### 2 Evaluating triple integrals

A triple integral is an integral of the form

$\phantom{\rule{2em}{0ex}}{\int }_{a}^{b}{\int }_{p\left(x\right)}^{q\left(x\right)}{\int }_{r\left(x,y\right)}^{s\left(x,y\right)}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dzdydx$

The evaluation can be split into an “inner integral" (the integral with respect to $z$ between limits which are functions of $x$ and $y$ ), an “intermediate integral" (the integration with respect to $y$ between limits which are functions of $x$ ) and an “outer integral" (the integration with respect to $x$ between limits which are constants. Note that there is nothing special about the variable names $x$ , $y$ and $z$ : other variable names could have been used instead.

Triple integrals can be represented in different ways. ${\int }_{V}f\phantom{\rule{0.3em}{0ex}}dV$ represents a triple integral where the $dV$ is replaced by $\phantom{\rule{0.3em}{0ex}}dxdydz$ (or equivalent) and the limit of $V$ on the integral is replaced by appropriate limits on the three integrals.

Note that the integral ${\int }_{V}\phantom{\rule{0.3em}{0ex}}dV$ (i.e. integrating the function $f\left(x,y,z\right)=1$ ) gives the volume of the relevant shape. Hence the alternative name of volume integral.

One special case is where the limits on all the integrals are constants (a constant is, of course, a special case of a function). This represents an integral over a cuboidal region.

##### Example 16

Consider a cube $V$ of side 1.

1. Express the integral ${\int }_{V}f\phantom{\rule{0.3em}{0ex}}dV$ (where $f$ is any function of $x$ , $y$ and $z$ ) as a

triple integral.

2. Hence evaluate ${\int }_{V}\left({y}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dV$

Figure 23

##### Solution
1. Consider a little element of length $\phantom{\rule{0.3em}{0ex}}dx$ , width $\phantom{\rule{0.3em}{0ex}}dy$ and height $\phantom{\rule{0.3em}{0ex}}dz$ . Then $\delta V$ (the volume of the small element) is the product of these lengths $\phantom{\rule{0.3em}{0ex}}dxdydz$ . The function is integrated three times. The first integration represents the integral over the vertical strip from $z=0$ to $z=1$ . The second integration represents this strip sweeping across from $y=0$ to $y=1$ and is the integration over the slice that is swept out by the strip. Finally the integration with respect to $x$ represents this slice sweeping from $x=0$ to $x=1$ and is the integration over the entire cube. The integral therefore becomes

$\phantom{\rule{2em}{0ex}}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dzdydx$

2. In the particular case where the function is $f\left(x,y,z\right)={y}^{2}+{z}^{2}$ , the integral becomes

$\phantom{\rule{2em}{0ex}}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}\left({y}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dzdydx$

The inner integral is

$\phantom{\rule{2em}{0ex}}{\int }_{0}^{1}\left({y}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dz={\left[{y}^{2}z+\frac{1}{3}{z}^{3}\right]}_{z=0}^{1}={y}^{2}×1+\frac{1}{3}×1-{y}^{2}×0-\frac{1}{3}×0={y}^{2}+\frac{1}{3}$

This inner integral is now placed into the intermediate integral to give

$\phantom{\rule{2em}{0ex}}{\int }_{0}^{1}\left({y}^{2}+\frac{1}{3}\right)\phantom{\rule{0.3em}{0ex}}dy={\left[\frac{1}{3}{y}^{3}+\frac{1}{3}y\right]}_{y=0}^{1}=\frac{1}{3}×{1}^{3}+\frac{1}{3}×1-\frac{1}{3}×{0}^{3}-\frac{1}{3}×0=\frac{2}{3}$

Finally, this intermediate integral can be placed into the outer integral to give

$\phantom{\rule{2em}{0ex}}{\int }_{0}^{1}\frac{2}{3}\phantom{\rule{0.3em}{0ex}}dx={\left[\frac{2}{3}x\right]}_{0}^{1}=\frac{2}{3}×1-\frac{2}{3}×0=\frac{2}{3}$

##### Example 17

Evaluate ${\int }_{0}^{1}{\int }_{0}^{2}{\int }_{0}^{3}8xyz\phantom{\rule{0.3em}{0ex}}dzdydx$ . This represents an integral over the cuboid given by $0\le x\le 1$ , $0\le y\le 2$ , $0\le z\le 3$ .

