# Section13.6Volume Between Surfaces and Triple Integration¶ permalink

We learned in Section 13.2 how to compute the signed volume $V$ under a surface $z=f(x,y)$ over a region $R\text{:}$ $V = \iint_R f(x,y)\ dA\text{.}$ It follows naturally that if $f(x,y)\geq g(x,y)$ on $R\text{,}$ then the volume between $f(x,y)$ and $g(x,y)$ on $R$ is \begin{equation*} V = \iint_R f(x,y)\ dA - \iint_R g(x,y)\ dA = \iint_R \big(f(x,y)-g(x,y)\big)\ dA. \end{equation*}

##### Example13.6.2Finding volume between surfaces

Find the volume of the space region bounded by the planes $z=3x+y-4$ and $z=8-3x-2y$ in the $1^\text{ st }$ octant. In Figure 13.6.3(a) the planes are drawn; in (b), only the defined region is given.

Solution

In the preceding example, we found the volume by evaluating the integral \begin{equation*} \ds \int_0^2\int_0^{4-2x} \big(8-3x-2y-(3x+y-4)\big)\ dy\ dx. \end{equation*}

Note how we can rewrite the integrand as an integral, much as we did in Section 13.1: \begin{equation*} 8-3x-2y-(3x+y-4) = \int_{3x+y-4}^{8-3x-2y}\ dz. \end{equation*}

Thus we can rewrite the double integral that finds volume as \begin{equation*} \int_0^2\int_0^{4-2x} \big(8-3x-2y-(3x+y-4)\big)\ dy\ dx = \int_0^2\int_0^{4-2x}\left(\int_{3x+y-4}^{8-3x-2y}\ dz\right)\ dy\ dx. \end{equation*}

This no longer looks like a “double integral,” but more like a “triple integral.” Just as our first introduction to double integrals was in the context of finding the area of a plane region, our introduction into triple integrals will be in the context of finding the volume of a space region.

To formally find the volume of a closed, bounded region $D$ in space, such as the one shown in Figure 13.6.6(a), we start with an approximation. Break $D$ into $n$ rectangular solids; the solids near the boundary of $D$ may possibly not include portions of $D$ and/or include extra space. In Figure 13.6.6(b), we zoom in on a portion of the boundary of $D$ to show a rectangular solid that contains space not in $D\text{;}$ as this is an approximation of the volume, this is acceptable and this error will be reduced as we shrink the size of our solids.

The volume $\Delta V_i$ of the $i^\text{ \,th }$ solid $D_i$ is $\Delta V_i = \dx_i\dy_i\ddz_i\text{,}$ where $\dx_i\text{,}$ $\dy_i$ and $\ddz_i$ give the dimensions of the rectangular solid in the $x\text{,}$ $y$ and $z$ directions, respectively. By summing up the volumes of all $n$ solids, we get an approximation of the volume $V$ of $D\text{:}$ \begin{equation*} V \approx \sum_{i=1}^n \Delta V_i = \sum_{i=1}^n \dx_i\dy_i\ddz_i. \end{equation*}

Let $\norm{\Delta D}$ represent the length of the longest diagonal of rectangular solids in the subdivision of $D\text{.}$ As $\norm{\Delta D}\to 0\text{,}$ the volume of each solid goes to 0, as do each of $\dx_i\text{,}$ $\dy_i$ and $\ddz_i\text{,}$ for all $i\text{.}$ Our calculus experience tells us that taking a limit as $\norm{\Delta D}\to 0$ turns our approximation of $V$ into an exact calculation of $V\text{.}$ Before we state this result in a theorem, we use a definition to define some terms.

##### Definition13.6.9Triple Integrals, Iterated Integration (Part I)

Let $D$ be a closed, bounded region in space. Let $a$ and $b$ be real numbers, let $g_1(x)$ and $g_2(x)$ be continuous functions of $x\text{,}$ and let $f_1(x,y)$ and $f_2(x,y)$ be continuous functions of $x$ and $y\text{.}$

1. The volume $V$ of $D$ is denoted by a triple integral, \begin{equation*} V = \iiint_D dV. \end{equation*}

2. The iterated integral $\ds \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} \ dz\ dy\ dx$ is evaluated as \begin{equation*} \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} \ dz\ dy\ dx=\int_a^b\int_{g_1(x)}^{g_2(x)}\left(\int_{f_1(x,y)}^{f_2(x,y)} \ dz\right)\ dy\ dx. \end{equation*} Evaluating the above iterated integral is triple integration.

