\(\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}\)

Triple Integrals

Introduction

Triple integrals are more difficult (in fact, usually impossible) to picture than double integrals, but the principle is much the same. Instead of integrals like \[\iint_R f(x,y) \ dA,\] now we'll handle integrals like \[\iiint_D f(x,y,z)\ dV\] where \(D\) is a 3-dimensional body.

Note:

The volume of \(D\) is given by the triple integral \[\textrm{Volume } = \iiint_D\ dz\ dy\ dx\]

Again, the challenge with triple integrals really lies with finding the limits of integration; once we have those, we simply have to carry out the integration, which usually isn't too complicated.

Evaluate the following triple integral.

\[\int_0^2 \int_0^3 \int_0^2\ dx\ dy\ dx\]

Solution

\[\begin{align} \int_0^2 \int_0^3 \int_0^2\ dx\ dy\ dx &= \int_0^2 \int_0^3 2\ dy\ dx\\ &= \int_0^2 6\ dx = \ans{12} \end{align}\]

Note that this gives the volume of a \(2 \times 3 \times 2\) cube.

Use a triple integral to find the volume of the prism \(D\) in the first octant bounded by the planes \(y=4-2x\) and \(z=6\).

Solution

Note that this prism is

\[D=\{(x,y,z)\ :\ 0 \leq x \leq 2,\ 0 \leq y \leq 4-2x,\ 0 \leq z \leq 6\}\] or \[D=\{(x,y,z)\ :\ 0 \leq x \leq 2-\dfrac{1}{2}y,\ 0 \leq y \leq 4,\ 0 \leq z \leq 6\}\]

We can write this integral in several ways, all of which give the same answer:

\[\begin{align} V &= \int_0^6 \int_0^2 \int_0^{4-2x} \ dy\ dx\ dz\\ &= \int_0^2 \int_0^{4-2x} \int_0^6 \ dx\ dy\ dx\\ &= \int_0^6 \int_0^4 \int_0^{2-y/2} \ dx\ dy\ dz\\ &= \ans{24} \end{align}\]

Notice that in each case, we have to make sure that we use the limits that involve one variable before we integrate with respect to that variable.

Try it yourself:

(click on the problem to show/hide the answer)

Find the volume of the solid in the first octant bounded by the plane \(2x+3y+6z=12\) and the coordinate planes.
\(\displaystyle\int_0^4 \displaystyle\int_0^{6-3y/2} \displaystyle\int_0^{12-2x-3y} \ dz\ dx\ dy = \displaystyle\int_0^6 \displaystyle\int_0^{4-2x/3} \displaystyle\int_0^{12-2x-3y} \ dz\ dy\ dx = 8\)

Simple Applications

Mass

If the density of a body is a function of position (\(\rho=f(x,y,z)\)), then the mass of the body is the integral

\[M = \iiint_D f(x,y,z)\ dV\]

If the density function is given by \(\rho=2-z\) kg/m\(^3\) for the cube \[D=\{(x,y,z)\ :\ 0 \leq x \leq 3,\ 0 \leq y \leq 2,\ 0 \leq z \leq 1\},\] find the mass of this cube.

Solution

\[\begin{align} M &= \int_0^1 \int_0^2 \int_0^3 2-z\ dx\ dy\ dz\\ &= \int_0^1 \int_0^2 6-3z\ dy\ dz\\ &= \int_0^1 12-6z\ dz\\ &= 12z-3z^2 \bigg|_0^1 = \ans{9\ kg} \end{align}\]

Average Value

The average value of \(f(x,y,z)\) is

\[\overline{f} = \dfrac{1}{\textrm{volume of } D} \iiint_D f(x,y,z)\ dV\]

Find the average value of \(f(x,y,z)=250 xy \sin (\pi z)\) over the cube

\[D=\{(x,y,z)\ :\ 0 \leq x \leq 2,\ 0 \leq y \leq 2,\ 0 \leq z \leq 1\}.\]

Solution

\[\begin{align} \overline{f} &= \dfrac{1}{4} \int_0^1 \int_0^2 \int_0^2 250 xy \sin (\pi z) \ dx\ dy\ dz\\ &= \dfrac{1}{4} \int_0^1 \int_0^2 500 y \sin (\pi z) \ dy\ dz\\ &= \dfrac{1}{4} \int_0^1 1000 \sin (\pi z)\ dz\\ &= \dfrac{1}{4} \bigg[ -1000 \cdot \dfrac{1}{\pi} \cos (\pi z) \bigg]_0^1\\ &= \ans{\dfrac{500}{\pi}} \end{align}\]

