Calculus III - Triple Integrals

Section 15.5 : Triple Integrals

8. Use a triple integral to determine the volume of 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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Okay, let’s start off with a quick sketch of the region we want the volume of so we can get a feel for what we’re dealing with. We’ll call this region $E$.

$A three dimensional sketch of the region $E$ drawn with the traditional axes. The orange surface across the top is the graph of $z = 4 - xy$, a gently sloping sheet that falls off a little as $x$ and $y$ increase. The blue and red surfaces are the two sides of the solid that are visible from this angle and the bottom of the region is the rectangle $0 \le x \le 2$, $0 \le y \le 1$ in the $xy$-plane, so the solid looks like a low rectangular block with a slightly tilted top.$

$A three dimensional sketch of the same region $E$, this time drawn with a set of “box” axes to make the solid easier to visualize. The tube is shaded red with the sloped yellow face on the front, and it runs out to about $x = 5$ with $z$ reaching up to roughly 2 near the $yz$-plane end.$

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.

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The volume of this solid is given by,

\[V = \iiint\limits_{E}{{dV}}\]

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So, we now need to get the limits set up for the integral. From the sketch above we know that we’ll have the following limits for $z$.

\[0 \le z \le 4 - xy\]

We’ll also need limits for $D$. 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 $x$.

Now, plugging all these limits into the integral the volume is,

\[V = \iiint\limits_{E}{{dV}} = \int_{0}^{2}{{\int_{0}^{1}{{\int_{0}^{{4 - xy}}{{dz}}\,dy}}\,dx}}\]

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Okay, now all we need to do is evaluate the integral. Here is the $z$ integration.

\[\begin{align*}V & = \int_{0}^{2}{{\int_{0}^{1}{{\left. z \right|_0^{4 - xy}\,dy}}\,dx}}\\ & = \int_{0}^{2}{{\int_{0}^{1}{{4 - xy\,dy}}\,dx}}\end{align*}\]

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Now let’s do the $y$ integration.

\[\begin{align*}V & = \int_{0}^{2}{{\left. {\left( {4y - \frac{1}{2}x{y^2}} \right)} \right|_0^1\,dx}}\\ & = \int_{0}^{2}{{4 - \frac{1}{2}x\,dx}}\end{align*}\]

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Finally, let’s do the $x$ integration to get the volume of the region.

\[V = \iiint\limits_{E}{{dV}} = \left. {\left( {4x - \frac{1}{4}{x^2}} \right)} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{7}\]