Our informal understanding of the notation $\iiint_D\ dV$ is “sum up lots of little volumes over $D\text{,}$” analogous to our understanding of $\iint_R\ dA$ and $\iint_R\ dm\text{.}$

We now state the major theorem of this section.

We evaluated the area of a plane region $R$ by iterated integration, where the bounds were “from curve to curve, then from point to point.” Theorem 13.6.10 allows us to find the volume of a space region with an iterated integral with bounds “from surface to surface, then from curve to curve, then from point to point.” In the iterated integral \begin{equation*} \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} \ dz\ dy\ dx, \end{equation*} the bounds $a\leq x\leq b$ and $g_1(x)\leq y\leq g_2(x)$ define a region $R$ in the $x$-$y$ plane over which the region $D$ exists in space. However, these bounds are also defining surfaces in space; $x=a$ is a plane and $y=g_1(x)$ is a cylinder. The combination of these 6 surfaces enclose, and define, $D\text{.}$

Examples will help us understand triple integration, including integrating with various orders of integration.

##### Example13.6.11Finding the volume of a space region with triple integration

Find the volume of the space region in the $1^{\,st}$ octant bounded by the plane $z=2-y/3-2x/3\text{,}$ shown in Figure 13.6.12(a), using the order of integration $dz\ dy\ dx\text{.}$ Set up the triple integrals that give the volume in the other 5 orders of integration.

Solution

In the previous example, we collapsed the surface into the $x$-$y\text{,}$ $x$-$z\text{,}$ and $y$-$z$ planes as we determined the “curve to curve, point to point” bounds of integration. Since the surface was a triangular portion of a plane, this collapsing, or projecting, was simple: the projection of a straight line in space onto a coordinate plane is a line.

The following example shows us how to do this when dealing with more complicated surfaces and curves.

##### Example13.6.18Finding the projection of a curve in space onto the coordinate planes

Consider the surfaces $z=3-x^2-y^2$ and $z=2y\text{,}$ as shown in Figure 13.6.19(a). The curve of their intersection is shown, along with the projection of this curve into the coordinate planes, shown dashed. Find the equations of the projections into the coordinate planes.

Solution
##### Example13.6.22Finding the volume of a space region with triple integration

Set up the triple integrals that find the volume of the space region $D$ bounded by the surfaces $x^2+y^2=1\text{,}$ $z=0$ and $z=-y\text{,}$ as shown in Figure 13.6.23(a), with the orders of integration $dz\ dy\ dx\text{,}$ $dy\ dx\ dz$ and $dx\ dz\ dy\text{.}$

Solution

The following theorem states two things that should make “common sense” to us. First, using the triple integral to find volume of a region $D$ should always return a positive number; we are computing volume here, not signed volume. Secondly, to compute the volume of a “complicated” region, we could break it up into subregions and compute the volumes of each subregion separately, summing them later to find the total volume.

We use this latter property in the next example.

##### Example13.6.30Finding the volume of a space region with triple integration

Find the volume of the space region $D$ bounded by the coordinate planes, $z=1-x/2$ and $z=1-y/4\text{,}$ as shown in Figure 13.6.31(a). Set up the triple integrals that find the volume of $D$ in all 6 orders of integration.

Solution

We give one more example of finding the volume of a space region.

##### Example13.6.37Finding the volume of a space region

Set up a triple integral that gives the volume of the space region $D$ bounded by $z= 2x^2+2$ and $z=6-2x^2-y^2\text{.}$ These surfaces are plotted in Figure 13.6.38(a) and (b), respectively; the region $D$ is shown in part (c) of the figure.

Solution

If all one wanted to do in Example 13.6.37 was find the volume of the region $D\text{,}$ one would have likely stopped at the first integration setup (with order $dz\ dy\ dx$) and computed the volume from there. However, we included the other two methods 1) to show that it could be done, “messy” or not, and 2) because sometimes we “have” to use a less desirable order of integration in order to actually integrate.

Triple Integration and Functions of Three Variables

There are uses for triple integration beyond merely finding volume, just as there are uses for integration beyond “area under the curve.” These uses start with understanding how to integrate functions of three variables, which is effectively no different than integrating functions of two variables. This leads us to a definition, followed by an example.

