So, from the sketch it is hopefully clear that the region \(D\) will be in the \(yz\)-plane and so we’ll need to integrate with respect to \(x\) first. That means that the \(x\) limits are,

\[0 \le x \le 2 - y - z\]

For the upper \(x\) limit all we need to do is solve the equation of the plane for \(x\).

Show Step 3

For this problem \(D\) is simply the disk \({y^2} + {z^2} \le 1\). Because \(D\) is in the \(yz\)-plane and is a disk we’ll need to use the following “modified” version of cylindrical coordinates.

\[\begin{align*}x & = x\\ y & = r\sin \theta \\ z & = r\cos \theta \end{align*}\]

This also matches up with the fact that we need to integrate \(x\) first (as we determined in Step 2) and the first variable of integration with cylindrical coordinates is always the “free” variable (i.e. not the one involving the trig functions).

So, we can easily describe the disk in terms of \(r\) and \(\theta \) so here are the cylindrical coordinates for this problem.

\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 1\\ 0 \le x \le 2 - r\sin \theta - r\cos \theta \end{array}\]

Don’t forget to convert the \(y\) and \(z\) in the \(x\) upper limit into cylindrical coordinate as well.

Show Step 4

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

\[\begin{align*}\iiint\limits_{E}{{z\,dV}}& = \int_{0}^{{2\pi }}{{\int_{0}^{1}{{\int_{0}^{{2 - r\sin \theta - r\cos \theta }}{{\left( {r\cos \theta } \right)r\,dx}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{1}{{\int_{0}^{{2 - r\sin \theta - r\cos \theta }}{{{r^2}\cos \theta \,dx}}\,dr}}\,d\theta }}\end{align*}\]

Don’t forget to convert the integrand to our modified cylindrical coordinates and also don’t forget that \(dV = r\,dx\,dr\,d\theta \) and so we pick up another \(r\) when converting the \(dV\) to cylindrical coordinates.

Show Step 5

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

\[\begin{align*}\iiint\limits_{E}{{z\,dV}} & = \int_{0}^{{2\pi }}{{\int_{0}^{1}{{\left. {\left( {{r^2}\cos \theta \,x} \right)} \right|_0^{2 - r\sin \theta - r\cos \theta }\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{1}{{{r^2}\cos \theta \,\left( {2 - r\sin \theta - r\cos \theta } \right)\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{1}{{2{r^2}\cos \theta - {r^3}\cos \theta \sin \theta - {r^3}{{\cos }^2}\theta \,dr}}\,d\theta }}\end{align*}\] Show Step 6

Next let’s do the \(r\) integration.

\[\begin{align*}\iiint\limits_{E}{{z\,dV}} & = \int_{0}^{{2\pi }}{{\left. {\left( {\frac{2}{3}{r^3}\cos \theta - \frac{1}{4}{r^4}\cos \theta \sin \theta - \frac{1}{4}{r^4}{{\cos }^2}\theta } \right)} \right|_0^1\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{2}{3}\cos \theta - \frac{1}{4}\cos \theta \sin \theta - \frac{1}{4}{{\cos }^2}\theta \,d\theta }}\end{align*}\] Show Step 7

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

\[\begin{align*}\iiint\limits_{E}{{z\,dV}} & = \int_{0}^{{2\pi }}{{\frac{2}{3}\cos \theta - \frac{1}{8}\sin \left( {2\theta } \right) - \frac{1}{8}\left( {1 + \cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \left. {\left( {\frac{2}{3}\sin \theta + \frac{1}{{16}}\cos \left( {2\theta } \right) - \frac{1}{8}\left( {\theta + \frac{1}{2}\sin \left( {2\theta } \right)} \right)} \right)} \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{\pi }{4}}}\end{align*}\]

Don’t forget to simplify the integrand before doing the final integration. In this case we used the sine double angle on the second term and the cosine half angle formula on the third term to simplify the integrand to allow us to quickly do this integration.