##### Solution

The inner integral is given by integrating the function with respect to $z$ while keeping $x$ and $y$ constant.

$\phantom{\rule{2em}{0ex}}{\int }_{0}^{3}8xyz\phantom{\rule{0.3em}{0ex}}dz={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}4xy{z}^{2}\right]}_{0}^{3}=4xy×9-0=36xy$

This result is now integrated with respect to $y$ while keeping $x$ constant:

$\phantom{\rule{2em}{0ex}}{\int }_{0}^{2}36xy\phantom{\rule{0.3em}{0ex}}dy={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}18xy\right]}_{0}^{2}=18x×4-0=72x$

Finally, this result is integrated with respect to $x$ :

$\phantom{\rule{2em}{0ex}}{\int }_{0}^{1}72x\phantom{\rule{0.3em}{0ex}}dx={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}36{x}^{2}\right]}_{0}^{1}=36×1-0=36$

Hence, ${\int }_{0}^{1}{\int }_{0}^{2}{\int }_{0}^{3}8xyz\phantom{\rule{0.3em}{0ex}}dzdydx=36$

More generally, the limits on the inner integral may be functions of the “intermediate" and “outer" variables and the limits on the intermediate integral may be functions of the “outer" variable.

##### Example 18

$V$ is the tetrahedron bounded by the planes $x=0$ , $y=0$ , $z=0$ and $x+y+z=4$ . (see Figure 24).

Figure 24

1. Express ${\int }_{V}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dV$ (where $f$ is a function of $x$ , $y$ and $z$ ) as a triple integral.
2. Hence find ${\int }_{V}x\phantom{\rule{0.3em}{0ex}}dV$ .
##### Solution

The tetrahedron is divided into a series of slices parallel to the $yz$ -plane and each slice is divided into a series of vertical strips. For each strip, the bottom is at $z=0$ and the top is on the plane $x+y+z=4$ i.e. $z=4-x-y$ . So the integral up each strip is given by ${\int }_{z=0}^{4-x-y}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dz$ and this (inner) integral will be a function of $x$ and $y$ .

This, in turn, is integrated over all strips which form the slice. For each value of $x$ , one end of the slice will be at $y=0$ and the other end at $y=4-x$ . So the integral over the slice is ${\int }_{y=0}^{4-x}{\int }_{z=0}^{4-x-y}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dzdy$ and this (intermediate) integral will be a function of $x$ .

Finally, integration is carried out over $x$ . The limits on $x$ are $x=0$ and $x=4$ . Thus the triple integral is ${\int }_{x=0}^{4}{\int }_{y=0}^{4-x}{\int }_{z=0}^{4-x-y}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dzdydx$ and this (outer) integral will be a constant.

Hence ${\int }_{V}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dV={\int }_{x=0}^{4}{\int }_{y=0}^{4-x}{\int }_{z=0}^{4-x-y}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dzdydx$ .

In the case where $f\left(x,y,z\right)=x$ , the integral becomes

$\begin{array}{rcll}{\int }_{V}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dV& =& {\int }_{x=0}^{4}{\int }_{y=0}^{4-x}{\int }_{z=0}^{4-x-y}x\phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & =& {\int }_{x=0}^{4}{\int }_{y=0}^{4-x}{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}xz\right]}_{z=0}^{4-x-y}\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& {\int }_{x=0}^{4}{\int }_{y=0}^{4-x}\left[\left(4-x-y\right)x-0\right]\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& {\int }_{x=0}^{4}{\int }_{y=0}^{4-x}\left[4x-{x}^{2}-xy\right]\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& {\int }_{x=0}^{4}{\left[4xy-{x}^{2}y-\frac{1}{2}x{y}^{2}\right]}_{y=0}^{4-x}\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\int }_{x=0}^{4}{\left[4x\left(4-x\right)-{x}^{2}\left(4-x\right)-\frac{1}{2}x{\left(4-x\right)}^{2}-0\right]}_{y=0}^{4-x}\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\int }_{x=0}^{4}\left[16x-4{x}^{2}-4{x}^{2}+{x}^{3}-8x+4{x}^{2}-\frac{1}{2}{x}^{3}\right]\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\int }_{x=0}^{4}\left[8x-4{x}^{2}+\frac{1}{2}{x}^{3}\right]\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\left[4{x}^{2}-\frac{4}{3}{x}^{3}+\frac{1}{8}{x}^{4}\right]}_{0}^{4}=4×{4}^{2}-\frac{4}{3}×{4}^{3}+\frac{1}{8}×{4}^{4}-0& \text{}\\ & =& 64-\frac{256}{3}+32& \text{}\\ & =& \frac{192-256+96}{3}& \text{}\\ & =& \frac{32}{3}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$
##### Key Point 7

Triple Integration

The procedure for carrying out a triple integral is very similar to that for a double integral except that the procedure requires three stages rather than two.

