Triple Integrals in Rectangular Coordinates, multivariable and vector calculus lesson

Picture a block of cheese whose density changes from one corner to the other. To weigh it on paper, cut it into tiny cubes, multiply each cube’s density by its volume, and add. A triple integral is the limit of that bookkeeping, and the hard part in practice is not the adding but describing the solid so the bounds come out right.

Key terms

Triple Riemann sums

Definition Let $f$ be defined on a closed box $B = [a, b] \times [c, d] \times [r, s]$ . Divide $[a, b]$ , $[c, d]$ and $[r, s]$ into $l$ , $m$ and $n$ subintervals of equal lengths $\Delta x$ , $\Delta y$ and $\Delta z$ . This cuts $B$ into $lmn$ subboxes $B_{ijk}$ , each of volume $\Delta V = \Delta x \, \Delta y \, \Delta z$ . Choose a sample point $P_{ijk}^* = (x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)$ in each $B_{ijk}$ . The triple Riemann sum is

\sum_{i=1}^{l} \sum_{j=1}^{m} \sum_{k=1}^{n} f(P_{ijk}^*) \, \Delta V.

The triple integral of $f$ over $B$ is

\begin{aligned} &\iiint_B f(x, y, z) \, dV \\[0.4em] &= \lim_{l, m, n \to \infty} \sum_{i=1}^{l} \sum_{j=1}^{m} \sum_{k=1}^{n} f(P_{ijk}^*) \, \Delta V \end{aligned}

if this limit exists and is the same for every choice of sample points. Then $f$ is called integrable on $B$ . Every continuous function on $B$ is integrable. For a bounded solid $E$ , enclose $E$ in a box $B$ and set $F = f$ on $E$ and $F = 0$ on the rest of $B$ . Then $\iiint_E f \, dV$ is defined as $\iiint_B F \, dV$ when that integral exists. This integral does not depend on the box chosen, and it exists in particular when $E$ is closed and bounded, $f$ is continuous on $E$ , and the boundary of $E$ consists of finitely many pieces, each the graph of a continuous function over a closed bounded region in one of the coordinate planes. Such a boundary has zero volume, so the jump of $F$ across it does no harm. All the solids below are of this kind.

In plain words Chop the solid into tiny boxes, multiply the value of $f$ in each box by the box’s volume, and add. The integral is what these sums settle on as the boxes shrink.

Example Take $f(x, y, z) = z^2$ on $B = [0, 1] \times [0, 1] \times [0, 2]$ and cut each edge in half, giving 8 subboxes with $\Delta V = \frac{1}{2} \cdot \frac{1}{2} \cdot 1 = \frac{1}{4}$ . At the midpoints, four boxes have $z = \frac{1}{2}$ and four have $z = \frac{3}{2}$ , so the sum is

4 \cdot \frac{1}{4} \cdot \frac{1}{4} + 4 \cdot \frac{9}{4} \cdot \frac{1}{4} = \frac{5}{2}.

The exact value, computed in the next section, is $\frac{8}{3} \approx 2.667$ , so even this coarse sum is close.

Iterated integrals and Fubini’s theorem

Definition Fubini’s theorem for boxes says that if $f$ is continuous on $B = [a, b] \times [c, d] \times [r, s]$ , then

\begin{aligned} &\iiint_B f \, dV \\[0.4em] &= \int_r^s \int_c^d \int_a^b f(x, y, z) \, dx \, dy \, dz, \end{aligned}

and the same holds for each of the six orders of integration. The iterated integral is evaluated from the inside out, holding the outer variables fixed while integrating the inner one.

In plain words On a box you can integrate one variable at a time, in any order you like, and the answer does not change.

Example For the Riemann sum example above,

\begin{aligned} \int_0^2 \int_0^1 \int_0^1 z^2 \, dx \, dy \, dz &= \int_0^2 z^2 \, dz \\[0.4em] &= \frac{8}{3}. \end{aligned}

A second one, over $B = [0, 1] \times [-1, 2] \times [0, 3]$ . Integrating $x$ , then $y$ , then $z$ gives

\begin{aligned} \iiint_B xyz^2 \, dV &= \int_0^3 \int_{-1}^{2} \frac{y z^2}{2} \, dy \, dz \\[0.4em] &= \int_0^3 \frac{3 z^2}{4} \, dz = \frac{27}{4}. \end{aligned}

