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Mathematics LibreTexts

4.5: Triple Integrals

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Learning Objectives

  • Recognize when a function of three variables is integrable over a rectangular box.
  • Evaluate a triple integral by expressing it as an iterated integral.
  • Recognize when a function of three variables is integrable over a closed and bounded region.
  • Simplify a calculation by changing the order of integration of a triple integral.
  • Calculate the average value of a function of three variables.

Previously, we discussed the double integral of a function f(x,y) of two variables over a rectangular region in the plane. In this section we define the triple integral of a function f(x,y,z) of three variables over a rectangular solid box in space, R3. Later in this section we extend the definition to more general regions in R3.

Integrable Functions of Three Variables

We can define a rectangular box B in R3 as

B={(x,y,z)|axb,cyd,ezf}.

We follow a similar procedure to what we did in previously. We divide the interval [a,b] into l subintervals [xi1,xi] of equal length Δx with

Δx=xixi1l,

divide the interval [c,d] into m subintervals [yi1,yi] of equal length Δy with

Δy=yjyj1m,

and divide the interval [e,f] into n subintervals [zi1,zi] of equal length Δz with

Δz=zkzk1n

Then the rectangular box B is subdivided into lmn subboxes:

Bijk=[xi1,xi]×[yi1,yi]×[zi1,zi],

as shown in Figure 4.5.1.

In x y z space, there is a box B with a subbox Bijk with sides of length Delta x, Delta y, and Delta z.
Figure 4.5.1: A rectangular box in R3 divided into subboxes by planes parallel to the coordinate planes.

For each i,j, and k, consider a sample point (xijk,yijk,zijk) in each sub-box Bijk. We see that its volume is ΔV=ΔxΔyΔz. Form the triple Riemann sum

li=1mj=1nk=1f(xijk,yijk,zijk)ΔxΔyΔz.

We define the triple integral in terms of the limit of a triple Riemann sum, as we did for the double integral in terms of a double Riemann sum.

Definition: The triple integral

The triple integral of a function f(x,y,z) over a rectangular box B is defined as

lim if this limit exists.

When the triple integral exists on B the function f(x,y,z) is said to be integrable on B. Also, the triple integral exists if f(x,y,z) is continuous on B. Therefore, we will use continuous functions for our examples. However, continuity is sufficient but not necessary; in other words, f is bounded on B and continuous except possibly on the boundary of B. The sample point (x_{ijk}^*, y_{ijk}^*, z_{ijk}^*) can be any point in the rectangular sub-box B_{ijk} and all the properties of a double integral apply to a triple integral. Just as the double integral has many practical applications, the triple integral also has many applications, which we discuss in later sections.

Now that we have developed the concept of the triple integral, we need to know how to compute it. Just as in the case of the double integral, we can have an iterated triple integral, and consequently, a version of Fubini’s theorem for triple integrals exists.

Fubini’s Theorem for Triple Integrals

If f(x,y,z) is continuous on a rectangular box B = [a,b] \times [c,d] \times [e,f], then

\iint_B f(x,y,z) \,dV = \int_e^f \int_c^d \int_a^b f(x,y,z) \,dx \, dy \, dz.

This integral is also equal to any of the other five possible orderings for the iterated triple integral.

For a, b, c, d, e and f real numbers, the iterated triple integral can be expressed in six different orderings:

\begin{align} \int_e^f \int_c^d \int_a^b f(x,y,z)\, dx \, dy \, dz = \int_e^f \left( \int_c^d \left( \int_a^b f(x,y,z) \,dx \right) dy \right) dz \\ = \int_c^d \left( \int_e^f \left( \int_a^b f(x,y,z) \,dx \right)dz \right) dy \\ = \int_a^b \left( \int_e^f \left( \int_c^d f(x,y,z) \,dy \right)dz \right) dx \\ = \int_e^f \left( \int_a^b \left( \int_c^d f(x,y,z) \,dy \right) dx \right) dz \\ = \int_c^d \left( \int_a^b \left( \int_c^d f(x,y,z) \,dz\right)dx \right) dy \\ = \int_a^b \left( \int_c^d \left( \int_e^f f(x,y,z) \,dz \right) dy \right) dx \end{align}

For a rectangular box, the order of integration does not make any significant difference in the level of difficulty in computation. We compute triple integrals using Fubini’s Theorem rather than using the Riemann sum definition. We follow the order of integration in the same way as we did for double integrals (that is, from inside to outside).

