete the square again:

\begin{align} &= (x-y)^4 + xy(2x^2 -3x y +2 y^2) \\ &= (x-y)^4 + xy(\sqrt{2}x-\sqrt{2}y)^2 + (xy)^2\\ &= (x-y)^4 + 2 xy(x-y)^2 + (xy)^2 \end{align}

This is a sum of squares:

\begin{align} &= ((x-y)^2 + xy)^2 \end{align}

Substituting in terms of $u$ and $v$ is now easy:

\begin{align} &= ((-2\sqrt{\frac{2}{3}}v)^2 + 2u^2 -\frac{2}{3}v^2)^2\\ &= 4(v^2 + u^2)^2 \end{align}

and now polar coordinates is easy to do.

Moral of the story: always look for repeated opportunities to complete the square when we have funny polynomials like the one you linked (ie. looks almost exactly like the polynomials appearing in the binomial theorem). Or maybe just in any case in multivariable calculus.


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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