I'm trying to evaluate the triple integral $\int\int\int_B\frac{dV}{\sqrt{x^2+y^2+z^2+3}}$, where $B$ is the ball of radius $2$ centered at the origin. Both the integrand and the nature of $B$ suggest a change to spherical coordinates. As we know, the volume elements for Cartesian and spherical coordinates are related by $dx dy dz=\rho^2\sin(\varphi)d\rho d\varphi d\theta$, so the original triple integral is equal to the following iterated integral in spherical coordinates:

$\int^{2\pi}_{0}\int^{\pi}_{0}\int^{2}_{0}\frac{\rho^2\sin(\varphi)}{\sqrt{\rho^2+3}}d\rho d\varphi d\theta$

The $\rho$-integral here isn't very pleasant to compute, though I suppose it's possible by means of a trigonometric substitution.

My questions are: Was it the right decision to switch to spherical coordinates in the first place (as opposed to Cartesian or cylindrical coordinates)? And if spherical coordinates are the best choice of coordinates, is there an easier way to do the $\rho$-integral than via the trig substitution $\rho=\sqrt{3}\tan(\alpha)$ for $-\pi/2<\alpha<\pi/2$ (assuming I've set up the iterated integral correctly)?


1 Answer 1


Just use the substitution $\rho = \sqrt{ 3 } \sinh(t)$ which simplifies the integral

$$ \int \frac{\rho^2}{\sqrt{\rho^2 +3}}d\rho = 3\int \sinh^2(t) =\dots.$$

Work out the limits of integration and finish the problem.


$$\sinh t = \frac{e^t-e^{-t}}{2}.$$


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