Use spherical coordinates to evaluate the triple integral:
$$\iiint_{E}{(x^2+y^2+z^2)dV}$$
Where E is the ball: $E=\left\{(x,y,z)\in \mathbb{R}^3 \mid x^2+y^2+z^2≤49\right\}$.
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$\begingroup$Well, although many would argue that the best is just to give a hint, I'll detail this a little bit. Sometimes the person needs a first example, and since this one is kinda trivial, it's a good example to get started. Look what I'll do and try some other exercises on your own.
Note that in spherical coordinates $(\rho, \theta, \phi)$ you have $x^2+y^2+z^2=\rho^2$. In this case, your ball becomes $E = \left\{(\rho, \theta, \phi)\in\mathbb{R}^3\mid\rho\leq7, \ \ 0\leq\theta\leq2\pi, \ \ 0\leq\phi\leq\pi\right\}$. Also, the volume element is given by $dV = \rho^2\sin \phi d\rho d\theta d\phi$.
In this case you get:
$$\iiint_E(x^2+y^2+z^2)dV=\int_0^\pi\int_0^{2\pi}\int_0^7\rho^2(\rho^2\sin\phi)d\rho d\theta d\phi$$
Now, you know that you can write this as:
$$\int_0^\pi\int_0^{2\pi}\int_0^7\rho^2(\rho^2\sin\phi)d\rho d\theta d\phi=\int_0^\pi \sin \phi d\phi \int_0^{2\pi}d\theta \int_0^7 \rho^4d\rho$$
Now it's simple. Those integrals are pretty straightforward and I'll let them to you.
I hope you've got the idea.
$\endgroup$ $\begingroup$Hint:
$$\iiint_E (x^2+y^2+z^2) \,dV = \int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi} \int_{r=0}^{7} r^2 \cdot r^2\sin\theta \,dr\,d\theta\,d\phi$$
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