Try it yourself:

(click on the problem to show/hide the answer)

Find the average value of \(f(x,y,z) = 128e^{-x-y-z}\) in the cube \(D = \{(x,y,z)\ :\ 0 \leq x \leq \ln 2, 0 \leq y \leq \ln 4, 0 \leq z \leq \ln 8\}\).
\(\dfrac{1}{(\ln 2)(\ln 4)(\ln 8)}\displaystyle\int_0^{\ln 8} \displaystyle\int_0^{\ln 4} \displaystyle\int_0^{\ln 2} 128e^{-x-y-z}\ dx\ dy\ dz = \dfrac{7}{(\ln 2)^3}\)

Examples

Evaluate \(\displaystyle\iiint_D 2x\ dV\), where

\[D=\{(x,y,z)\ :\ 0 \leq y \leq 2,\ 0 \leq x \leq \sqrt{4-y^2},\ 0 \leq z \leq y\}.\]

Solution

\[\begin{align} \int_0^2 \int_0^{\sqrt{4-y^2}} \int_0^y &2x\ dz\ dx\ dy\\ &= \int_0^2 \int_0^{\sqrt{4-y^2}} 2xy \ dx\ dy\\ &= \int_0^2 (4-y^2)y\ dy\\ &= 2y-\dfrac{1}{4}y^4 \bigg|_0^2 = \ans{4} \end{align}\]

Find the volume of the solid in the first octant formed when the cylinder \(z=\sin y\) for \(0 \leq y \leq \pi\) is sliced by the planes \(y=x\) and \(x=0\).

Solution

Note that

\[D=\{(x,y,z)\ :\ 0 \leq x \leq y,\ 0 \leq y \leq \pi,\ 0 \leq z \leq \sin y\}.\]

The volume, then, is given by

\[\begin{align} V &= \int_0^\pi \int_0^y \int_0^{\sin y} \ dz\ dx\ dy\\ &= \int_0^\pi \int_0^y \sin y\ dx\ dy\\ &= \int_0^\pi y \sin y\ dy\\ &= -y\cos y + \sin y \bigg|_0^\pi \ \ \ \textrm{ (using Integration by Parts)}\\ &= \ans{\pi} \end{align}\]

Find the volume of the prism in the first octant bounded by \(z=2-4x\) and \(y=8\).


Solution

Note that

\[D=\{(x,y,z)\ :\ 0 \leq x \leq 1/2,\ 0 \leq y \leq 8,\ 0 \leq z \leq 2-4x\}.\]

The volume, then, is given by

\[\begin{align} V &= \int_0^8 \int_0^{1/2} \int_0^{2-4x} \ dz\ dx\ dy\\ &= \int_0^8 \int_0^{1/2} 2-4x\ dx\ dy\\ &= \int_0^8 \bigg[ 2x-2x^2 \bigg]_0^{1/2}\ dy\\ &= \int_0^8 \dfrac{1}{2}\ dy = \ans{4} \end{align}\]

Find the volume of the wedge bounded by the parabolic cylinder \(y=x^2\) and the planes \(z=3-y\) and \(z=0\).


Solution

Note that

\[D=\{(x,y,z)\ :\ -\sqrt{3} \leq x \leq \sqrt{3},\ x^2 \leq y \leq 3,\ 0 \leq z \leq 3-y\}.\]

The volume, then, is given by

\[\begin{align} V &= \int_{-\sqrt{3}}^{\sqrt{3}} \int_{x^2}^{3} \int_0^{3-y} \ dz\ dy\ dx\\ &= \int_{-\sqrt{3}}^{\sqrt{3}} \int_{x^2}^{3} 3-y\ dy\ dx\\ &= \int_{-\sqrt{3}}^{\sqrt{3}} \dfrac{9}{2}-3x^2+\dfrac{1}{2}x^4\ dx\\ &= \dfrac{9}{2}x-x^3+\dfrac{1}{10}x^5 \bigg|_{-\sqrt{3}}^{\sqrt{3}} = \ans{\dfrac{24\sqrt{3}}{5}} \end{align}\]