Region descriptions

Definition A solid $E$ is of type 1 if

\begin{gathered} E = \{(x, y, z) \mid (x, y) \in D, \\ u_1(x, y) \le z \le u_2(x, y)\}, \end{gathered}

where $D$ is the projection of $E$ onto the $xy$ -plane, $D$ is a closed bounded region of type I or type II, and $u_1 \le u_2$ are continuous on $D$ . If $f$ is continuous on $E$ , then

\begin{aligned} &\iiint_E f \, dV \\[0.4em] &= \iint_D \left[ \int_{u_1(x, y)}^{u_2(x, y)} f \, dz \right] dA. \end{aligned}

Types 2 and 3 are defined the same way with $x$ or $y$ as the inner variable, projecting onto the $yz$ -plane or the $xz$ -plane.

In plain words Look at the shadow of the solid on a coordinate plane. Above each point of the shadow, a segment runs from a bottom surface to a top surface. The inner integral covers that segment, and a double integral covers the shadow.

Example Let $E$ be the tetrahedron bounded by the coordinate planes and the plane $x + y + z = 1$ . Its shadow on the $xy$ -plane is the triangle $0 \le x \le 1$ , $0 \le y \le 1 - x$ , and above each shadow point $z$ runs from $0$ to $1 - x - y$ . Then

\begin{aligned} &\iiint_E z \, dV \\[0.4em] &= \int_0^1 \int_0^{1 - x} \int_0^{1 - x - y} z \, dz \, dy \, dx \\[0.4em] &= \int_0^1 \int_0^{1 - x} \frac{(1 - x - y)^2}{2} \, dy \, dx \\[0.4em] &= \int_0^1 \frac{(1 - x)^3}{6} \, dx = \frac{1}{24}. \end{aligned}

Changing the order of integration

Definition For $f$ continuous on a solid $E$ that can be described in more than one of the three types, every valid description gives the same value of $\iiint_E f \, dV$ . To change the order, first write $E$ as a set of inequalities, then read off a new projection and new inner bounds from those same inequalities. The new limits must describe exactly the same set of points.

In plain words Never shuffle the limits around directly. Go back to the solid, sketch or list its inequalities, and rebuild the bounds for the new order from scratch.

Example The integral

\int_0^1 \int_0^{x^2} \int_0^y f \, dz \, dy \, dx

covers the solid $0 \le x \le 1$ , $0 \le y \le x^2$ , $0 \le z \le y$ . For the order $dy \, dz \, dx$ , fix $x$ . The inequalities $0 \le z \le y \le x^2$ say that $z$ runs from $0$ to $x^2$ and then $y$ from $z$ to $x^2$ , so the integral equals

\int_0^1 \int_0^{x^2} \int_z^{x^2} f \, dy \, dz \, dx.

For the order $dx \, dy \, dz$ , the same inequalities read $0 \le z \le y \le 1$ and $\sqrt{y} \le x \le 1$ , so the integral equals

\int_0^1 \int_z^1 \int_{\sqrt{y}}^1 f \, dx \, dy \, dz.

With $f = 1$ all three give $\frac{1}{10}$ , a good check that nothing was lost.

Volume

Definition The volume of a solid $E$ for which the integral exists is

V(E) = \iiint_E 1 \, dV.

In plain words Integrating the constant $1$ just adds up the volumes of the tiny boxes.

Example The tetrahedron cut from the first octant by $2x + y + z = 4$ has intercepts $2$ , $4$ and $4$ . Its shadow is $0 \le x \le 2$ , $0 \le y \le 4 - 2x$ , and $z$ runs from $0$ to $4 - 2x - y$ , so

\begin{aligned} V &= \int_0^2 \int_0^{4 - 2x} (4 - 2x - y) \, dy \, dx \\[0.4em] &= \int_0^2 \frac{(4 - 2x)^2}{2} \, dx = \frac{16}{3}. \end{aligned}

This agrees with the formula one sixth times the product of the intercepts, $\frac{1}{6} \cdot 2 \cdot 4 \cdot 4 = \frac{16}{3}$ .

Mass

Definition If a solid $E$ has continuous density $\delta(x, y, z) \ge 0$ , measured in mass per unit volume, its mass is

m = \iiint_E \delta(x, y, z) \, dV.

In plain words Each tiny box has mass about density times volume. Add them all.