Example \PageIndex{1}: Evaluating a Triple Integral

Evaluate the triple integral \int_{z=0}^{z=1} \int_{y=2}^{y=4} \int_{x=-1}^{x=5} (x + yz^2)\, dx \, dy \, dz. \nonumber

Solution

The order of integration is specified in the problem, so integrate with respect to x first, then y, and then z.

\begin{align*} \int_{z=0}^{z=1} \int_{y=2}^{y=4} \int_{x=-1}^{x=5} (x + yz^2) \, dx \, dy \, dz \\ = \int_{z=0}^{z=1} \int_{y=2}^{y=4} \left. \left[ \dfrac{x^2}{2} + xyz^2\right|_{x=-1}^{x=5}\right]\, dy \, dz \text{Integrate with respect to $x$.} \\ = \int_{z=0}^{z=1} \int_{y=2}^{y=4} \left[12+6yz^2\right] \,dy \, dz \text{Evaluate.} \\ = \int_{z=0}^{z=1} \left[ \left.12y+6\dfrac{y^2}{2}z^2 \right|_{y=2}^{y=4} \right] dz \text{Integrate with respect to $y$.} \\ = \int_{z=0}^{z=1} [24+36z^2] \, dz \text{Evaluate.} \\ = \left[ 24z+36\dfrac{z^3}{3} \right]_{z=0}^{z=1} \text{Integrate with respect to $z$.}\\ =36. \text{Evaluate.} \end{align*}

Example \PageIndex{2}: Evaluating a Triple Integral

Evaluate the triple integral

\iiint_B x^2 yz \,dV

where B = \big\{(x,y,z)\,|\, - 2 \leq x \leq 1, \, 0 \leq y \leq 3, \, 1 \leq z \leq 5 \big\} as shown in Figure \PageIndex{2}.

In x y z space, there is a box given with corners (1, 0, 5), (1, 0, 1), (1, 3, 1), (1, 3, 5), (negative 2, 0, 5), (negative 2, 0, 1), (negative 2, 3, 1), and (negative 2, 3, 5).
Figure \PageIndex{2}: Evaluating a triple integral over a given rectangular box.

Solution

The order is not specified, but we can use the iterated integral in any order without changing the level of difficulty. Choose, say, to integrate y first, then x, and then z.

\begin{align*} \iiint\limits_{B} x^2 yz \,dV = \int_1^5 \int_{-2}^1 \int_0^3 [x^2 yz] \,dy \, dx \, dz \\ = \int_1^5 \int_{-2}^1 \left[ \left. x^2 \dfrac{y^3}{3} z\right|_0^3 \right] dx \, dz \\ = \int_1^5 \int_{-2}^1 \dfrac{y}{2} x^2 z \,dx \, dz \\ = \int_1^5 \left[ \left. \dfrac{9}{2} \dfrac{x^3}{3} z \right|_{-2}^1 \right] dz = \int_1^5 \dfrac{27}{2} z \, dz \\ = \left. \dfrac{27}{2} \dfrac{z^2}{2} \right|_1^5 = 162. \end{align*}

Now try to integrate in a different order just to see that we get the same answer. Choose to integrate with respect to x first, then z, then y

\begin{align*} \iiiint\limits_{B} x^2yz \,dV = \int_0^3 \int_1^5 \int_{-2}^1 [x^2yz] \,dx\, dz\, dy \\ = \int_0^3 \int_1^5 \left[ \left. \dfrac{x^3}{3} yz \right|_{-2}^1 \right] dz \,dy \\ =\int_0^3 \int_1^5 3yz \; dz \,dy \\ = \int_0^3 \left.\left[ 3y\dfrac{z^2}{2} \right|_1^5 \right] \,dy \\ = \int_0^3 36y \; dy \\ = \left. 36\dfrac{y^2}{2} \right|_0^3 =18(9-0) =162. \end{align*}

