The region in the \(xz\)-plane bounded by \(z = 2x\), \(z = 5\) and \(x = 0\)that is referenced in the problem statement is the region \(D\) and so we know that, for this problem, we’ll need to be integrating \(y\) first (since \(D\) is in the \(xz\)-plane).

Therefore, we have the following limits for \(y\).

\[0 \le y \le 10 - 2z\]

With these limits we can also get the triple integral at least partially set up as follows.

\[\iiint\limits_{E}{{12y - 8x\,dV}} = \iint\limits_{D}{{\left[ {\int_{0}^{{10 - 2z}}{{12y - 8x\,dy}}} \right]\,dA}}\] Show Step 3

Next, we’ll need limits for \(D\) so we can finish setting up the integral. For this problem \(D\) is the region in the \(xz\)-plane from the problem statement as we noted above.

Here is a quick sketch of \(D\).

Another way to think this region is that it is the “back” of the solid sketched in Step 1 and the solid we sketched in Step 1 is all behind this region. In other words the positive y‑axis is goes directly back into the page while the negative \(y\)-axis comes directly out of the page.

As noted in the sketch of \(D\) we can easily defined by either of the following sets of limits.

\[\begin{array}{*{20}{c}}\begin{aligned}& \,\,\,0 \le x \le \frac{5}{2}\\ & 2x \le z \le 5\end{aligned}&{\,\,\,\,\,\,\,\,\,\,\,{\mbox{OR}}\,\,\,\,\,\,\,\,\,\,}&\begin{aligned} & \,\, 0 \le z \le 5\\ & 0 \le x \le \frac{1}{2}z\end{aligned}\end{array}\]

The integrand doesn’t really suggest one of these would be easier than the other so we’ll use the 2^nd set of limits for no other reason than the lower limits for both are zero which might make things a little nicer in the integration process.

With these limits plugged into the integral we now have,

\[\iiint\limits_{E}{{12y - 8x\,dV}} = \int_{0}^{5}{{\int_{0}^{{\frac{1}{2}z}}{{\int_{0}^{{10 - 2z}}{{12y - 8x\,dy}}\,dx}}\,dz}}\] Show Step 4

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

\[\begin{align*}\iiint\limits_{E}{{12y - 8x\,dV}} & = \int_{0}^{5}{{\int_{0}^{{\frac{1}{2}z}}{{\left. {\left( {6{y^2} - 8xy} \right)} \right|_0^{10 - 2z}\,dx}}\,dz}}\\ & = \int_{0}^{5}{{\int_{0}^{{\frac{1}{2}z}}{{6{{\left( {10 - 2z} \right)}^2} - 8x\left( {10 - 2z} \right)\,dx}}\,dz}}\end{align*}\]

Note that we did no simplification here because it is not yet clear that we need to do any simplification. The next integration is with respect to \(x\) and the first term is a constant as far as that integration is concerned and the second term is multiplied by an \(x\) and so it will might well be easier to leave it in that form for the \(x\) integration.

Show Step 5

Now let’s do the \(x\) integration.

\[\begin{align*}\iiint\limits_{E}{{12y - 8x\,dV}} & = \int_{0}^{5}{{\left. {\left[ {6{{\left( {10 - 2z} \right)}^2}x - 4{x^2}\left( {10 - 2z} \right)} \right]} \right|_0^{\frac{1}{2}z}\,dz}}\\ & = \int_{0}^{5}{{3z{{\left( {10 - 2z} \right)}^2} - {z^2}\left( {10 - 2z} \right)\,dz}}\\ & = \int_{0}^{5}{{14{z^3} - 130{z^2} + 300z\,dz}}\end{align*}\] Show Step 6

Finally, let’s do the \(z\) integration.

\[\iiint\limits_{E}{{12y - 8x\,dV}} = \left. {\left( {\frac{7}{2}{z^4} - \frac{{130}}{3}{z^3} + 150{z^2}} \right)} \right|_0^5 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{3125}}{6}}}\]

So, once we got the limits all set up, the integration for this problem wasn’t too bad. That will often be the case with these problems. Getting the limits for the integrals set up will often, but not always, be the hardest part of the problem. Once they get set up the integration is often pretty simple.