Triple Integrals

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.

Integrable Functions of Three Variables

We can define a rectangular box $B$ in ${{\mathbb{R}}^3}$ as ${B} = {\left \{{(x,y,z)}{\mid}{a} \ {\leq} \ {x} \ {\leq} \ {b,c} \ {\leq} \ {y} \ {\leq} \ {d,e} \ {\leq} \ {z} \ {\leq} \ {f} \right \}}$ . We follow a similar procedure to what we did in Double Integrals over Rectangular Regions. We divide the interval $[a,b]$ into $l$ subintervals $[{{x}_{i-1}},{x_i}]$ of equal length ${{\Delta}{x}} = {\frac{{x_i}-{x_{i-1}}}{l}}$ , divide the interval $[c,d]$ into $m$ subintervals $[{{y}_{i-1}},{y_i}]$ of equal length ${{\Delta}{y}} = {\frac{{y_j}-{y_{j-1}}}{m}}$ , and divide the interval $[e,f]$ into $n$ subintervals $[{{z}_{i-1}},{z_i}]$ of equal length ${{\Delta}{z}} = {\frac{{z_k}-{z_{k-1}}}{n}}$ . Then the rectangular box $B$ is subdivided into $lmn$ subboxes ${{B}_{ijk}} = [{{x}_{i-1}},{x_i}] \ {\times} \ [{{y}_{i-1}},{y_i}] \ {\times} \ [{{z}_{i-1}},{z_i}]$ , as shown in Figure 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 1. A rectangular box in $\mathbb{R}^{3}$ divided into subboxes by planes parallel to the coordinate planes.

For each $i$ , $j$ , and $k$ , consider a sample point ${({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}$ in each sub-box ${{B}_{ijk}}$ . We see that its volume is ${{\Delta}{V}} = {{\Delta}{x}}{{\Delta}{y}}{{\Delta}{z}}$ . Form the triple Riemann sum

$\Large{\displaystyle\sum_{i=1}^{l}}{\displaystyle\sum_{j=1}^{m}}{\displaystyle\sum_{k=1}^{n}}{f}{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}{{\Delta}{x}}{{\Delta}{y}}{{\Delta}{z}}$ .

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

$\Large{\displaystyle\lim_{{l,m,n}{\rightarrow}{\infty}}} \ {\displaystyle\sum_{i=1}^{l}}{\displaystyle\sum_{j=1}^{m}}{\displaystyle\sum_{k=1}^{n}} \ {f}{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}{{\Delta}{x}}{{\Delta}{y}}{{\Delta}{z}} = \underset{B}{\displaystyle\iiint}{f}{(x,y,z)}{dV}$

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 thereom for triple integrals exists.

theorem: 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

$\Large{\underset{B}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = {\displaystyle\int_{e}^{f}}{\displaystyle\int_{c}^{d}}{\displaystyle\int_{a}^{b}}{f}{(x,y,z)}{dx} \ {dy} \ {dz}}$ .

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:

$\large{\begin{align} {\displaystyle\int_{e}^{f}}{\displaystyle\int_{c}^{d}}{\displaystyle\int_{a}^{b}}{f}{(x,y,z)}{dx} \ {dy} \ {dz}&={\displaystyle\int_{e}^{f}}\left({\displaystyle\int_{c}^{d}}\left({\displaystyle\int_{a}^{b}}{f}{(x,y,z)}{dx}\right){dy}\right){dz}={\displaystyle\int_{c}^{d}}\left({\displaystyle\int_{e}^{f}}\left({\displaystyle\int_{a}^{b}}{f}{(x,y,z)}{dx}\right){dz}\right){dy} \\ &={\displaystyle\int_{a}^{b}}\left({\displaystyle\int_{e}^{f}}\left({\displaystyle\int_{c}^{d}}{f}{(x,y,z)}{dy}\right){dz}\right){dx}={\displaystyle\int_{e}^{f}}\left({\displaystyle\int_{a}^{b}}\left({\displaystyle\int_{c}^{d}}{f}{(x,y,z)}{dy}\right){dx}\right){dz} \\ &={\displaystyle\int_{c}^{e}}\left({\displaystyle\int_{a}^{b}}\left({\displaystyle\int_{e}^{f}}{f}{(x,y,z)}{dz}\right){dx}\right){dy}={\displaystyle\int_{a}^{b}}\left({\displaystyle\int_{c}^{e}}\left({\displaystyle\int_{e}^{f}}{f}{(x,y,z)}{dz}\right){dy}\right){dx}. \end{align}}$
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: evaluating a triple integral 1

