## Section4.3Triple Integrals: Volume and Average Value

### Subsection4.3.1

It will come as no surprise that we can also do triple integrals—integrals over a three-dimensional region. The simplest application allows us to compute volumes in an alternate way.

We follow the same method as we have done when we defined a single integral for functions of one variable and a double integral for functions of two variables. Suppose that $f(x,y,z)$ is a continuous function on a closed bounded region $S$ in space. If the boundaries of $S$ are “relatively smooth”, then we can divide the three-dimensional region into small rectangular boxes with dimensions $\Delta x\times\Delta y\times\Delta z$ and with volume $dV = \Delta x\Delta y\Delta z\text{.}$ Then we add them all up and take the limit, to get an integral:

\begin{equation*} \iiint_S f(x,y,z)\,dV\text{.} \end{equation*}

Note:

1. Fubini's Theorem also holds for triple integrals, which means that the order of integration does not matter and we can choose from setting up a triple integral with any of the following six choices for the order of integration: $dx\,dy\,dz\text{,}$ $dx\,dz\,dy\text{,}$ $dy\,dx\,dz\text{,}$ $dy\,dz\,dx\text{,}$ $dz\,dx\,dy\text{,}$ and $dz\,dy\,dx\text{.}$ However, the same word of caution holds here as well, as some orders may lead to a more readily computable triple integral, while others may simply be too difficult to compute.

2. If the three-variable function $f$ is the constant 1, then the triple integral $\ds\iiint_S dV$ evaluates to the volume of the closed bounded region $S\text{.}$

3. If the three-variable function $f$ is the constant 1 and $S$ is bounded by constants, then we are simply computing the volume of a rectangular box.

###### Example4.19. Volume of a Box.

Compute the volume of the box with opposite corners at $(0,0,0)$ and $(1,2,3)\text{.}$

Solution

We begin by drawing an outline of the rectangular box as shown below.

Since the faces of the rectangular box are parallel to the coordinate planes, we deduce that the integration bounds are given by

\begin{equation*} 0\leq x\leq 1, \ 0 \leq y\leq2, \ 0\leq z \leq 3\text{.} \end{equation*}

Hence, the following triple integral computes the volume of the rectangular box:

\begin{equation*} \begin{split} \int_0^1\int_0^2\int_0^3 dz\,dy\,dx\amp=\int_0^1\int_0^2z\bigg\vert_0^3 \,dy\,dx \\ \amp=\int_0^1\int_0^2 3\,dy\,dx\\ \amp =\int_0^1 3y\bigg\vert_0^2 \,dx \\ \amp=\int_0^1 6\,dx = 6.\end{split} \end{equation*}

Note that any of the following triple integrals would have resulted in the volume of the box:

Of course, this is more interesting and useful when the limits are not constant.

###### Example4.20. Order of Integration.

Calculate the volume of the prism shown using the order of integration

1. $dx\,dy\,dz$

2. $dy\,dz\,dx$

Solution

Let $V$ be the volume of the prism. We begin by identifying the plane for each face of the prism:

1. For the triple integral with order of integration $dx\,dy\,dz\text{,}$ we begin by drawing a line parallel to the $x$-axis as shown below to the left that cuts through the prism. We notice that any such line stays between the values $x=0$ and $x=3$ and is thus constant. Hence, the inner most integral is

\begin{equation*} \int_{?}^{?}\int_{?}^{?}\int_0^3 \,dx\,dy\,dz\text{.} \end{equation*}

Next, we are now extending the line in the $x$-$y$-plane to create a cross-sectional area that slices through the prism perpendicularly to the $z$-axis as shown above in the centre. These horizontal cross-sections are not constant and vary depending on the $z$-value that is chosen as shown above to the right. The $z$-values themselves are bounded by $z=0$ and $z=2\text{.}$ Hence, the triple integral needed to evaluate the volume of the prism is

\begin{equation*} V = \int_0^2\int_0^{1-0.5z}\int_0^3 dx\,dy\,dz\text{.} \end{equation*}

