Now, since we are integrating over a portion of a sphere it makes sense to use spherical coordinate for the integral and the limits are,

\[\begin{array}{c}0 \le \varphi \le \frac{\pi }{2}\\ 0 \le \theta \le 2\pi \\ 0 \le \rho \le 4\end{array}\]

Remember that \(\varphi \) is the angle from the positive \(z\)-axis that we rotate through as we cover the region and \(\theta \) is the angle we rotate around the \(z\)-axis as we cover the region.

In this case we have the full upper half of the sphere and so \(\theta \) will range from 0 to \(2\pi \) while \(\varphi \) will range from 0 to \(\frac{\pi }{2}\).

Show Step 3

Plugging these limits into the integral and converting to spherical coordinates gives,

\[\begin{align*}\iiint\limits_{E}{{10xz + 3\,dV}}& = \int_{0}^{{\frac{\pi }{2}}}{{\int_{0}^{{2\pi }}{{\int_{0}^{4}{{\left[ {10\left( {\rho \sin \varphi \cos \theta } \right)\left( {\rho \cos \varphi } \right) + 3} \right]\left( {{\rho ^2}\sin \varphi } \right)\,d\rho }}\,d\theta }}\,d\varphi }}\\ & = \int_{0}^{{\frac{\pi }{2}}}{{\int_{0}^{{2\pi }}{{\int_{0}^{4}{{10{\rho ^4}{{\sin }^2}\varphi \cos \varphi \cos \theta + 3{\rho ^2}\sin \varphi \,d\rho }}\,d\theta }}\,d\varphi }}\end{align*}\]

Don’t forget to convert the \(x\) and \(z\) into spherical coordinates and also don’t forget that \(dV = {\rho ^2}\sin \varphi \,d\rho \,d\theta \,d\varphi \) and so we’ll pick up a couple of extra terms when converting the \(dV\) to spherical coordinates.

Show Step 4

Okay, now all we need to do is evaluate the integral. Here is the \(\rho \) integration.

\[\begin{align*}\iiint\limits_{E}{{10xz + 3\,dV}} & = \int_{0}^{{\frac{\pi }{2}}}{{\int_{0}^{{2\pi }}{{\left. {\left( {2{\rho ^5}{{\sin }^2}\varphi \cos \varphi \cos \theta + {\rho ^3}\sin \varphi } \right)} \right|_0^4\,d\theta }}\,d\varphi }}\\ & = \int_{0}^{{\frac{\pi }{2}}}{{\int_{0}^{{2\pi }}{{2048{{\sin }^2}\varphi \cos \varphi \cos \theta + 64\sin \varphi \,\,d\theta }}\,d\varphi }}\end{align*}\] Show Step 5

Next let’s do the \(\theta \) integration.

\[\begin{align*}\iiint\limits_{E}{{10xz + 3\,dV}} & = \int_{0}^{{\frac{\pi }{2}}}{{\left. {\left( {2048{{\sin }^2}\varphi \cos \varphi \sin \theta + 64\theta \sin \varphi } \right)} \right|_0^{2\pi }\,d\varphi }}\\ & = \int_{0}^{{\frac{\pi }{2}}}{{128\pi \sin \varphi \,d\varphi }}\end{align*}\] Show Step 6

Finally, we’ll do the \(\varphi \) integration.

\[\iiint\limits_{E}{{10xz + 3\,dV}} = \left. {\left( { - 128\pi \cos \varphi } \right)} \right|_0^{\frac{\pi }{2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{128\pi }}\]

Note that, in this case, because the limits of each of the integrals were all constants we could have done the integration in any order we wanted to. In this case, it might have been “simpler” to do the \(\varphi \) first or second as that would have greatly reduced the integrand for the remaining integral(s).