Evaluate the triple integral ${\displaystyle\int^{z=1}_{z=0}}{\displaystyle\int^{y=4}_{y=2}}{\displaystyle\int^{x=5}_{x=-1}}{(x+yz^2)}{dx} \ {dy} \ {dz}$ .

Show Solution

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

$\hspace{3cm}\begin{align} &\hspace{0.75cm}\displaystyle\int^{z=1}_{z=0}\displaystyle\int^{y=4}_{y=2}\displaystyle\int^{x=5}_{x=-1}(x+yz^2)dx \ dy \ dz \\ &=\displaystyle\int^{z=1}_{z=0}\displaystyle\int^{y=4}_{y=2}\left[\frac{x^2}2+xyz^2\bigg|^{x=5}_{x=-1}\right]dy \ dz &\quad &\text{Integrate with respect to }x. \\ &=\displaystyle\int^{z=1}_{z=0}\displaystyle\int^{y=4}_{y=2}[12+6yz^2] \ dy \ dz &\quad &\text{Evaluate.} \\ &=\displaystyle\int^{z=1}_{z=0}\left[12y+6\frac{y^2}2x^2\bigg|^{y=4}_{y=2}\right]{dz} &\quad &\text{Integrate with respect to }y. \\ &={\displaystyle\int^{z=1}_{z=0}}[24+36z^2]{dz} &\quad &\text{Evaluate}. \\ &=\left[24z+36\frac{z^3}3\right]_{z=0}^{z=1}=36. \end{align}$

Example: evaluating a triple integral 2

Evaluate the triple integral $\underset{B}{\displaystyle\iiint}{x^2}{yz}{dV}$ where

${B} = {\left \{{(x,y,z)}{\mid}{-2} \ {\leq} \ {x} \ {\leq} \ {1,0} \ {\leq} \ {y} \ {\leq} \ {3,1} \ {\leq} \ {z} \ {\leq} \ {5} \right \}}$ as shown in the following figure.

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 2. Evaluating a triple integral over a given rectangular box.

Show 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 $z$ .

$\hspace{2cm}\begin{align} \underset{B}{\displaystyle\iiint}{x^2}{yz}{dV}&=\displaystyle\int_1^5\displaystyle\int_{-2}^1\displaystyle\int_0^3[x^2yz]dy \ dx \ dz =\displaystyle\int_1^5\displaystyle\int_{-2}^1\left[x^2\frac{y^2}2z\bigg|_0^3\right]dx \ dz \\ &=\displaystyle\int_1^5\displaystyle\int_{-2}^1\frac92x^2z \ dx \ dz = \displaystyle\int_1^5\left[\frac92\frac{x^3}3z\bigg|_{-2}^1\right]dz=\displaystyle\int_1^5\frac{27}2\frac{z^2}2\bigg|_1^5=162. \end{align}$

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 $z$ , and then $y$ .

$\hspace{2cm}\begin{align} \underset{B}{\displaystyle\iiint}{x^2}{yz}{dV}&=\displaystyle\int_0^3\displaystyle\int_1^5\displaystyle\int_{-2}^1[x^2yz]dx \ dz \ dy =\displaystyle\int_0^3\displaystyle\int_1^5\left[\frac{x^3}3yz\bigg|_{-2}^1\right]dz \ dy \\ &=\displaystyle\int_0^3\displaystyle\int_1^53yz \ dz \ dy =\displaystyle\int_0^3\left[3y\frac{z^2}2\bigg|_1^5\right]dy=\displaystyle\int_0^336y \ dy = 36\frac{y^2}2\bigg|_0^3=18(9-0)=162. \end{align}$

try it

Evaluate the triple integral $\underset{B}{\displaystyle\iiint}{z} \ {\sin \ x} \ {\cos \ y} \ {dV}$ where

${B} = {\left \{{(x,y,z)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {{\pi},{\frac{3{\pi}}{2}}} \ {\leq} \ {y} \ {\leq} \ {2{\pi},1} \ {\leq} \ {z} \ {\leq} \ {3} \right \}}$ .

