5.4 Triple Integrals

Big idea

A double integral adds up values over a flat region, weighting each point by a little piece of area. A triple integral adds up values over a solid, weighting each point by a little piece of volume. The construction is the same one again: chop the solid into small boxes, sample the function in each box, multiply by the box’s volume, add, and refine. For a continuous function on a reasonable solid the sums converge to a single number written $\iiint_E f(x,y,z)\, dV$.

What changes is the interpretation. With a double integral of a positive function you could see the answer as a volume, because the graph lived one dimension up and there was room to draw it. A triple integral of a positive function has no such picture; the graph of $f$ would need a fourth dimension. Instead the number is an accumulated total: if $f$ is a density in mass per unit volume, the integral is a mass. The one case with a geometric reading is $f = 1$, where the integral returns the volume of $E$ itself.

Evaluation follows the same outward-peeling logic as before, one level deeper. Choose a variable to integrate first, say $z$. For each point $(x,y)$ in the shadow of the solid, the vertical line through it enters $E$ at a bottom surface and leaves at a top surface, so the inner limits are those two surfaces written as functions of $x$ and $y$. What remains is a double integral over the shadow, and you already know how to handle that. Three integrals, each an ordinary one-variable problem.

There are six possible orders, and they are all correct. Choosing well matters more here than in two dimensions, because a poor choice can force you to split the solid into pieces or to face an antiderivative you cannot produce. The question to ask is which coordinate plane the solid casts the simplest shadow on, and which pair of surfaces is easiest to describe as functions of the other two variables.

Decoder

A triple integral over a solid equals an iterated integral whose innermost limits are the entering and leaving surfaces of the solid, whose middle limits describe the shadow of the solid on a coordinate plane, and whose outer limits are two constants.

The sentence describes a ladder of decreasing generality. Innermost limits may involve both other variables. Middle limits may involve the outer variable only. Outer limits may involve nothing. Each integration removes a variable, so each level of limits can mention one fewer than the level inside it. If a limit ever mentions a variable that has already been integrated away, or the one currently being integrated, the setup is wrong.

The word doing the real work is “shadow”. After the inner integral is done, what is left is a double integral over the projection of $E$ onto a coordinate plane, and that projection is a plain two-dimensional region of the kind handled in the previous sections. Finding it is a genuine step, not a formality: for a solid bounded by two surfaces that meet, the shadow is bounded by the curve where they intersect, and you find that curve by setting the two surface equations equal.

The most common failure is to guess the shadow from the picture of one surface rather than from the intersection. For the solid caught between a paraboloid and a horizontal plane, the shadow is the disk where the paraboloid meets that plane, not the whole plane and not the whole paraboloid. Set the surfaces equal, solve, and the boundary of the shadow falls out.

Definitions and results

Triple Riemann sum. Partition a box $B = [a,b] \times [c,d] \times [p,q]$ into subboxes of volume $\Delta V = \Delta x\, \Delta y\, \Delta z$, sample $f$ once in each, and sum the products $f \cdot \Delta V$.

The triple integral. If the Riemann sums approach one number as all the subboxes shrink, independently of the sample points, that number is $\iiint_E f(x,y,z)\, dV$. Continuous functions on closed bounded solids are integrable.

Volume. Setting $f = 1$ gives $V(E) = \iiint_E dV$.

Iterated integral over a box. Over $B$ as above, the integral is the triple iterated integral in any of the six orders, all limits constant. If the integrand factors as a product of a function of $x$, one of $y$ and one of $z$, the integral is the product of three single integrals.

Solids between two surfaces. If $E$ consists of the points above $z = u_1(x,y)$ and below $z = u_2(x,y)$ for $(x,y)$ in a plane region $D$, then

$$ \iiint_E f\, dV = \iint_D \left[ \int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\, dz \right] dA $$

and the outer double integral is then set up as a type I or type II region.

Projection. The region $D$ is the shadow of $E$ on the $xy$-plane. When $E$ is bounded by two surfaces that meet, the boundary of $D$ is the projection of their curve of intersection, found by setting $u_1 = u_2$.

Changing the order. The same solid can be sliced along any axis, giving six iterated integrals with the same value. To change order, redescribe the solid: name the new entering and leaving surfaces, then the new shadow.

Average value. The average of $f$ over $E$ is $\frac{1}{V(E)}\iiint_E f\, dV$. Comparing that average with the range of $f$ on $E$ is a fast check on a finished answer.

