Evaluate the integral using spherical coordinates

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Given the integral $\int^{1}_{0}\int^{\sqrt{1-x^{2}}}_{0}\int^{\sqrt{1-x^{2}-y^{2}}}_{0} \dfrac{1}{x^{2}+y^{2}+z^{2}}dzdxdy$

I need to evaluate this using spherical coordinates.
So far I have that $0\leq r \leq 1$ and I understand that $\theta$ is the angle made in the xy plane and has to be less than or equal to $2\pi$ and $\varphi$ is the angle made revolving around the z-axis and is less than or equal to $\pi$ however I am not sure on how to workout the limits of $\theta$ and $\varphi$ for this question.

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In spherical coordinates the function you are integrating over is $$\frac{1}{x^2+y^2+z^2}=\frac{1}{r^2},$$ and this function is constant over shells of radius $r$. Since the region of integration is the ball of radius $1$ in the first octant, and since the surface area of a shell of radius $r$ is $4\pi r^2$, your integral equals $$\frac{1}{8}\int_0^1\frac{1}{r^2}\cdot 4\pi r^2dr=\frac{\pi}{2}.$$

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I will work on this problem when radius $r$ is constant in general.

Step 1. Spherical coordinates means:

$x = r \sin(\varphi) \cdot \cos(\theta)$

$y = r \sin(\varphi) \cdot \sin(\theta)$

$z = r \cos(\varphi)$

(Convince your self by drawing pictures)

Step 2. Change of coordinates needs Jacobian

$|J| = r^2 \cdot \sin(\varphi)$ in this case.

So, $dxdydz = r^2 \cdot \sin(\varphi) drd \varphi d\theta$

Step 3. Calculate

$\int_0^r \int_0^\sqrt{r^2-x^2} \int_0^\sqrt{r^2-x^2-y^2} \frac{1}{x^2 + y^2 + z^2} dzdydx$

= $\int_0^{\pi/2} \int_0^{\pi/2} \int_0^r \frac{1}{r^2} r^2 \sin(\varphi) drd \varphi d\theta$

= $\int_0^{\pi/2} \int_0^{\pi/2} \int_0^r \sin(\varphi) drd \varphi d\theta$

=$r \cdot \pi/2$.

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Jacobian is

$$ = \cos \phi\, r^2 \,dr \, d \phi \, d \theta $$ for volume element conversion to spherical coordinates from rectangular.

The integrand

$$= \dfrac{1}{r^2}$$ which simplifies integral to

$$\int^{\pi/2}_{0} \, \int ^{\pi/2}_{0} \int^1_{0} dr \, \cos \phi \, d\phi \, d \theta $$

$$ = \pi/2, $$

as the area under half sine wave is unity.