Following through with the integration yields

\begin{equation*} \begin{split} V \amp = \int_0^2\int_0^{1-0.5z}\int_0^3 dx\,dy\,dz \\ \amp= \int_0^2\int_0^{1-0.5z} x\big\vert_0^3 \,dy\,dz\\ \amp = \int_0^2\int_0^{1-0.5z} 3\,dy\,dz \\ \amp = \int_0^2 3y\big\vert_0^{1-0.5z}\,dz\\ \amp = \int_0^2 3(1-0.5z)\,dz = \left[3z-\frac{3}{4}z^2\right]_0^2 = 3. \end{split} \end{equation*}
2. For the triple integral with order of integration $dy\,dz\,dx\text{,}$ we begin by drawing a line parallel to the $y$-axis as shown below to the left that cuts through the prism, since we are integrating with respect to $y\text{.}$ We notice that any such line stays between the values $y=0$ and $y=1-0.5z\text{,}$ and is thus variant depending on the $z$-value that is chosen. Hence, the inner most integral is

\begin{equation*} \int_{?}^{?}\int_{?}^{?}\int_0^{1-0.5z} \,dy\,dz\,dx\text{.} \end{equation*}

Next, we integrate with respect to z, which means we are summing up all such lines from above that are perpendicular to the $z$-axis in the $y$-$z$-plane of the prism as shown above in the centre, which of course sums to the area of the triangle in the $y$-$z$-plane. These lines are bounded by $z=0$ and $z=2\text{.}$ Hence, the inner two integrals needed to evaluate the volume of the prism are

\begin{equation*} \int_{?}^{?}\int_{0}^{2}\int_0^{1-0.5z} \,dy\,dz\,dx\text{.} \end{equation*}

Lastly, we integrate with respect to $x\text{,}$ which means we are summing up all such areas from above that are perpendicular to the $x$-axis. These areas are constant and are bounded by $x=0$ and $x=3$ as shown above to the right. Hence, the outer integral is

\begin{equation*} \int_{0}^{3}\int_{0}^{2}\int_0^{1-0.5z} \,dy\,dz\,dx\text{.} \end{equation*}

Following through with the integration yields the volume $V$ to be

\begin{equation*} \begin{split} V \amp = \int_{0}^{3}\int_{0}^{2}\int_0^{1-0.5z} \,dy\,dz\,dx \\ \amp= \int_0^3\int_0^2 y\big\vert_0^{1-0.5z}\,dz\,dz\\ \amp = \int_0^3\int_0^2(1-0.5z)\,dz\,dx\\ \amp = \int_0^3\left[z-\frac{z^2}{4}\right]_0^2\,dx\\ \amp = \int_0^3 1 \,dx = x\big\vert_0^3 = 3, \end{split} \end{equation*}

###### Example4.21. Volume of a Tetrahedron.

Find the volume of the tetrahedron with corners at $(0,0,0)\text{,}$ $(0,3,0)\text{,}$ $(2,3,0)\text{,}$ and $(2,3,5)\text{.}$

Solution

The whole problem comes down to correctly describing the region by inequalities: $0\le x\le 2\text{,}$ $3x/2\le y\le 3\text{,}$ $0\le z\le 5x/2\text{.}$ The lower $y$ limit comes from the equation of the line $y=3x/2$ that forms one edge of the tetrahedron in the $x$-$y$-plane; the upper $z$ limit comes from the equation of the plane $z=5x/2$ that forms the “upper” side of the tetrahedron as shown below.

Now the volume is

\begin{align*} \int_0^2\int_{3x/2}^3\int_0^{5x/2}dz\,dy\,dx \amp =\int_0^2\int_{3x/2}^3\left.z\right|_0^{5x/2} \,dy\,dx\\ \amp =\int_0^2\int_{3x/2}^3 {5x\over2}\,dy\,dx\\ \amp =\int_0^2 \left.{5x\over2}y\right|_{3x/2}^3 \,dx\\ \amp =\int_0^2 {15x\over2}-{15x^2\over4}\,dx\\ \amp =\left. {15x^2\over4}-{15x^3\over12}\right|_0^2\\ \amp =15-10=5\text{.} \end{align*}

Pretty much just the way we did for two dimensions we can use triple integration in a variety of different physical, social and biological applications, and in computing various average quantities.