Exercise \PageIndex{1}

Evaluate the triple integral

\iint_B z \, \sin \, x \, \cos \, y \, dV\nonumber

where B = \big\{(x,y,z)\,|\,0 \leq x \leq \pi, \, \dfrac{3\pi}{2} \leq y \leq 2\pi, \, 1 \leq z \leq 3 \big\}.

Hint

Follow the steps in the previous example.

Answer

\iint_B z \, \sin \, x \, \cos \, y \, dV = 8 \nonumber

Triple Integrals over a General Bounded Region

We now expand the definition of the triple integral to compute a triple integral over a more general bounded region E in \mathbb{R}^3. The general bounded regions we will consider are of three types. First, let D be the bounded region that is a projection of E onto the xy-plane. Suppose the region E in \mathbb{R}^3 has the form

E = \big\{(x,y,z)\,|\,(x,y) \in D, u_1(x,y) \leq z \leq u_2(x,y) \big\}.

For two functions z = u_1(x,y) and u_2(x,y), such that u_1(x,y) \leq u_2(x,y) for all (x,y) in D as shown in the following figure.

In x y z space, there is a shape E with top surface z = u2(x, y) and bottom surface z = u1(x, y). The bottom projects onto the x y plane as region D.
Figure \PageIndex{3}: We can describe region E as the space between u_1(x,y) and u_2(x,y) above the projection D of E onto the xy-plane.

Triple Integral over a General Region

The triple integral of a continuous function f(x,y,z) over a general three-dimensional region

E = \big\{(x,y,z)\,|\,(x,y) \in D, \, u_1(x,y) \leq z \leq u_2(x,y) \big\}

in \mathbb{R}^3, where D is the projection of E onto the xy-plane, is

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

Similarly, we can consider a general bounded region D in the xy-plane and two functions y = u_1(x,z) and y = u_2(x,z) such that u_1(x,z) \leq u_2(x,z) for all 9x,z) in D. Then we can describe the solid region E in \mathbb{R}^3 as

E = \big\{(x,y,z)\,|\,(x,z) \in D, \, u_1(x,z) \leq z \leq u_2(x,z) \big\} where D is the projection of E onto the xy-plane and the triple integral is

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

Finally, if D is a general bounded region in the xy-plane and we have two functions x = u_1(y,z) and x = u_2(y,z) such that u_1(y,z) \leq u_2(y,z) for all (y,z) in D, then the solid region E in \mathbb{R}^3 can be described as

E = \big\{(x,y,z)\,|\,(y,z) \in D, \, u_1(y,z) \leq z \leq u_2(y,z) \big\} where D is the projection of E onto the xy-plane and the triple integral is

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

Note that the region D in any of the planes may be of Type I or Type II as described in previously. If D in the xy-plane is of Type I (Figure \PageIndex{4}), then

E = \big\{(x,y,z)\,|\,a \leq x \leq b, \, g_1(x) \leq y \leq g_2(x), \, u_1(x,y) \leq z \leq u_2(x,y) \big\}.

In x y z space, there is a complex shape E with top surface z = u2(x, y) and bottom surface z = u1(x, y). The bottom projects onto the xy plane as region D with boundaries x = a, x = b, y = g1(x), and y = g2(x).
Figure \PageIndex{4}: A box E where the projection D in the xy-plane is of Type I.

Then the triple integral becomes

\iiint_E f(x,y,z) \,dV = \int_a^b \int_{g_1(x)}^{g_2(x)} \int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) \,dz \, dy \, dx.

If D in the xy-plane is of Type II (Figure \PageIndex{5}), then

E = \big\{(x,y,z)\,|\,c \leq x \leq d, h_1(x) \leq y \leq h_2(x), \, u_1(x,y) \leq z \leq u_2(x,y) \big\}.

