calculus - A triple definite integral from Cartesian coordinates to Spherical coordinates. Help!

I am trying to solve evaluate this triple definite integral. The integral is given below.

$$

\Lambda =\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)^2} \, dz\,dy\,dx
$$

I know that this problem has to be done in spherical coordinates, since we have a triple integral, and we have the following relationship :

$$
x^2+y^2+z^2
$$

and I know that this equals to:

$$
\rho=\sqrt{x^2+y^2+z^2}
$$

I know that for spherical coordinates my x,y,z variables have the following form:

$$
x = \rho\cos(\theta)\sin(\phi)
\\

y = \rho\sin(\theta)\sin(\phi)
\\
z = \rho\cos(\phi)
$$

Once I make the substitutions my integral expression gets really complicated, and I have no idea how to proceed from there. Could anybody post a step by step solution to this triple definite integral?And also showing how the change of variables are done within the coordinates and the integral. I do believe that using spherical coordinates for this problem is the best way to solve evaluate this triple integral. Thank You!

Answer

we need to deal the sphere in the first octant. so $\theta$ varies from $o$ to $\pi/2$, $\rho$ from $0$ to $3$, $\phi$ from $0$ to $\pi/2$.
$$
\Lambda =\int_0^3\int_0^{\sqrt{9-x^2}}\int_0^{\sqrt{9-x^2-y^2}} \frac{\sqrt{x^2+y^2+z^2}}{1+(x^2+y^2+z^2)^2} \, dz \, dy \, dx$$

$$=\int_0^{\pi/2}d\theta\int_{0}^{\phi=\pi/2}\sin(\phi) \, d\phi \int_0^{\rho=3} \frac{{\rho^3}}{1+(\rho)^4} \, d\rho \, d\phi
$$
Here you can integrate.