I was doing the next integral and I have some questions about to solve it:
$$\int\int\int_{S}\frac{1}{(x^{2}+y^{2}+z^{2})^{2}}dV$$ where S is determined by $x^{2}+y^{2}+z^{2}\geq1$ this implies that we are integrating outside the unit sphere. Whith this we can find out quickly the limits of integration: $1\leq x,y,z\leq\infty$, then:
$$\int_{1}^{\infty}\int_{1}^{\infty}\int_{1}^{\infty}\frac{1}{(x^{2}+y^{2}+z^{2})^{2}}dxdydz$$
In this case I'll use spherical coordinates because I think is a little more easy reduce the expression, so the integration limits in my point of view are: $0\leq\theta\leq2\pi$ and $0\leq\phi\leq\pi$ for the angles, because is technically all the sphere the integration volume, but the radius will be $1\leq r\leq\infty$, the reason is we are outside of the unitary sphere. Applying it, using the differential changes and solving we have:
$$\int_{1}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{r^{2}\sin\phi}{(r^{2})^{2}}d\theta d\phi dr=\int_{1}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{\sin\phi}{r^{2}}d\theta d\phi dr=4\pi$$
My problem begins when I've put the integral in cartesian coordinates in Wolfram Alpha, the result is 0.329662 and there is not procedure, only that number. Why is that result? Where is my mistake? Is Actually more easy use in this case rectangular or cylindrical form?
Thank u for reading, I hope u have a good day :D
Not a big deal, but I just want to write some of my observations. This is an hell of an integral. If we make a change of variables, $x\rightarrow x+1, y\rightarrow y+1, z\rightarrow z+1$ then the integral becomes $$I=\int_0^\infty\int_0^\infty\int_0^\infty\frac{1}{((x+1)^2+(y+1)^2+(z+1)^2)^2}dxdydz$$ and in spherical coordinates $$I=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_{0}^{\infty}\frac{r^2\sin\phi}{(r^2+2(\cos\theta\sin\phi+\sin\theta\sin\phi+\cos\phi)r+3)^2}dr d\theta d\phi.$$ If we approximate the "denominator" of the integrand by half angles, $\theta\approx\frac{\pi}{4}$, $\phi\approx\frac{\pi}{4}$ then $$I\approx \frac{\pi}{2}\int_{0}^{\infty}\frac{r^2}{(r^2+(2+\sqrt{2})r+3)^2}dr\approx 0.305$$