Section 15.5 : Triple Integrals
4. Evaluate \(\displaystyle \iiint\limits_{E}{{3 - 4x\,dV}}\) where \(E\) is the region below \(z = 4 - xy\) and above the region in the \(xy\)-plane defined by \(0 \le x \le 2\), \(0 \le y \le 1\).
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Start SolutionOkay, let’s start off with a quick sketch of the region \(E\) so we can get a feel for what we’re dealing with.
We’ve given the sketches with a set of “traditional” axes as well as a set of “box” axes to help visualize the surface and region.
The top portion of the region (the orange colored surface) is the graph of \(z = 4 - xy\). The two sides shown (the blue and red surfaces) show the two sides of the region that we can see given the orientation of the region. The bottom of the region is the \(xy\)-plane.
Show Step 2So, from the sketch above we know that we’ll have the following limits for \(z\).
\[0 \le z \le 4 - xy\]With these limits we can also get the triple integral at least partially set up as follows.
\[\iiint\limits_{E}{{3 - 4x\,dV}} = \iint\limits_{D}{{\left[ {\int_{0}^{{4 - xy}}{{3 - 4x\,dz}}} \right]\,dA}}\] Show Step 3Next, we’ll need limits for \(D\) so we can finish setting up the integral. In this case that is really simple as we can see from the problem statement that \(D\) is just a rectangle in the \(xy\)-plane and in fact the limits are given in the problem statement as,
\[\begin{array}{c}0 \le x \le 2\\ 0 \le y \le 1\end{array}\]There really isn’t any advantage to doing one order vs. the other so, in this case, we’ll integrate \(y\) and then \(x\).
With these limits plugged into the integral we now have,
\[\iiint\limits_{E}{{3 - 4x\,dV}} = \int_{0}^{2}{{\int_{0}^{1}{{\int_{0}^{{4 - xy}}{{3 - 4x\,dz}}\,dy}}\,dx}}\] Show Step 4Okay, now all we need to do is evaluate the integral. Here is the \(z\) integration.
\[\begin{align*}\iiint\limits_{E}{{3 - 4x\,dV}} & = \int_{0}^{2}{{\int_{0}^{1}{{\left. {\left( {3 - 4x} \right)z} \right|_0^{4 - xy}\,dy}}\,dx}}\\ & = \int_{0}^{2}{{\int_{0}^{1}{{\left( {3 - 4x} \right)\left( {4 - xy} \right)\,dy}}\,dx}}\\ & = \int_{0}^{2}{{\int_{0}^{1}{{4{x^2}y - 3xy - 16x + 12\,dy}}\,dx}}\end{align*}\]Note that because the integrand had no \(z\)’s in it we treated the whole integrand as a constant and just added a single \(z\) as shown above. We could just as easily integrated each term individually but it seemed easier to just deal with the integrand as a single term.
Show Step 5Now let’s do the \(y\) integration.
\[\begin{align*}\iiint\limits_{E}{{3 - 4x\,dV}} & = \int_{0}^{2}{{\left. {\left( {2{x^2}{y^2} - \frac{3}{2}x{y^2} - 16xy + 12y} \right)} \right|_0^1\,dx}}\\ & = \int_{0}^{2}{{12 - \frac{{35}}{2}x + 2{x^2}\,dx}}\end{align*}\] Show Step 6Finally, let’s do the \(x\) integration.
\[\iiint\limits_{E}{{3 - 4x\,dV}} = \left. {\left( {12x - \frac{{35}}{4}{x^2} + \frac{2}{3}{x^3}} \right)} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{17}}{3}}}\]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.