In x y z space, there is a complex shape E with top surface z = u2(x, y) and bottom surface z = u1(x, y). The bottom projects onto the xy plane as region D with boundaries y = c, y = d, x = h1(y), and x = h2(y).
Figure \PageIndex{5}: A box E where the projection D in the xy-plane is of Type II.

Then the triple integral becomes

\iiint_E f(x,y,z) \,dV = \int_{y=c}^{y=d} \int_{x=h_1(y)}^{x=h_2(y)} \int_{z=u_1(x,y)}^{z=u_2(x,y)} f(x,y,z)\,dz \, dx \, dy.

Example \PageIndex{3A}: Evaluating a Triple Integral over a General Bounded Region

Evaluate the triple integral of the function f(x,y,z) = 5x - 3y over the solid tetrahedron bounded by the planes x = 0, \, y = 0, \, z = 0, and x + y + z = 1.

Solution

Figure \PageIndex{6} shows the solid tetrahedron E and its projection D on the xy-plane.

In x y z space, there is a solid E with boundaries being the x y, z y, and x z planes and z = 1 minus x minus y. The points are the origin, (1, 0, 0), (0, 0, 1), and (0, 1, 0). Its surface on the x y plane is shown as being a rectangle marked D with line y = 1 minus x. Additionally, there is a vertical line shown on D.
Figure \PageIndex{6}: The solid E has a projection D on the xy-plane of Type I.

We can describe the solid region tetrahedron as

E = \big\{(x,y,z)\,|\,0 \leq x \leq 1, \, 0 \leq y \leq 1 - x, \, 0 \leq z \leq 1 - x - y \big\}. \nonumber

Hence, the triple integral is

\iiint_E f(x,y,z) \,dV = \int_{x=0}^{x=1} \int_{y=0}^{y=1-x} \int_{z=0}^{z=1-x-y} (5x - 3y) \,dz \, dy \, dx. \nonumber

To simplify the calculation, first evaluate the integral \displaystyle \int_{z=0}^{z=1-x-y} (5x - 3y) \,dz. We have

\int_{z=0}^{z=1-x-y} (5x - 3y) \,dz = (5x - 3y)z \bigg|_{z=0}^{z=1-x-y} = (5x - 3y)(1 - x - y).\nonumber

Now evaluate the integral

\int_{y=0}^{y=1-x} (5x - 3y)(1 - x - y) \,dy, \nonumber

obtaining

\int_{y=0}^{y=1-x} (5x - 3y)(1 - x - y)\,dy = \dfrac{1}{2}(x - 1)^2 (6x - 1).\nonumber

Finally evaluate

\int_{x=0}^{x=1} \dfrac{1}{2}(x - 1)^2 (6x - 1)\,dx = \dfrac{1}{12}.\nonumber

Putting it all together, we have

\iiint_E f(x,y,z)\,dV = \int_{x=0}^{x=1} \int_{y=0}^{y=1-x} \int_{z=0}^{z=1-x-y}(5x - 3y)\,dz \, dy \, dx = \dfrac{1}{12}.\nonumber

Just as we used the double integral \iint_D 1 \,dA to find the area of a general bounded region D we can use \iiint_E 1\,dV to find the volume of a general solid bounded region E. The next example illustrates the method.

Example \PageIndex{3B}: Finding a Volume by Evaluating a Triple Integral

Find the volume of a right pyramid that has the square base in the xy-plane [-1,1] \times [-1,1] and vertex at the point (0, 0, 1) as shown in the following figure.

In x y z space, there is a pyramid with a square base centered at the origin. The apex of the pyramid is (0, 0, 1).
Figure \PageIndex{7}: Finding the volume of a pyramid with a square base.

