Calculating $\iint_R \frac{x^2 y^2}{(x^2+y^2)^{5/2}}\,\mathrm{d}x\,\mathrm{d}y$ with possible coordinate change

Question

I have to calculate the given integral: $$\iint_R \frac{x^2 \cdot y^2}{(x^2+y^2)^{5/2}}\,\mathrm{d}x\,\mathrm{d}y$$

where the region $R$ is the rectangle $[0,\sqrt{3}]\times[0,1]$.

I tried making the usual substitutions with no apparent success. So I concluded that changing to polar coordinates could help.

• The problem is that when I try to find out how the rectangle $R$ would transform in polar coordinates, the fact that on side rests in coordinates $y=0$ seems to impossibilitate the process. What I mean by this is that, for the transformation, I would have to consider $\tan(\theta)=\frac{x}{y}$. I could just say that when $y=0$ we have $\theta=\pi/2$ but in my sketch of the region in polar coordinates it doesn't seem to "close" the region.

Am I doing something wrong? Is there a change of coordinates I'm missing? Thanks in advance!

Answer 1

Let $R_1$ be the part of $R$ below that line $x=\sqrt3y$ and let $R_2$ be the part of $R$ above that line.

The integral$$\iint_{R_1}\frac{x^2y^2}{(x^2+y^2)^{5/2}}\,\mathrm dx\,\mathrm dy$$can be computed in polar coordinates:\begin{align}\iint_{R_1}\frac{x^2y^2}{(x^2+y^2)^{5/2}}\,\mathrm dx\,\mathrm dy&=\int_0^{\pi/6}\,\int_0^{\sqrt3/\cos\theta}\cos^2(\theta)\sin^2(\theta)\,\mathrm dr\,\mathrm d\theta\\&=\sqrt3\int_0^{\pi/6}\cos(\theta)\sin^2(\theta)\,\mathrm d\theta\\&=\frac{\sqrt3}{24}.\end{align}

The integral$$\iint_{R_2}\frac{x^2y^2}{(x^2+y^2)^{5/2}}\,\mathrm dx\,\mathrm dy$$can also be computed in polar coordinates; it's equal to$$\int_{\pi/6}^{\pi/2}\int_0^{1/\sin\theta}\cos^2(\theta)\sin^2(\theta)\,\mathrm dr\,\mathrm d\theta.$$And, of course,$$\iint_{R}\frac{x^2y^2}{(x^2+y^2)^{5/2}}\,\mathrm dx\,\mathrm dy=\iint_{R_1}\frac{x^2y^2}{(x^2+y^2)^{5/2}}\,\mathrm dx\,\mathrm dy+\iint_{R_2}\frac{x^2y^2}{(x^2+y^2)^{5/2}}\,\mathrm dx\,\mathrm dy.$$

Answer 2

You can also do it without polar coordinates:$$\begin{align}\int_{0}^{1}\mathrm{d}y\:y^{2}\int_{0}^{\sqrt{3}}\frac{x^{2}\mathrm{d}x}{\left(x^{2}+y^{2}\right)^{5/2}}&=\frac{1}{3}\int_{0}^{1}\mathrm{d}y\left[\frac{x^{3}}{\left(x^{2}+y^{2}\right)^{3/2}}\right]_{0}^{\sqrt{3}}\\&=\sqrt{3}\int_{0}^{1}\frac{\mathrm{d}y}{\left(3+y^{2}\right)^{3/2}}\\&=\frac{1}{\sqrt{3}}\left[y\left(3+y^{2}\right)^{-1/2}\right]_{0}^{1}\\&=\frac{1}{\sqrt{6}}.\end{align}$$