##### Example 19

Find the integral of $x$ over the shape shown in Figure 25. It represents half (positive $x$ ) of a cylinder centered at $x=y=0$ with radius 1 and vertical extent from $z=0$ to $z=1$ .

Figure 25

##### Solution

In terms of $x$ , the shape goes from $x=0$ to $x=1$ . For each value of $x$ , $y$ goes from $-\sqrt{1-{x}^{2}}$ to $\sqrt{1-{x}^{2}}$ . The variable $z$ varies from $z=0$ to $z=1$ . Hence the triple integral is

$\begin{array}{rcll}I& =& {\int }_{x=0}^{1}{\int }_{y=-\sqrt{1-{x}^{2}}}^{\sqrt{1-{x}^{2}}}{\int }_{z=0}^{1}x\phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & & & \text{}\\ & =& {\int }_{x=0}^{1}{\int }_{y=-\sqrt{1-{x}^{2}}}^{\sqrt{1-{x}^{2}}}{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}xz\right]}_{z=0}^{1}\phantom{\rule{0.3em}{0ex}}dydx={\int }_{x=0}^{1}{\int }_{y=-\sqrt{1-{x}^{2}}}^{\sqrt{1-{x}^{2}}}\left[x-0\right]\phantom{\rule{0.3em}{0ex}}dydx={\int }_{x=0}^{1}{\int }_{y=-\sqrt{1-{x}^{2}}}^{\sqrt{1-{x}^{2}}}x\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & & & \text{}\\ & =& {\int }_{x=0}^{1}{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}xy\right]}_{y=-\sqrt{1-{x}^{2}}}^{\sqrt{1-{x}^{2}}}\phantom{\rule{0.3em}{0ex}}dx={\int }_{x=0}^{1}2x\sqrt{1-{x}^{2}}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

This outer integral can be evaluated by means of the substitution $U=1-{x}^{2}$ i.e. $dU=-2x\phantom{\rule{0.3em}{0ex}}dx$ and noting that $U=1$ when $x=0$ and $U=0$ when $x=1$ i.e.

$\begin{array}{rcll}I={\int }_{x=0}^{1}2x\sqrt{1-{x}^{2}}\phantom{\rule{0.3em}{0ex}}dx=-{\int }_{1}^{0}{U}^{1∕2}dU={\int }_{0}^{1}{U}^{1∕2}dU={\left[\frac{2}{3}{U}^{3∕2}\right]}_{0}^{1}=\frac{2}{3}-0=\frac{2}{3}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& & & \text{}\end{array}$

It is important to note that the three integrations can be carried out in whatever order is most convenient. The result does not depend on the order in which the integrals are carried out. However, when the order of the integrations is changed, it is necessary to consider carefully what the limits should be on each integration. Simply moving the limits from one integration to another will only work in the case of integration over a cuboid (i.e. where all limits are constants).

##### Key Point 8

Order of Integration for Triple Integrals

1. The three integrations can be carried out in whichever order is most convenient.
2. When changing the order of the integrations, it is important to reconsider the limits on each

integration; a diagram can often help.

##### Example 20

For the triangular prism in Figure 26, with ends given by the planes $y=0$ and $y=2$ and remaining faces given by the planes $x=0$ , $z=0$ and $x+4z=4$ , find the integral of $x$ over the prism, by

1. Integrating first with respect to $z$ , then $y$ and finally $x$ , and
2. Changing the order of the integrations to $x$ first, then $y$ , then $z$ .