Solution

In this pyramid the value of z changes from 0 to 1 and at each height z the cross section of the pyramid for any value of z is the square

[-1 + z, \, 1 - z] \times [-1 + z, \, 1 - z].\nonumber

Hence, the volume of the pyramid is \iiint_E 1\,dV\nonumber where

E = \big\{(x,y,z)\,|\,0 \leq z \leq 1, \, -1 + z \leq y \leq 1 - z, \, -1 + z \leq x \leq 1 - z \big\}.\nonumber

Thus, we have

\begin{align*} \iiint_E 1\,dV = \int_{z=0}^{z=1} \int_{y=1+z}^{1-z} \int_{x=1+z}^{1-z} 1\,dx \, dy \, dz \\ = \int_{z=0}^{z=1} \int_{y=1+z}^{1-z} (2 - 2z)\, dy \, dz \\ = \int_{z=0}^{z=1}(2 - 2z)^2 \,dz = \dfrac{4}{3}. \end{align*}

Hence, the volume of the pyramid is \dfrac{4}{3} cubic units.

Exercise \PageIndex{3}

Consider the solid sphere E = \big\{(x,y,z)\,|\,x^2 + y^2 + z^2 = 9 \big\}. Write the triple integral \iiint_E f(x,y,z) \,dV\nonumber for an arbitrary function f as an iterated integral. Then evaluate this triple integral with f(x,y,z) = 1. Notice that this gives the volume of a sphere using a triple integral.

Hint

Follow the steps in the previous example. Use symmetry.

Answer

\begin{align*} \iiint_E 1\,dV = 8 \int_{x=-3}^{x=3} \int_{y=-\sqrt{9-z^2}}^{y=\sqrt{9-z^2}}\int_{z=-\sqrt{9-x^2-y^2}}^{z=\sqrt{9-x^2-y^2}} 1\,dz \, dy \, dx \\ = 36 \pi \,\text{cubic units}. \end{align*}

Changing the Order of Integration

As we have already seen in double integrals over general bounded regions, changing the order of the integration is done quite often to simplify the computation. With a triple integral over a rectangular box, the order of integration does not change the level of difficulty of the calculation. However, with a triple integral over a general bounded region, choosing an appropriate order of integration can simplify the computation quite a bit. Sometimes making the change to polar coordinates can also be very helpful. We demonstrate two examples here.

Example \PageIndex{4}: Changing the Order of Integration

Consider the iterated integral

\int_{x=0}^{x=1} \int_{y=0}^{y=x^2} \int_{z=0}^{z=y} f(x,y,z)\,dz \, dy \, dx.

The order of integration here is first with respect to z, then y, and then x. Express this integral by changing the order of integration to be first with respect to x, then z, and then y. Verify that the value of the integral is the same if we let f (x,y,z) =xyz.

Solution

The best way to do this is to sketch the region E and its projections onto each of the three coordinate planes. Thus, let

E = \big\{(x,y,z)\,|\,0 \leq x \leq 1, \, 0 \leq y \leq x^2, \, 0 \leq z \leq y \big\}.\nonumber

and

\int_{x=0}^{x=1} \int_{y=0}^{y=x^2} \int_{z=0}^{z=x^2} f(x,y,z) \,dz \, dy \, dx = \iiint_E f(x,y,z)\,dV.\nonumber

We need to express this triple integral as

\int_{y=c}^{y=d} \int_{z=v_1(y)}^{z=v_2(y)} \int_{x=u_1(y,z)}^{x=u_2(y,z)} f(x,y,z)\,dx \, dz \, dy.\nonumber

Knowing the region E we can draw the following projections (Figure \PageIndex{8}):

on the xy-plane is D_1 = \big\{(x,y)\,|\, 0 \leq x \leq 1, \, 0 \leq y \leq x^2 \big\} = \{ (x,y) \,|\, 0 \leq y \leq 1, \, \sqrt{y} \leq x \leq 1 \big\},

on the yz-plane is D_2 = \big\{(y,z) \,|\, 0 \leq y \leq 1, \, 0 \leq z \leq y^2 \big\}, and

on the xz-plane is D_3 = \big\{(x,z) \,|\, 0 \leq x \leq 1, \, 0 \leq z \leq x^2 \big\}.