###### Example4.22. Average Temperature in a Cube.

Suppose the temperature at a point is given by $T=xyz\text{.}$ Find the average temperature in the cube with opposite corners at $(0,0,0)$ and $(2,2,2)\text{.}$

Solution

In two dimensions:

1. Add up the temperature at each point in a region.

2. Divide by the area.

In three dimensions:

1. Add up the temperature at each point in space.

2. Divide by the volume.

Therefore, the average temperature in the cube is

\begin{align*} {1\over8}\int_{0}^2\int_{0}^2\int_{0}^2 xyz\,dz\,dy\,dx \amp ={1\over8}\int_{0}^2\int_{0}^2\left.{xyz^2\over2}\right|_0^2\,dy\,dx\\ \amp ={1\over16}\int_{0}^2\int_{0}^2 xy\,dy\,dx\\ \amp ={1\over4}\int_{0}^2\left.{xy^2\over2}\right|_0^2\,dx\\ \amp ={1\over8}\int_{0}^2 4x\,dx\\ \amp ={1\over2}\left.{x^2\over2}\right|_0^2 =1\text{.} \end{align*}
##### Exercises for Section 4.3.

Evaluate the following triple integrals.

1. $\ds\int_{0}^{1}\int_{0}^{x}\int_{0}^{x+y} (2x+y-1) \,dz\,dy\,dx$

$11/24$
Solution
\begin{equation*} \begin{aligned}\int_0^1 \int_0^x \int_0^{x+y} (2x+y-1)\,dz\,dy\,dx\amp= \int_0^1 \int_0^x (2x+y-1)(x+y)\,dy\,dx \\ \amp= \int_0^1 \int_0^x \left[2 x^2 + 3 x y - x + y^2 - y \right]\,dy\,dx \\ \amp= \int_0^1 \left[2x^2 y + \frac{3}{2}xy^2 -xy + \frac{y^3}{3} - \frac{y^2}{2}\right]_0^x\,dx \\ \amp= \int_0^1 \left[\frac{-9}{6} x^2 + \frac{23}{6} x^3\right]\,dx \\ \amp= \frac{11}{24}\end{aligned} \end{equation*}
2. $\ds\int_{0}^{2}\int_{-1}^{x^2}\int_{1}^{y} xyz \,dz\,dy\,dx$

$623/60$
Solution
\begin{equation*} \begin{aligned}\int_0^2\int_{-1}^{x^2}\int_1^y xyz \,dz\,dy\,dx \amp = \int_0^2 \int_{-1}^{x^2} \frac{xyz^2}{2} \bigg\vert_{z=1}^y\,dy\,dx\\ \amp= \int_0^2 \int_{-1}^{x^2} \left(\frac{xy^3}{2}-\frac{xy}{2}\right)\,dy\,dx \\ \amp =\int_0^2 \left[\frac{xy^4}{8} - \frac{xy^2}{4}\right]_{y=-1}^{x^2}\,dx\\ \amp = \int_0^2 \left(\frac{x^9}{8}-\frac{x^5}{4}+\frac{x}{8}\right)\,dx \\ \amp = \left[\frac{x^{10}}{8\cdot 10} - \frac{x^6}{4\cdot 6} + \frac{x^2}{8\cdot 2}\right]_{x=0}^2 = \frac{623}{30} \end{aligned} \end{equation*}
3. $\ds\int_{0}^{1}\int_{0}^{x}\int_{0}^{\ln y} e^{x+y+z}\,dz\,dy\,dx$

$-3e^2/4+2e-3/4$
Solution
\begin{equation*} \begin{aligned}\int_0^1 \int_0^x \int_0^{\ln y} e^{x+y+z}\,dz\,dy\,dx \amp = \int_0^1 \int_0^x \int_0^{\ln y} e^xe^ye^z\,dz\,dy\,dx \\ \amp = \int_0^1 \int_0^x e^xe^ye^z \bigg\vert_{z=0}^{\ln y} \,dy\,dx \\ \amp= \int_0^1\int_0^x e^xe^y(y-1)\,dy\,dx \end{aligned} \end{equation*}

