Using spherical coordinates for triple integral

Question

$ \int_{0}^3 \int_{0}^{\sqrt{9-x^2}} \int_{0}^{\sqrt{9-x^2-y^2}} \frac{\sqrt{x^2+y^2+z^2}}{1+x^2+y^2+z^2} \ dz \ dy \ dx$

Using spherical co-ord's this becomes :

$ \int_{0}^{2\pi} \int_{0}^\pi \int_{0}^3 \frac{r^3\sin(\theta)}{1+r^2} \ dr \ d\theta \ d\phi $

Is this correct? If it is how do I carry on from here?

Answer 1

It looks fine to me, except for the limits of integration. The next step is to write it as$$\frac\pi2\left(\int_0^{\frac\pi2}\sin(\theta)\,\mathrm d\theta\right)\left(\int_0^3\frac{r^3}{1+r^2}\,\mathrm dr\right).$$

Answer 2

Your limits aren't quite correct. The spherical region you are integrating over is only in the first octant. Therefore $\theta$ should range from 0 to $\pi/2$ and $\phi$ should range from $0$ to $\pi/2$.