Three similar versions of the following graph are shown: In the x y plane, a region D1 is bounded by the x axis, the line x = 1, and the curve y = x squared. In the second version, region D2 on the z y plane is shown with equation z = y squared. And in the third version, region D3 on the x z plane is shown with equation z = x squared.
Figure \PageIndex{8}. The three cross sections of E on the three coordinate planes.

Now we can describe the same region E as \big\{(x,y,z) \,|\, 0 \leq y \leq 1, \, 0 \leq z \leq y^2, \, \sqrt{y} \leq x \leq 1 \big\}, and consequently, the triple integral becomes

\int_{y=c}^{y=d} \int_{z=v_1(y)}^{z=v_2(y)} \int_{x=u_1(y,z)}^{x=u_2(y,z)} f(x,y,z)\,dx \, dz \, dy = \int_{y=0}^{y=1} \int_{z=0}^{z=x^2} \int_{x=\sqrt{y}}^{x=1} f(x,y,z)\,dx \, dz \, dy

Now assume that f (x,y,z) = xyz in each of the integrals. Then we have

\begin{align*} \int_{x=0}^{x=1} \int_{y=0}^{y=x^2} \int_{z=0}^{z=y^2} xyz \, dz \, dy \, dx \\ = \int_{x=0}^{x=1} \int_{y=0}^{y=x^2} \left. \left[xy \dfrac{z^2}{2} \right|_{z=0}^{z=y^2} \right] \, dy \, dx = \int_{x=0}^{x=1} \int_{y=0}^{y=x^2} \left( x \dfrac{y^5}{2}\right) dy \, dx = \int_{x=0}^{x=1} \left. \left[ x\dfrac{y^6}{12} \right|_{y=0}^{y=x^2}\right] dx = \int_{x=0}^{x=1} \dfrac{x^{13}}{12} dx = \left. \dfrac{x^{14}}{168}\right|_{x=0}^{x=1} = \dfrac{1}{168}, \end{align*}

\begin{align*} \int_{y=0}^{y=1} \int_{z=0}^{z=y^2} \int_{x=\sqrt{y}}^{x=1} xyz \, dx \, dz \, dy \\ = \int_{y=0}^{y=1} \int_{z=0}^{z=y^2} \left.\left[yz \dfrac{x^2}{2} \right|_{\sqrt{y}}^{1} \right] dz \, dy = \int_{y=0}^{y=1} \int_{z=0}^{z=y^2} \left( \dfrac{yz}{2} - \dfrac{y^2z}{2} \right) dz \, dy = \int_{y=0}^{y=1} \left. \left[ \dfrac{yz^2}{4} - \dfrac{y^2z^2}{4} \right|_{z=0}^{z=y^2} \right] dy = \int_{y=0}^{y=1} \left(\dfrac{y^5}{4} - \dfrac{y^6}{4} \right) dy = \left. \left(\dfrac{y^6}{24} - \dfrac{y^7}{28} \right) \right|_{y=0}^{y=1} = \dfrac{1}{168}. \end{align*}

The answers match.

Exercise \PageIndex{4}

Write five different iterated integrals equal to the given integral

\int_{z=0}^{z=4} \int_{y=0}^{y=4-z} \int_{x=0}^{x=\sqrt{y}} f(x,y,z) \, dx \, dy \, dz.\nonumber

Hint

Follow the steps in the previous example, using the region E as \big\{(x,y,z) \,|\, 0 \leq z \leq 4, \, 0 \leq y \leq 4 - z, \, 0 \leq x \leq \sqrt{y} \big\}, and describe and sketch the projections onto each of the three planes, five different times.