Show Solution

$\underset{B}{\displaystyle\iiint}z\sin{x}\cos{y} \ dV=8$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.23” here (opens in new window).

Try It

Triple Integrals over a General Bounded Region

We now expand the definition of the triple integral to compute a triple integral over a more 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

$\Large{E} = \left \{{(x,y,z)}{\mid}{(x,y)} \ {\in} \ {{D},{u_1}{(x,y)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(x,y)} \right \}$ .

For two functions $z=u_1(x, y)$ and $z=u_2(x,y)$ , 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 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.

theorem: triple integral over a general region

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

$\large{E} = \left \{{(x,y,z)}{\mid}{(x,y)} \ {\in} \ {{D},{u_1}{(x,y)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(x,y)} \right \}$

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

$\large{\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = \underset{D}{\displaystyle\iint}{\left [{\displaystyle\int^{{u_2}{(x,y)}}_{{u_1}{(x,y)}}}{f}{(x,y,z)}{dz} \right ]}{dA}}$ .

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 $(x, z)$ in $D$ . Then we can describe the solid region $E$ in ${{\mathbb{R}}^3}$ as

${E} = \left \{{(x,y,z)}{\mid}{(x,z)} \ {\in} \ {{D},{u_1}{(x,z)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(x,z)} \right \}$

where $D$ is the projection of $E$ onto the $xy$ -plane and the triple integral is

$\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = \underset{D}{\displaystyle\iint}{\left [{\displaystyle\int^{{u_2}{(x,z)}}_{{u_1}{(x,z)}}}{f}{(x,y,z)}{dy} \right ]}{dA}$ .

Finally, if $D$ is a general bounded region in the $yz$ -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

${E} = \left \{{(x,y,z)}{\mid}{(y,z)} \ {\in} \ {{D},{u_1}{(y,z)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(y,z)} \right \}$

where $D$ is the projection of $E$ onto the $yz$ -plane and the triple integral is

$\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = \underset{D}{\displaystyle\iint}{\left [{\displaystyle\int^{{u_2}{(y,z)}}_{{u_1}{(y,z)}}}{f}{(x,y,z)}{dx} \right ]}{dA}$ .

Note that the region $D$ in any of the planes may be of Type I or Type II as described in Double Integrals over General Regions. If $D$ in the $xy$ -plane is of Type I (Figure 4), then

${E} = \left \{{(x,y,z)}{\mid}{a} \ {\leq} \ {x} \ {\leq} \ {b,{g_1}{(x)}} \ {\leq} \ {y} \ {\leq} \ {g_2{(x)}},{{u_1}{(x,y)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(x,y)} \right \}$ .

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 4. A box $E$ where the projection $D$ in the $xy$ -plane is of Type I.

Then the triple integral becomes

$\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = {\displaystyle\int_{a}^{b}} \ {\displaystyle\int_{{g_1}{(x)}}^{{g_2}{(x)}}} \ {\displaystyle\int_{{u_1}{(x,y)}}^{{u_2}{(x,y)}}}{f}{(x,y,z)}{dz}{dy}{dx}$ .

If $D$ in the $xy$ -plane is of Type II (Figure 5), then

${E} = \left \{{(x,y,z)}{\mid}{c} \ {\leq} \ {x} \ {\leq} \ {d,{h_1}{(x)}} \ {\leq} \ {y} \ {\leq} \ {h_2{(x)}},{{u_1}{(x,y)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(x,y)} \right \}$ .

Figure 5. A box $E$ where the projection $D$ in the $xy$ -plane is of Type II.

Then the triple integral becomes

$\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = {\displaystyle\int^{y=d}_{y=c}}{\displaystyle\int^{x={{h_2}{(y)}}}_{x={h_1}{(y)}}}{\displaystyle\int^{z={{u_2}{(x,y)}}}_{z={u_1}{(x,y)}}}{f}{(x,y,z)}{dz}{dx}{dy}$ .