Worked examples

A product over a box

Evaluate $\iiint_B xyz\, dV$ over $B = [0,1] \times [0,2] \times [0,3]$. The limits are constants and the integrand factors, so the integral splits:

$$ \left( \int_0^1 x\, dx \right)\left( \int_0^2 y\, dy \right)\left( \int_0^3 z\, dz \right) = \frac{1}{2} \cdot 2 \cdot \frac{9}{2} = \frac{9}{2} $$

The box has volume $6$, so the average of $xyz$ on it is $\tfrac{3}{4}$. The integrand runs from $0$ to $6$ on the box, so that average is plausible.

The volume of a tetrahedron, two ways

Let $E$ be the solid in the first octant under the plane $2x + y + z = 4$. The plane crosses the axes at $(2,0,0)$, $(0,4,0)$ and $(0,0,4)$, so $E$ is a tetrahedron with those three points and the origin as vertices.

Slice vertically first. The solid sits between $z = 0$ and $z = 4 - 2x - y$, and its shadow on the $xy$-plane is the triangle $x \ge 0$, $y \ge 0$, $2x + y \le 4$, found by setting the two $z$ surfaces equal:

$$ \int_0^2 \int_0^{4-2x} \int_0^{4-2x-y} dz\, dy\, dx = \int_0^2 \int_0^{4-2x} \big( 4 - 2x - y \big)\, dy\, dx = \int_0^2 \frac{(4-2x)^2}{2}\, dx = \frac{16}{3} $$

The last step used $u = 4 - 2x$, turning the integral into $\tfrac{1}{4}\int_0^4 u^2\, du$.

Now slice along $x$ instead. For fixed $y$ and $z$ the solid runs from $x = 0$ to $x = \tfrac{4-y-z}{2}$, and the shadow on the $yz$-plane is the triangle $y \ge 0$, $z \ge 0$, $y + z \le 4$:

$$ \int_0^4 \int_0^{4-z} \frac{4-y-z}{2}\, dy\, dz = \int_0^4 \frac{(4-z)^2}{4}\, dz = \frac{1}{4}\cdot\frac{64}{3} = \frac{16}{3} $$

The inner step used $\int_0^a \tfrac{a-y}{2}\, dy = \tfrac{a^2}{4}$ with $a = 4 - z$. Both orders give $\tfrac{16}{3}$, and the elementary formula for a tetrahedron cut off by a plane with intercepts $a$, $b$, $c$ is $\tfrac{abc}{6} = \tfrac{2 \cdot 4 \cdot 4}{6} = \tfrac{16}{3}$.

Average height of that tetrahedron

Keep the same solid and evaluate $\iiint_E z\, dV$. The setup is unchanged; only the innermost integrand differs:

$$ \int_0^{4-2x-y} z\, dz = \frac{(4-2x-y)^2}{2} $$

$$ \int_0^{4-2x} \frac{(4-2x-y)^2}{2}\, dy = \frac{(4-2x)^3}{6} $$

$$ \int_0^2 \frac{(4-2x)^3}{6}\, dx = \frac{1}{12}\int_0^4 u^3\, du = \frac{64}{12} = \frac{16}{3} $$

Dividing by the volume $\tfrac{16}{3}$ gives an average height of exactly $1$. That is the check: the centroid of a tetrahedron is the average of its four vertices, whose $z$ coordinates are $0$, $0$, $0$ and $4$, giving $\bar{z} = 1$.

A solid trapped between two surfaces

Find the volume of the solid bounded below by the paraboloid $z = x^2 + y^2$ and above by the plane $z = 4$.

First the shadow. Setting the surfaces equal gives $x^2 + y^2 = 4$, so $D$ is the disk of radius $2$. The inner integral in $z$ runs from the paraboloid to the plane:

$$ V = \iint_D \int_{x^2+y^2}^{4} dz\, dA = \iint_D \big( 4 - x^2 - y^2 \big)\, dA $$

The remaining double integral is over a disk with $x^2+y^2$ in the integrand, so switch to polar:

$$ \int_0^{2\pi} \int_0^2 \big( 4 - r^2 \big) r\, dr\, d\theta = 2\pi \left[ 2r^2 - \frac{r^4}{4} \right]_0^2 = 2\pi\big( 8 - 4 \big) = 8\pi $$

The circumscribing cylinder has radius $2$ and height $4$, so volume $16\pi$, and a paraboloid splits its cylinder into two equal halves. The answer $8\pi$ is that half, and it is positive as a volume must be.

Practice

Work outward. Decide which variable to integrate first, name the entering and leaving surfaces for that variable, then project the solid to find the two-dimensional region the remaining double integral lives on. Some of these ask you to rewrite a given triple iterated integral in another order.

Quiz

Five items on triple integrals over boxes and general solids, projections onto coordinate planes, volumes, and reordering the variables.