Answer

(i) \, \int_{z=0}^{z=4} \int_{x=0}^{x=\sqrt{4-z}} \int_{y=x^2}^{y=4-z} f(x,y,z) \, dy \, dx \, dz, \, (ii) \, \int_{y=0}^{y=4} \int_{z=0}^{z=4-y} \int_{x=0}^{x=\sqrt{y}} f(x,y,z) \,dx \, dz \, dy, \,(iii) \, \int_{y=0}^{y=4} \int_{x=0}^{x=\sqrt{y}} \int_{z=0}^{Z=4-y} f(x,y,z) \,dz \, dx \, dy, \, \nonumber

(iv) \, \int_{x=0}^{x=2} \int_{y=x^2}^{y=4} \int_{z=0}^{z=4-y} f(x,y,z) \,dz \, dy \, dx, \, (v) \int_{x=0}^{x=2} \int_{z=0}^{z=4-x^2} \int_{y=x^2}^{y=4-z} f(x,y,z) \,dy \, dz \, dx \nonumber

Example \PageIndex{5}: Changing Integration Order and Coordinate Systems

Evaluate the triple integral

\iiint_{E} \sqrt{x^2 + z^2} \,dV, \nonumber

where E is the region bounded by the paraboloid y = x^2 + z^2 (Figure \PageIndex{9}) and the plane y = 4.

The paraboloid y = x squared + z squared is shown opening up along the y axis to y = 4.
Figure \PageIndex{9}. Integrating a triple integral over a paraboloid.

Solution

The projection of the solid region E onto the xy-plane is the region bounded above by y = 4 and below by the parabola y = x^2 as shown.

In the x y plane, the graph of y = x squared is shown with the line y = 4 intersecting the graph at (negative 2, 4) and (2, 4).
Figure \PageIndex{10}. Cross section in the xy-plane of the paraboloid in Figure \PageIndex{9}.

Thus, we have

E = \big\{(x,y,z) \,|\, -2 \leq x \leq 2, \, x^2 \leq y \leq 4, \, -\sqrt{y - x^2} \leq z \sqrt{y - x^2} \big\}.\nonumber

The triple integral becomes

\iiint_E \sqrt{x^2 + z^2} \,dV = \int_{x=-2}^{x=2} \int_{y=x^2}^{y=4} \int_{z=-\sqrt{y-x^2}}^{z=\sqrt{y-x^2}} \sqrt{x^2 + z^2} \,dz \, dy \, dx.\nonumber

This expression is difficult to compute, so consider the projection of E onto the xz-plane. This is a circular disc x^2 + z^2 \leq 4. So we obtain

\iiint_E \sqrt{x^2 + z^2} \,dV = \int_{x=-2}^{x=2} \int_{y=x^2}^{y=4} \int_{z=-\sqrt{y-x^2}}^{z=\sqrt{y-x^2}} \sqrt{x^2 + z^2} \,dz \, dy \, dx = \int_{x=-2}^{x=2} \int_{z=-\sqrt{4-x^2}}^{z=\sqrt{4-x^2}} \int_{y=x^2+z^2}^{y=4} \sqrt{x^2 + z^2} \,dy \, dz \, dx.\nonumber

Here the order of integration changes from being first with respect to z then y and then x to being first with respect to y then to z and then to x. It will soon be clear how this change can be beneficial for computation. We have

\int_{x=-2}^{x=2} \int_{z=\sqrt{4-x^2}}^{z=\sqrt{4-x^2}} \int_{y=x^2+z^2}^{y=4} \sqrt{x^2 + z^2} \,dy \, dz \, dx = \int_{x=-2}^{x=2} \int_{z=-\sqrt{4-x^2}}^{z=\sqrt{4-x^2}} (4 - x^2 - z^2) \sqrt{x^2 + z^2} \,dz \, dx.\nonumber

Now use the polar substitution x = r \, \cos \, \theta, \, z = r \, \sin \, \theta, and dz \, dx = r \, dr \, d\theta in the xz-plane. This is essentially the same thing as when we used polar coordinates in the xy-plane, except we are replacing y by z. Consequently the limits of integration change and we have, by using r^2 = x^2 + z^2,

\int_{x=-2}^{x=2} \int_{z=-\sqrt{4-x^2}}^{z=\sqrt{4-x^2}} (4 - x^2 - z^2) \sqrt{x^2 + z^2}\,dz \, dx = \int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=2} (4 - r^2) rr \, dr \, d\theta = \int_0^{2\pi} \left. \left[ \dfrac{4r^3}{3} - \dfrac{r^5}{5} \right|_0^2 \right] \, d\theta = \int_0^{2\pi} \dfrac{64}{15} \,d\theta = \dfrac{128\pi}{15}\nonumber

Average Value of a Function of Three Variables

Recall that we found the average value of a function of two variables by evaluating the double integral over a region on the plane and then dividing by the area of the region. Similarly, we can find the average value of a function in three variables by evaluating the triple integral over a solid region and then dividing by the volume of the solid.