Example: 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 $x+y+z=1$ .

Show Solution

Figure 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 6. The solid $E$ has a projection $D$ on the $xy$ -plane of Type I.

We can describe the solid region tetrahedron as

${E} = \left \{{(x,y,z)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {1,0} \ {\leq} \ {y} \ {\leq} \ {1-x},{0} \ {\leq} \ {z} \ {\leq} \ {1-x-y} \right \}$ .

Hence, the triple integral is

$\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = {\displaystyle\int^{x=1}_{x=0}}{\displaystyle\int^{y={1-x}}_{y=0}}{\displaystyle\int^{z={1-x-y}}_{z=0}}{(5x-3y)}{dz}{dy}{dx}$ .

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

${\displaystyle\int^{z=1-x-y}_{z=0}}{(5x-3y)}{dz} = {(5x-3y)}{(1-x-y)}$ .

Now evaluate the integral ${\displaystyle\int^{y=1-x}_{y=0}}{(5x-3y)}{(1-x-y)}{dy}$ , obtaining

${\displaystyle\int^{y=1-x}_{y=0}}{(5x-3y)}{(1-x-y)}{dy} = {\frac{1}{2}}{(x-1)^2}{(6x-1)}$ .

Finally, evaluate

${\displaystyle\int^{x=1}_{x=0}}{\frac{1}{2}}{(x-1)^2}{(6x-1)}{dx} = {\frac{1}{12}}$ .

Putting it all together, we have

$\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = {\displaystyle\int^{x=1}_{x=0}}{\displaystyle\int^{y={1-x}}_{y=0}}{\displaystyle\int^{z={1-x-y}}_{z=0}}{(5x-3y)}{dz}{dy}{dx} = {\frac{1}{12}}$ .

Just as we used the double integral $\underset{D}{\displaystyle\iint}{1}{dA}$ to find the area of a general bounded region $D$ , we can use $\underset{E}{\displaystyle\iiint}{1}{dV}$ to find the volume of a general solid bounded region $E$ . The next example illustrates the method.

Example: 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 7. Finding the volume of a pyramid with a square base.

Show 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 $[-1+z,1-z] \ {\times} \ [-1+z,1-z]$ . Hence, the volume of the pyramid is $\underset{E}{\displaystyle\iiint}{1}{dV}$ where

${E} = \left \{{(x,y,z)}{\mid}{0} \ {\leq} \ {z} \ {\leq} \ {1,-1+z} \ {\leq} \ {y} \ {\leq} \ {1-z,-1+z} \ {\leq} \ {x} \ {\leq} \ {1-z} \right \}$ .

Thus, we have

$\underset{E}{\displaystyle\iiint}{1}{dV} = {\displaystyle\int^{z=1}_{z=0}}{\displaystyle\int^{y={1-z}}_{y=-1+z}}{\displaystyle\int^{x={1-z}}_{x=1+z}}{1}{dx}{dy}{dz} = {\displaystyle\int^{z=1}_{z=0}}{\displaystyle\int^{y={1-z}}_{y=-1+z}}{(2-2z)}{dy}{dz} = {\displaystyle\int^{z=1}_{z=0}}{(2-2z)^2}{dz} = {\frac{4}{3}}$ .

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

try it

Consider the solid sphere ${E} = {(x,y,z)}{\mid}\left \{{{x^2}+{y^2}+{z^2}} = {9} \right \}$ . Write the triple integral $\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV}$ 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.

Show Solution

$\underset{E}{\displaystyle\iiint}1 \ dV=8\displaystyle\int_{x=-3}^{x=3}\displaystyle\int_{y=-\sqrt{9-x^2}}^{y=\sqrt{9-x^2}}\displaystyle\int_{z=-\sqrt{9-x^2-y^2}}^{z=\sqrt{9-x^2-y^2}}1 \ dz \ dy \ dx = 36\pi.$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.24” here (opens in new window).

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: changing the order of integration

Consider the iterated integral

${\displaystyle\int^{x=1}_{x=0}} \ {\displaystyle\int^{y=x^2}_{y=0}} \ {\displaystyle\int^{z=y^2}_{z=0}}{f}{(x,y,z)}{dz}{dy}{dx}$ .