We now use the fact that $\ds\int y e^y\,dy = e^y(y-1):$

\begin{equation*} \begin{aligned}\int_0^1\int_0^x e^xe^y(y-1)\,dy\,dx \amp = \int_0^1 \left[-e^xe^y+e^xe^y(y-1)\right]_{y=0}^x \,dx \\ \amp = \int_0^1 \left[e^xe^y(y-2)\right]_{y=0}^x \,dx \\ \amp= \int_0^1 e^{2x}(x-2)+2e^x \,dx \\ \amp = \left[\frac{1}{4}e^{2x}(2x-5) + 2e^x\right]_{x=0}^1 \\ \amp = \frac{-3}{4}e^2+2e - \frac{3}{4} \approx -0.855 \end{aligned} \end{equation*}

Thus, $\ds\int_0^1 \int_0^x \int_0^{\ln y} e^{x+y+z}\,dz\,dy\,dx \approx -0.855.$

4. $\ds\int_{0}^{\pi/2}\int_{0}^{\sin\theta}\int_{0}^{r\cos\theta} r^2\,dz\,dr\,d\theta$

$1/20$
Solution
Solution
\begin{equation*} \begin{aligned}\int_{0}^{\pi/2}\int_0^{\sin\theta}\int_0^{r\cos\theta} r^2\,dz\,dr\,d\theta \amp = \int_{0}^{\pi/2}\int_0^{\sin\theta} r^2 z \big\vert_{z=0}^{r\cos\theta} \,dr\,d\theta \\ \amp = \int_0^{\pi/2}\int_0^{\sin\theta} r^3\cos\theta \,dr\,d\theta \\ \amp =\int_0^{\pi/2} \frac{r^4}{4}\cos\theta \bigg\vert_{r=0}^{\sin\theta} \,d\theta \\ \amp = \int_0^{\pi/2} \frac{\sin^4\theta \cos\theta}{4}\,d\theta \end{aligned} \end{equation*}

Let $u=\sin\theta$ and $du = \cos\theta d\theta\text{.}$ Then

\begin{equation*} \int_0^{\pi/2} \frac{\sin^4\theta \cos\theta}{4}\,d\theta = \frac{1}{4}\int_0^1 u^4 \,du = \frac{1}{20} u^5 \bigg\vert_0^1 = \frac{1}{20}\text{.} \end{equation*}

Therefore, $\displaystyle{\int_{0}^{\pi/2}\int_0^{\sin\theta}\int_0^{r\cos\theta} r^2\,dz\,dr\,d\theta = \frac{1}{20}.}$

5. $\ds\int_{0}^{\pi}\int_{0}^{\sin\theta}\int_{0}^{r\sin\theta} r\cos^2\theta \,dz\,dr\,d\theta$

$\pi/48$
Solution
\begin{equation*} \begin{aligned} \int_0^{\pi} \int_0^{\sin\theta} \int_0^{r\sin\theta} r\cos^2\theta \,dz\,dr \,d\theta \amp= \int_0^{\pi} \int_0^{\sin\theta} r^2 \cos^2\theta\sin\theta \,dr\, d\theta\\ \amp=\int_0^{\pi} \frac{r^3}{3} \cos^2\theta \sin\theta \big\vert_0^{\sin\theta}\,d\theta\\ \amp= \int_0^{\pi} \frac{1}{3} \cos^2\theta \sin^{4}\theta\,d\theta \amp= \frac{\pi}{16(3)} = \frac{\pi}{48}.\end{aligned} \end{equation*}
6. $\ds\int_{0}^{1}\int_{0}^{y^2}\int_{0}^{x+y} x\,dz\,dx\,dy$

$11/84$
Solution
\begin{equation*} \begin{aligned} \int_0^1 \int_0^{y^2} \int_0^{x+y} x\,dz\,dx\,dy \amp= \int_0^1 \int_0^{y^2} \left(x^2+xy\right)\,dx\,dy\\ \amp= \int_0^1 \left[\frac{x^3}{3} + \frac{x^2y}{2}\right]_0^{y^2}\,dy\\ \amp= \frac{1}{6}\int_0^1 \left(2y^6+3y^5\right)\,dy\\ \amp=\frac{11}{84}.\end{aligned} \end{equation*}
7. $\ds\int_{1}^{2}\int_{y}^{y^2}\int_{0}^{\ln(y+z)} e^x\,dx\,dz\,dy$

