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How do you integrate the triple integral of $\sqrt(x^2 + y^2 + z^2)$ where $x^2 + y^2 + z^2 \leq 1$, using spherical coordinates. Attempt: Intuitively, I know $0 < \phi < \pi$, $0 < \theta < 2\pi$, $0 < p < 1$. I know $dV = p^2\sin(\phi)dp\,d\theta \,d\phi$. I am stuck here.

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up vote 1 down vote accepted

you have set up $$ \int_0^1\int_0^{2\pi}\int_0^{\pi}\rho^3\sin\phi \ d\phi d\theta d\rho $$ (since you are integrating the function $\rho=\sqrt{x^2+y^2+z^2}$). integrate with respect to $\phi$ noting that $\rho^3$ is constant to get $$ \int_0^1\int_0^{2\pi}2\rho^3d\theta d\rho $$ since $$ \int_0^\pi \sin\phi d\phi=-\cos\phi\Bigg|_0^{\pi}=2 $$ next integrate with respect to $\theta$ nothing that $2\rho^3$ is constant to get $$ \int_0^14\pi\rho^3 d\rho $$ finally integrate with respect to $\rho$ to get $\pi$

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