Average Value of a Function of Three Variables

If f(x,y,z) is integrable over a solid bounded region E with positive volume V \, (E), then the average value of the function is

f_{ave} = \dfrac{1}{V \, (E)} \iiint_E f(x,y,z) \, dV.

Note that the volume is

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

Example \PageIndex{6}: Finding an Average Temperature

The temperature at a point (x,y,z) of a solid E bounded by the coordinate planes and the plane x + y + z = 1 is T(x,y,z) = (xy + 8z + 20) \, \text{°}\text{C} . Find the average temperature over the solid.

Solution

Use the theorem given above and the triple integral to find the numerator and the denominator. Then do the division. Notice that the plane x + y + z = 1 has intercepts (1,0,0), \, (0,1,0), and (0,0,1). The region E looks like

E = \big\{(x,y,z) \,|\, 0 \leq x \leq 1, \, 0 \leq y \leq 1 - x, \, 0 \leq z \leq 1 - x - y \big\}.\nonumber

Hence the triple integral of the temperature is

\iiint_E f(x,y,z) \,dV = \int_{x=0}^{x=1} \int_{y=0}^{y=1-x} \int_{z=0}^{z=1-x-y} (xy + 8z + 20) \, dz \, dy \, dx = \dfrac{147}{40}. \nonumber

The volume evaluation is

V \, (E) = \iiint_E 1\,dV = \int_{x=0}^{x=1} \int_{y=0}^{y=1-x} \int_{z=0}^{z=1-x-y} 1 \,dz \, dy \, dx = \dfrac{1}{6}. \nonumber

Hence the average value is

T_{ave} = \dfrac{147/40}{1/6} = \dfrac{6(147)}{40} = \dfrac{441}{20} \, \text{°}\text{C} \nonumber.

Exercise \PageIndex{6}

Find the average value of the function f(x,y,z) = xyz over the cube with sides of length 4 units in the first octant with one vertex at the origin and edges parallel to the coordinate axes.

Hint

Follow the steps in the previous example.

Answer

f_{ave} = 8

Key Concepts

  • To compute a triple integral we use Fubini’s theorem, which states that if f(x,y,z) is continuous on a rectangular box B = [a,b] \times [c,d] \times [e,f], then \iiint_B f(x,y,z) \,dV = \int_e^f \int_c^d \int_a^b f(x,y,z) \, dx \, dy \, dz \nonumber and is also equal to any of the other five possible orderings for the iterated triple integral.
  • To compute the volume of a general solid bounded region E we use the triple integral V \, (E) = \iiint_E 1 \,dV. \nonumber
  • Interchanging the order of the iterated integrals does not change the answer. As a matter of fact, interchanging the order of integration can help simplify the computation.
  • To compute the average value of a function over a general three-dimensional region, we use f_{ave} = \dfrac{1}{V \, (E)} \iiint_E f(x,y,z) \,dV. \nonumber

Key Equations

  • Triple integral

\lim_{l,m,n \rightarrow \infty} \sum_{i=1}^l \sum_{j=1}^m \sum_{k=1}^n f(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*) \,\Delta x \Delta y \Delta z = \iiint_B f(x,y,z) \,dV \nonumber

Glossary

triple integral
the triple integral of a continuous function f(x,y,z) over a rectangular solid box B is the limit of a Riemann sum for a function of three variables, if this limit exists

Contributors

  • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


This page titled 4.5: Triple Integrals is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.

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