##### Definition13.6.45Iterated Integration, (Part II)

Let $D$ be a closed, bounded region in space, over which $g_1(x)\text{,}$ $g_2(x)\text{,}$ $f_1(x,y)\text{,}$ $f_2(x,y)$ and $h(x,y,z)$ are all continuous, and let $a$ and $b$ be real numbers.

The iterated integral $\ds \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} h(x,y,z)\ dz\ dy\ dx$ is evaluated as \begin{equation*} \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{f_1(x,y)}^{f_2(x,y)} h(x,y,z)\ dz\ dy\ dx = \int_a^b\int_{g_1(x)}^{g_2(x)}\left(\int_{f_1(x,y)}^{f_2(x,y)} h(x,y,z)\ dz\right) dy\ dx. \end{equation*}

##### Example13.6.46Evaluating a triple integral of a function of three variables

Evaluate $\ds \int_0^1\int_{x^2}^x\int_{x^2-y}^{2x+3y} \big(xy+2xz\big)\ dz\ dy\ dx\text{.}$

Solution

We now know how to evaluate a triple integral of a function of three variables; we do not yet understand what it means. We build up this understanding in a way very similar to how we have understood integration and double integration.

Let $h(x,y,z)$ a continuous function of three variables, defined over some space region $D\text{.}$ We can partition $D$ into $n$ rectangular–solid subregions, each with dimensions $\dx_i\times\dy_i\times\ddz_i\text{.}$ Let $(x_i,y_i,z_i)$ be some point in the $i^{\,\text{ th } }$ subregion, and consider the product $h(x_i,y_i,z_i)\dx_i\dy_i\ddz_i\text{.}$ It is the product of a function value (that's the $h(x_i,y_i,z_i)$ part) and a small volume $\Delta V_i$ (that's the $\dx_i\dy_i\ddz_i$ part). One of the simplest understanding of this type of product is when $h$ describes the density of an object, for then $h\times\text{ volume } =\text{ mass }\text{.}$

We can sum up all $n$ products over $D\text{.}$ Again letting $\norm{\Delta D}$ represent the length of the longest diagonal of the $n$ rectangular solids in the partition, we can take the limit of the sums of products as $\norm{\Delta D}\to 0\text{.}$ That is, we can find \begin{equation*} S = \lim_{\norm{\Delta D}\to 0} \sum_{i=1}^n h(x_i,y_i,z_i)\Delta V_i=\lim_{\norm{\Delta D}\to 0} \sum_{i=1}^n h(x_i,y_i,z_i)\dx_i\dy_i\ddz_i. \end{equation*}

While this limit has lots of interpretations depending on the function $h\text{,}$ in the case where $h$ describes density, $S$ is the total mass of the object described by the region $D\text{.}$

We now use the above limit to define the triple integral, give a theorem that relates triple integrals to iterated iteration, followed by the application of triple integrals to find the centers of mass of solid objects.

##### Definition13.6.47Triple Integral

Let $w=h(x,y,z)$ be a continuous function over a closed, bounded space region $D\text{,}$ and let $\Delta D$ be any partition of $D$ into $n$ rectangular solids with volume $\Delta V_i\text{.}$ The triple integral of $h$ over $D$ is \begin{equation*} \iiint_Dh(x,y,z)\ dV = \lim_{\norm{\Delta D}\to 0}\sum_{i=1}^n h(x_i,y_i,z_i)\Delta V_i. \end{equation*}

In the marginal note on <<Unresolved xref, reference "note_doubleint"; check spelling or use "provisional" attribute>>, we showed how the summation of rectangles over a region $R$ in the plane could be viewed as a double sum, leading to the double integral. Likewise, we can view the sum $\ds \sum_{i=1}^nh(x_i,y_i,z_i)\dx_i\dy_i\ddz_i$ as a triple sum, \begin{equation*} \sum_{k=1}^p\sum_{j=1}^n\sum_{i=1}^mh(x_i,y_j,z_k)\dx_i\dy_j\ddz_k, \end{equation*} which we evaluate as \begin{equation*} \sum_{k=1}^p\left(\sum_{j=1}^n\left(\sum_{i=1}^mh(x_i,y_j,z_k)\dx_i\right)\dy_j\right)\ddz_k. \end{equation*}

Here we fix a $k$ value, which establishes the $z$-height of the rectangular solids on one “level” of all the rectangular solids in the space region $D\text{.}$ The inner double summation adds up all the volumes of the rectangular solids on this level, while the outer summation adds up the volumes of each level.

