Integrating $ \frac{1}{\sqrt {x^{2}+y^{2}}}-\frac{1}{\sqrt {y^{2}+z^{2}}}\ $

Question

I am trying to solve this integral over a Cube:

$$\frac{1}{\sqrt {x^{2}+y^{2}}}-\frac{1}{\sqrt {y^{2}+z^{2}}}\ $$

I can see that this that this will turn into zero because of symmetry since I am integrating the difference of a "radius" over the whole cube.

But is there a concrete way to solve this integral ?
I tried to simplify or using cylinder coordinate, but I couldn't get an easy or straightforward way to do it.
Any hint would be more than appreciated. Thank you.

Answer 1

You're looking at: $$ I \equiv \int\limits_0^1dx\int\limits_0^1dy\int\limits_0^1dz\left( \frac{1}{\sqrt{x^2+y^2}} - \frac{1}{\sqrt{y^2+z^2}} \right) $$

Split apart the integral: $$ I = \int\limits_0^1dx\int\limits_0^1dy\int\limits_0^1dz \frac{1}{\sqrt{x^2+y^2}} - \int\limits_0^1dx\int\limits_0^1dy\int\limits_0^1dz \frac{1}{\sqrt{y^2+z^2}} $$

In the left integral, you can integrate over $z$ to get a factor of $1$. In the right integral you can integrate over $x$ to get a factor of $1$. $$ I = (1) \int\limits_0^1dx\int\limits_0^1dy \frac{1}{\sqrt{x^2+y^2}} - (1) \int\limits_0^1dy\int\limits_0^1dz \frac{1}{\sqrt{y^2+z^2}} $$

You've got the same integral added and subtracted now. To see this more clearly, you can re-label the variables in each like: $$ I = \int\limits_0^1dA\int\limits_0^1dB \frac{1}{\sqrt{A^2+B^2}} - \int\limits_0^1dA\int\limits_0^1dB \frac{1}{\sqrt{A^2+B^2}} $$

This is of course $I=0$.

Answer 2

how about: $$I_1=\int_0^1\int_0^1\int_0^1\frac{1}{\sqrt{x^2+y^2}}dzdydx=\int_0^1\int_0^1\frac{1}{\sqrt{x^2+y^2}}dxdy$$ then we can look at the second integral: $$I_2=\int_0^1\int_0^1\int_0^1\frac{1}{\sqrt{y^2+z^2}}dzdydx=\int_0^1\int_0^1\frac{1}{\sqrt{y^2+z^2}}dydz$$ and these two integrals are clearly equal so: $$I_1-I_2=0 \therefore\,I=0$$

Answer 3

Doing things the hard way by calculating $I=\int_0^1\int_0^1\frac{dy\ dx}{\sqrt{x^2+y^2}}$ is feasible with polar coordinates. The integral transforms to $$I=2\int_0^{\frac{\pi}4}\int_0^{\sec \theta}1\ dr\ d\theta =2 \int_0^{\frac{\pi}4}{\sec \theta}\;d\theta=2\log(\sec \theta + \tan \theta)\Biggr|_0^\frac{\pi}4=2\log(1+\sqrt 2)$$