Figure 26

##### Solution

For every value of $x$ and $y$ , the vertical coordinate $z$ varies from $z=0$ to $z=1-x∕4$ . Hence the limits on $z$ are $z=0$ and $z=1-x∕4$ . For every value of $x$ , the limits on $y$ are $y=0$ to $y=2$ . The limits on $x$ are $x=0$ and $x=4$ (the limits on the figure). Hence the triple integral is ${\int }_{0}^{4}{\int }_{0}^{2}{\int }_{0}^{1-x∕4}x\phantom{\rule{0.3em}{0ex}}dzdydx$ which can be evaluated as follows

$\begin{array}{rcll}I& =& {\int }_{0}^{4}{\int }_{0}^{2}{\int }_{0}^{1-x∕4}x\phantom{\rule{0.3em}{0ex}}dzdydx={\int }_{0}^{4}{\int }_{0}^{2}{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}xz\right]}_{0}^{1-x∕4}\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& {\int }_{0}^{4}{\int }_{0}^{2}\left[x\left(1-\frac{x}{4}\right)-0\right]\phantom{\rule{0.3em}{0ex}}dydx={\int }_{0}^{4}{\int }_{0}^{2}\left(x-\frac{1}{4}{x}^{2}\right)\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& {\int }_{0}^{4}{\left[\left(x-\frac{1}{4}{x}^{2}\right)y\right]}_{0}^{2}\phantom{\rule{0.3em}{0ex}}dx={\int }_{0}^{4}\left[\left(x-\frac{1}{4}{x}^{2}\right)×2-\left(x-\frac{1}{4}{x}^{2}\right)×0\right]\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\int }_{0}^{4}\left(2x-\frac{1}{2}{x}^{2}\right)\phantom{\rule{0.3em}{0ex}}dx={\left[{x}^{2}-\frac{1}{6}{x}^{3}\right]}_{0}^{4}={4}^{2}-\frac{1}{6}×{4}^{3}-0=16-\frac{32}{3}=\frac{16}{3}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Now, if the order of the integrations is changed, it is necessary to re-derive the limits on the integrals. For every combination of $y$ and $z$ , $x$ varies between $x=0$ (left) and $x=4-4z$ (right). Hence the limits on $x$ are $x=0$ and $x=4-4z$ . The limits on $y$ are $y=0$ and $y=2$ (for all $z$ ). The limits of $z$ are $z=0$ (bottom) and $z=1$ (top).

So the triple integral becomes ${\int }_{0}^{1}{\int }_{0}^{2}{\int }_{0}^{4-4z}x\phantom{\rule{0.3em}{0ex}}dxdydz$ which can be evaluated as follows

$\begin{array}{rcll}I& =& {\int }_{0}^{1}{\int }_{0}^{2}{\int }_{0}^{4-4z}x\phantom{\rule{0.3em}{0ex}}dxdydz={\int }_{0}^{1}{\int }_{0}^{2}{\left[\frac{1}{2}{x}^{2}\right]}_{0}^{4-4z}\phantom{\rule{0.3em}{0ex}}dydz& \text{}\\ & =& {\int }_{0}^{1}{\int }_{0}^{2}\left[\frac{1}{2}{\left(4-4z\right)}^{2}\right]\phantom{\rule{0.3em}{0ex}}dydz={\int }_{0}^{1}{\int }_{0}^{2}\left(8-16z+8{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dydz& \text{}\\ & =& {\int }_{0}^{1}{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\left(8-16z+8{z}^{2}\right)y\right]}_{0}^{2}\phantom{\rule{0.3em}{0ex}}dz=2{\int }_{0}^{1}\left(8-16z+8{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dz& \text{}\\ & =& 2{\left[8z-8{z}^{2}+\frac{8}{3}{z}^{3}\right]}_{0}^{1}=2\left(8-8+\frac{8}{3}-0\right)=\frac{16}{3}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$
##### Key Point 9

Limits of Integration

While for different orders of integration the integral will always evaluate to the same value, the limits of integration will in general be different.

Evaluate the triple integral: ${\int }_{0}^{2}{\int }_{0}^{3}{\int }_{0}^{2}{x}^{3}{y}^{2}z\phantom{\rule{0.3em}{0ex}}dxdydz$

The inner integral is

$\phantom{\rule{2em}{0ex}}{\int }_{0}^{2}{x}^{3}{y}^{2}z\phantom{\rule{0.3em}{0ex}}dx={\left[\frac{1}{4}{x}^{4}{y}^{2}z\right]}_{0}^{2}=\frac{1}{4}{2}^{4}{y}^{2}z-0=4{y}^{2}z$

This is put into the intermediate integral i.e.