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 $f(x, y, z)=xyz$ .

Show 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

${E} = \left \{{(x,y,z)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {1,0} \ {\leq} \ {y} \ {\leq} \ {x^2,0} \ {\leq} \ {z} \ {\leq} \ {y^2} \right \}$ .

and

${\displaystyle\int^{x=1}_{x=0}} \ {\displaystyle\int^{y=x^2}_{y=0}} \ {\displaystyle\int^{z=y^2}_{z=0}}{f}{(x,y,z)}{dz}{dy}{dx} = \underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV}$ .

We need to express this triple integral as

${\displaystyle\int^{y=d}_{y=c}} \ {\displaystyle\int^{z={{v_2}{(y)}}}_{z={{v_1}{(y)}}}} \ {\displaystyle\int^{x={{u_2}{(y,z)}}}_{x={{u_1}{(y,z)}}}}{f}{(x,y,z)}{dx}{dz}{dy}$ .

Knowing the region $E$ we can draw the following projections (Figure 8):

on the $xy$ -plane is ${D_1} = {\left \{{(x,y)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {1,0} \ {\leq} \ {y} \ {\leq} \ {x^2} \right \}} = {\left \{{(x,y)}{\mid}{0} \ {\leq} \ {y} \ {\leq} \ {1,{\sqrt{y}}} \ {\leq} \ {x} \ {\leq} \ {1} \right \}}$ ,

on the $yz$ -plane is ${D_2} = {\left \{{(y,z)}{\mid}{0} \ {\leq} \ {y} \ {\leq} \ {1,0} \ {\leq} \ {z} \ {\leq} \ {y^2} \right \}}$ , and

on the $xz$ -plane is ${D_3} = {\left \{{(x,z)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {1,0} \ {\leq} \ {z} \ {\leq} \ {x^2} \right \}}$ .

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 cubed.

Figure 8. The three cross sections of $E$ on the three coordinate planes.

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

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

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

[latex]\begin{align}

\displaystyle\int^{x=1}_{x=0}\displaystyle\int^{y=x^2}_{y=0}\displaystyle\int^{z=y^2}_{z=0}xyz{dz}{dy}{dx}&=\displaystyle\int^{x=1}_{x=0}\displaystyle\int^{y=x^2}_{y=0}\left[xy\frac{z^2}2\bigg|_{z=0}^{z=y^2}\right]{dy}{dx} \\ &=\displaystyle\int^{x=1}_{x=0}\displaystyle\int^{y=x^2}_{y=0}\left(x\frac{y^5}2\right)dy \ dx \\ &=\displaystyle\int^{x=1}_{x=0}\left[x\frac{y^6}{12}\bigg|^{y=x^2}_{y=0}\right]dx \\ & =\displaystyle\int^{x=1}_{x=0}\frac{x^{13}}{12}dx =\frac1{168}, \\ \displaystyle\int^{y=1}_{y=0}\displaystyle\int^{z=y^2}_{z=0}\displaystyle\int^{x=1}_{x=\sqrt{y}}xyz{dx}{dz}{dy} &=\displaystyle\int^{y=1}_{y=0}\displaystyle\int^{z=y^2}_{z=0}\left[yz\frac{x^2}2\bigg|^{1}_{\sqrt{y}}\right]dz \ dy\\ &=\displaystyle\int^{y=1}_{y=0}\displaystyle\int^{z=y^2}_{z=0}\left(\frac{yz}2-\frac{y^2z}2\right)dz \ dy \\ &=\displaystyle\int^{y=1}_{y=0}\left[\frac{yz^2}4-\frac{y^2z^2}4\bigg|^{z=y^2}_{z=0}\right]dy \\ &=\displaystyle\int^{y=1}_{y=0}\left(\frac{y^5}4-\frac{y^6}4\right)dy=\frac1{168}. \end{align}[/latex]

The answers match.

try it

Write five different iterated integrals equal to the given integral ${\displaystyle\int^{z=4}_{z=0}} \ {\displaystyle\int^{y={4-z}}_{y=0}} \ {\displaystyle\int^{x={\sqrt{y}}}_{x=0}}{f}{(x,y,z)}{dx}{dy}{dz}$ .