Example Give the tetrahedron bounded by the coordinate planes and $x + y + z = 1$ the density $\delta = y$ , so it gets heavier away from the $xz$ -plane. Then

\begin{aligned} m &= \int_0^1 \int_0^{1 - x} y(1 - x - y) \, dy \, dx \\[0.4em] &= \int_0^1 \frac{(1 - x)^3}{6} \, dx = \frac{1}{24}. \end{aligned}

Moments and center of mass

Definition For a solid $E$ with continuous density $\delta$ and mass $m > 0$ , the first moments about the three coordinate planes are

\begin{gathered} M_{yz} = \iiint_E x \, \delta \, dV, \\[0.4em] M_{xz} = \iiint_E y \, \delta \, dV, \\[0.4em] M_{xy} = \iiint_E z \, \delta \, dV, \end{gathered}

and the center of mass is

(\bar{x}, \bar{y}, \bar{z}) = \left( \frac{M_{yz}}{m}, \frac{M_{xz}}{m}, \frac{M_{xy}}{m} \right).

The moment of inertia about the $z$ -axis is $I_z = \iiint_E (x^2 + y^2) \, \delta \, dV$ , and $I_x$ , $I_y$ are defined the same way with the squared distance to that axis.

In plain words The center of mass is the balance point. Each coordinate is a weighted average of that coordinate over the solid, with the density as the weight.

Example For the tetrahedron with $\delta = y$ from the mass example, the same bounds give

\begin{gathered} M_{yz} = \frac{1}{120}, \qquad M_{xz} = \frac{1}{60}, \\[0.4em] M_{xy} = \frac{1}{120}. \end{gathered}

Dividing by $m = \frac{1}{24}$ gives the center of mass $\left( \frac{1}{5}, \frac{2}{5}, \frac{1}{5} \right)$ . With constant density the center would be at $\left( \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \right)$ . The heavier side pulls the balance point toward larger $y$ and away from the other two directions, as you would expect.

Average value

Definition If $f$ is integrable on a solid $E$ with $V(E) > 0$ , the average value of $f$ on $E$ is

f_{\text{avg}} = \frac{1}{V(E)} \iiint_E f(x, y, z) \, dV.

In plain words Total up $f$ over the solid and divide by the volume, just as an ordinary average divides a total by a count.

Example On the tetrahedron bounded by the coordinate planes and $x + y + z = 1$ , which has volume $\frac{1}{6}$ , take $f = x^2 + y^2 + z^2$ , the squared distance from the origin. Each of the three terms integrates to $\frac{1}{60}$ , so

f_{\text{avg}} = 6 \cdot \frac{3}{60} = \frac{3}{10}.

Check yourself

Evaluate the integral of $x + y + z$ over the box $[0, 1] \times [0, 2] \times [0, 3]$ .

Show the answer

The box has volume $6$ . Integrating term by term gives $3 + 6 + 9$ , so

\iiint_B (x + y + z) \, dV = 18.

For example, the $x$ term is $\frac{1}{2} \cdot 2 \cdot 3 = 3$ .

Rewrite the integral below in the order $dx \, dz \, dy$ .

\int_0^1 \int_0^{1 - x} \int_0^{1 - x - y} f \, dz \, dy \, dx

Show the answer

The solid is $x, y, z \ge 0$ with $x + y + z \le 1$ . Its shadow on the $yz$ -plane is $0 \le y \le 1$ , $0 \le z \le 1 - y$ , and for each such point $x$ runs from $0$ to $1 - y - z$ , so the integral is

\int_0^1 \int_0^{1 - y} \int_0^{1 - y - z} f \, dx \, dz \, dy.

Find the center of mass of the cube $[0, 1]^3$ with density $\delta = z$ .

Show the answer

The mass is $m = \int_0^1 z \, dz = \frac{1}{2}$ and $M_{xy} = \int_0^1 z^2 \, dz = \frac{1}{3}$ , so $\bar{z} = \frac{2}{3}$ . The density does not depend on $x$ or $y$ and the cube is symmetric about $x = \frac{1}{2}$ and $y = \frac{1}{2}$ , so the center of mass is $\left( \frac{1}{2}, \frac{1}{2}, \frac{2}{3} \right)$ .

Triple Integrals in Rectangular Coordinates

Key terms

Triple Riemann sums

Iterated integrals and Fubini’s theorem

Region descriptions

Changing the order of integration

Volume

Mass

Moments and center of mass

Average value

Check yourself

How this lesson connects