$151/60$
Solution
\begin{equation*} \begin{aligned} \int_1^2 \int_y^{y^2} \int_0^{\ln(y+z)} e^x\,dx\,dz\,dy \amp= \int_1^2 \int_y^{y^2} (y+z-1)\,dz\,dy\\ \amp= \int_1^2 \left[yz+\frac{z^2}{2} - z\right]_y^{y^2}\,dy\\ \amp= \int_1^2 \left[\frac{y^4}{2} + y^3 - \frac{5y^2}{2} + y\right]\,dy\\ \amp= \frac{151}{60}.\end{aligned \end{equation*}
8. $\ds \int_0^\pi\int_0^{\pi/2}\int_0^1 (z\sin x+z\cos y)\,dz\,dy\,dx$

$\pi$
Solution
\begin{equation*} \begin{aligned} \int_0^{\pi}\int_0^{\pi/2} \int_0^1 (z\sin z+ z\cos y)\,dz\,dy\,dx \amp= \int_0^{\pi} \int_0^{\pi/2} (\sin x+ \cos y) \frac{z^2}{2}\,dy\,dx\\ \amp= \frac{1}{2} \int_0^{\pi} \int_0^{\pi/2} \left(\sin x+\cos y\right)\,dy \,dx \\ \amp= \frac{1}{2} \int_0^{\pi} \left[y\sin x+\sin y\right]_0^{\pi/2} \,dx \\ \amp= \frac{1}{2}\int_0^{\pi} \left(\frac{\pi}{2}\sin x + 1\right)\,dx\\ \amp= \frac{\pi}{4} \left[-\cos x\right]_0^{\pi} + \frac{x}{2} \bigg\vert_0^{\pi}\\ \amp= \pi. \end{aligned} \end{equation*}

Setup $\ds\iiint \left(x+y+z\right) \,dV$ over the region inside $x^2+y^2+z^2\le 1$ in the first octant, but do not follow through on the integration as it requires a special technique that we have not introduced.

\begin{equation*} \iiint_{V} \left(x+y+z\right) \,dV = \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} \int_0^{\sqrt{1-z^2-x^2}} \left(x+y+z\right) \,dy\,dx\,dz\text{.} \end{equation*}
Solution

We wish setup the integral $\displaystyle{\iiint_{V} \left(x+y+z\right) \,dV}\text{,}$ where $V$ is the interior of the sphere centred at the origin with radius 1 in the first quadrant. That is, $V = \{(x,y,z) : x^2+y^2+z^2 \leq 1 \text{ and } x \geq 0, y\geq 0, z\geq 0\}\text{.}$

Consider fixing $z\text{.}$ Then in the $x$-$y$-plane, the desired region is the quarter circle shown below:

where $r$ is now a function of $z\text{.}$ We see that if we let $x$ vary from 0 to $r\text{,}$ then $y$ must vary from $0$ to $\sqrt{r^2-x^2}\text{.}$

To find $r\text{,}$ let $x=r$ and $y=0\text{.}$ Then use the fact that on the boundary of the region, we have

\begin{equation*} 1 = x^2+y^2+z^2 \implies 1 = r^2 + z^2 \implies r = \sqrt{1-z^2}\text{.} \end{equation*}

All together, we see that

\begin{equation*} \iiint_{V} \left(x+y+z\right) \,dV = \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} \int_0^{\sqrt{1-z^2-x^2}} \left(x+y+z\right) \,dy\,dx\,dz\text{.} \end{equation*}

Find the region $E$ for which $\ds\iiint_E (1-x^2-y^2-z^2) \; dV$ is a maximum.

$E = \{x^2+y^2+z^2 \leq 1\}$
Notice that $x^2+y^2+z^2=1$ is the equation of the sphere of radius 1 centred at the origin. Hence, the integral will be maximized when $E$ is the interior of this sphere (it does not make a difference whether or not we include the boundary).