Show Solution

(i) $\displaystyle\int_{z=0}^{z=4}\displaystyle\int_{x=0}^{x=\sqrt{x-4}}\displaystyle\int_{y=x^2}^{y=4-z}f(x,y,z)dy \ dx \ dz$ ,

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

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

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

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

Example: changing integration order and coordinate systems

Evaluate the triple integral $\underset{E}{\displaystyle\iiint}{\sqrt{{x^2}+{z^2}}}{dV}$ , where $E$ is the region bounded by the paraboloid $y=x^{2}+z^{2}$ (Figure 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 9. Integrating a triple integral over a paraboloid.

Show 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 10. Cross section in the $xy$ -plane of the paraboloid in Figure 9.

Thus, we have

${E} = {\left \{ {(x,y,z)}{\mid}{-2} \ {\leq} \ {x} \ {\leq} \ {2,{x^2}} \ {\leq} \ {y} \ {\leq} \ {{4},-{\sqrt{y-x^2}}} \ {\leq} \ {z} \ {\leq} \ {\sqrt{y-x^2}} \right \}}$ .

The triple integral becomes

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

$\underset{E}{\displaystyle\iiint}{\sqrt{x^2+z^2}dV} = {\displaystyle\int^{x=2}_{x=-2}} \ {\displaystyle\int^{y=4}_{y=x^2}} \ {\displaystyle\int^{z={\sqrt{y-x^2}}}_{z=-{\sqrt{y-x^2}}}}{\sqrt{{x^2}+{z^2}}}{dz}{dy}{dx} = {\displaystyle\int^{x=2}_{x=-2}} \ {\displaystyle\int^{z={\sqrt{4-x^2}}}_{z=-{\sqrt{4-x^2}}}} \ {\displaystyle\int^{y=4}_{y=x^2+z^2}}{\sqrt{{x^2}+{z^2}}}{dy}{dz}{dx}$ .

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

${\displaystyle\int^{x=2}_{x=-2}} \ {\displaystyle\int^{z={\sqrt{4-x^2}}}_{z=-{\sqrt{4-x^2}}}} \ {\displaystyle\int^{y=4}_{y=x^2+z^2}}{\sqrt{{x^2}+{z^2}}}{dy}{dz}{dx} = {\displaystyle\int^{x=2}_{x=-2}} \ {\displaystyle\int^{z={\sqrt{4-x^2}}}_{z=-{\sqrt{4-x^2}}}}{(4-x^2-z^2)}{\sqrt{{x^2}+{z^2}}}{dz}{dx}$ .

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 $r^{2}=x^{2}+z^{2}$ ,

${\displaystyle\int^{x=2}_{x=-2}} \ {\displaystyle\int^{z={\sqrt{4-x^2}}}_{z=-{\sqrt{4-x^2}}}}{(4-x^2-z^2)}{\sqrt{{x^2}+{z^2}}}{dz}{dx} = {\displaystyle\int^{{\theta} = 2{\pi}}_{{\theta} = 0}} \ {\displaystyle\int^{r=2}_{r=0}}{(4-{r^2})}{rr}{dr}{{d}{\theta}}$

$= {\displaystyle\int^{2{\pi}}_{0}} \ {\left [ {\frac{4r^3}{3}}-{\frac{r^5}{5}}{{\bigg\vert}^2_0} \right ]} \ {{d}{\theta}} = {\displaystyle\int^{2{\pi}}_{0}} \ {\frac{64}{15}}{{d}{\theta}} = {\frac{{128}{\pi}}{15}}$ .

Module 5: Multiple Integration

Learning Objectives

Integrable Functions of Three Variables

definition

theorem: fubini’s theorem for triple integrals

Example: evaluating a triple integral 1

Example: evaluating a triple integral 2

try it

Try It

Triple Integrals over a General Bounded Region

theorem: triple integral over a general region

Example: evaluating a triple integral over a general bounded region

Example: finding a volume by evaluating a triple integral

try it

Changing the Order of Integration

Example: changing the order of integration

try it

Example: changing integration order and coordinate systems

Candela Citations