This triple summation understanding leads to the $\iiint_D$ notation of the triple integral, as well as the method of evaluation shown in Theorem 13.6.48.

The following theorem assures us that the above limit exists for continuous functions $h$ and gives us a method of evaluating the limit.

We now apply triple integration to find the centers of mass of solid objects.

# Subsection13.6.1Mass and Center of Mass

One may wish to review Section 13.4 for a reminder of the relevant terms and concepts.

##### Definition13.6.49Mass, Center of Mass of Solids

Let a solid be represented by a region $D$ in space with variable density function $\delta(x,y,z)\text{.}$

1. The mass of the object is $\ds M= \iiint_D \ dm=\iiint_D \delta(x,y,z)\ dV\text{.}$

2. The moment about the $x$-$y$ plane is $\ds M_{xy}=\iiint_D z\delta(x,y,z)\ dV\text{.}$

3. The moment about the $x$-$z$ plane is $\ds M_{xz}=\iiint_D y\delta(x,y,z)\ dV\text{.}$

4. The moment about the $y$-$z$ plane is $\ds M_{yz}=\iiint_D x\delta(x,y,z)\ dV\text{.}$

5. The center of mass of the object is \begin{equation*} \big(\overline{x},\overline{y},\overline{z}\big) = \left(\frac{M_{yz}}M,\frac{M_{xz}}M,\frac{M_{xy}}M\right). \end{equation*}

##### Example13.6.50Finding the center of mass of a solid

Find the mass and center of mass of the solid represented by the space region bounded by the coordinate planes and $z=2-y/3-2x/3\text{,}$ shown in Figure 13.6.51, with constant density $\delta(x,y,z)=3$gm/cm$^3\text{.}$ (Note: this space region was used in Example 13.6.11.)

Solution
##### Example13.6.52Finding the center of mass of a solid

Find the center of mass of the solid represented by the region bounded by the planes $z=0$ and $z=-y$ and the cylinder $x^2+y^2=1\text{,}$ shown in Figure 13.6.53, with density function $\delta(x,y,z) = 10+x^2+5y-5z\text{.}$ (Note: this space region was used in Example 13.6.22.)

Solution

As stated before, there are many uses for triple integration beyond finding volume. When $h(x,y,z)$ describes a rate of change function over some space region $D\text{,}$ then $\ds \iiint_D h(x,y,z)\ dV$ gives the total change over $D\text{.}$ Our one specific example of this was computing mass; a density function is simply a “rate of mass change per volume” function. Integrating density gives total mass.

While knowing how to integrate is important, it is arguably much more important to know how to set up integrals. It takes skill to create a formula that describes a desired quantity; modern technology is very useful in evaluating these formulas quickly and accurately.

This chapter investigated the natural follow–on to partial derivatives: iterated integration. We learned how to use the bounds of a double integral to describe a region in the plane using both rectangular and polar coordinates, then later expanded to use the bounds of a triple integral to describe a region in space. We used double integrals to find volumes under surfaces, surface area, and the center of mass of lamina; we used triple integrals as an alternate method of finding volumes of space regions and also to find the center of mass of a region in space.

Integration does not stop here. We could continue to iterate our integrals, next investigating “quadruple integrals” whose bounds describe a region in 4–dimensional space (which are very hard to visualize). We can also look back to “regular” integration where we found the area under a curve in the plane. A natural analogue to this is finding the “area under a curve,” where the curve is in space, not in a plane. These are just two of many avenues to explore under the heading of “integration.”

# Subsection13.6.2Exercises

Terms and Concepts

In the following exercises, two surfaces $f_1(x,y)$ and $f_2(x,y)$ and a region $R$ in the $x\text{,}$ $y$ plane are given. Set up and evaluate the double integral that finds the volume between these surfaces over $R\text{.}$

In the following exercises, a domain $D$ is described by its bounding surfaces, along with a graph. Set up the triple integrals that give the volume of $D$ in all 6 orders of integration, and find the volume of $D$ by evaluating the indicated triple integral.

In the following exercises, evaluate the triple integral.

In the following exercises, find the center of mass of the solid represented by the indicated space region $D$ with density function $\delta(x,y,z)\text{.}$