# Evaluate the Triple integral Using spherical coordinates

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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-1 You haven't said what you can/can't do, nor shown your attempt. –  Daryl Mar 24 '13 at 22:33
@Daryl Though that isn't compulsory, it is customary. Maybe, since the user is new, you can first point out what we prefer the OP shows or does when asking questions rather than downvoting. There are some nice templates to welcome users to the site, also. Behold: –  Pedro Tamaroff Mar 24 '13 at 22:49

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.

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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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