$\phantom{\rule{2em}{0ex}}{\int }_{0}^{3}4{y}^{2}z\phantom{\rule{0.3em}{0ex}}dy={\left[\frac{4}{3}{y}^{3}z\right]}_{0}^{3}=\frac{4}{3}{3}^{3}z-0=36z$

Finally, this is put in the outer integral to give

$\phantom{\rule{2em}{0ex}}I={\int }_{0}^{2}36z\phantom{\rule{0.3em}{0ex}}dz={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}18{z}^{2}\right]}_{0}^{2}=18×{2}^{2}-0=72$

##### Exercises

Evaluate the following triple integrals

1. ${\int }_{0}^{2}{\int }_{0}^{x}{\int }_{0}^{x+z}\left(x+y+z\right)\phantom{\rule{0.3em}{0ex}}dydzdx$
2. ${\int }_{2}^{4}{\int }_{-1}^{3}{\int }_{x∕2-2}^{2-x∕2}\left(x+y\right)\phantom{\rule{0.3em}{0ex}}dzdydx$
1. $14$
2. $\frac{88}{3}$

Find the volume of the solid prism shown in the diagram below. Check that when the order of integration is changed, the volume remains unaltered.

The volume is given by the triple integral $\int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dV$ .

Putting $z$ on the outer integral, $y$ on the intermediate integral and $x$ on the inner integral, the limits on $z$ are $z=0$ to $z=1$ . For each value of $z$ , $y$ varies from $y=0$ (base) to $y=1-z$ on the sloping face. For each combination of $y$ and $z$ , $x$ varies from $x=0$ to $x=3$ . Thus, the volume is given by

$\begin{array}{rcll}V& =& \int \int \int dV={\int }_{z=0}^{1}{\int }_{y=0}^{1-z}{\int }_{x=0}^{3}\phantom{\rule{0.3em}{0ex}}dxdydz& \text{}\\ & =& {\int }_{z=0}^{1}{\int }_{y=0}^{1-z}{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}x\right]}_{0}^{3}\phantom{\rule{0.3em}{0ex}}dydz={\int }_{z=0}^{1}{\int }_{y=0}^{1-z}3\phantom{\rule{0.3em}{0ex}}dydz& \text{}\\ & =& {\int }_{z=0}^{1}{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}3y\right]}_{y=0}^{1-z}\phantom{\rule{0.3em}{0ex}}dz={\int }_{z=0}^{1}\left(3\left(1-z\right)-0\right)\phantom{\rule{0.3em}{0ex}}dz={\int }_{z=0}^{1}\left(3-3z\right)\phantom{\rule{0.3em}{0ex}}dz& \text{}\\ & =& {\left[3z-\frac{3}{2}{z}^{2}\right]}_{0}^{1}=3-\frac{3}{2}-\left(0-0\right)=\frac{3}{2}=1.5\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Now, the three integrations can be carried out in a different order. For example, with $x$ on the outer integral, $z$ on the intermediate integral and $y$ on the inner integral, the limits on $x$ are $x=0$ to $x=3$ ; for each value of $x$ , $z$ varies from $z=0$ to $z=1$ and for each combination of $x$ and $z$ , $y$ varies from $y=0$ to $y=1-z$ . The volume is therefore given by

$\begin{array}{rcll}V& =& \int \int \int dV={\int }_{x=0}^{3}{\int }_{z=0}^{1}{\int }_{y=0}^{1-z}\phantom{\rule{0.3em}{0ex}}dydzdx\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& {\int }_{x=0}^{3}{\int }_{z=0}^{1}{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}y\right]}_{y=0}^{1-z}\phantom{\rule{0.3em}{0ex}}dzdx& \text{}\\ & =& {\int }_{x=0}^{3}{\int }_{z=0}^{1}\left[1-z\right]\phantom{\rule{0.3em}{0ex}}dzdx& \text{}\\ & =& {\int }_{x=0}^{3}{\left[z-\frac{{z}^{2}}{2}\right]}_{z=0}^{1}\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\int }_{x=0}^{3}\left[1-\frac{{1}^{2}}{2}-0\right]\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\int }_{x=0}^{3}\frac{1}{2}\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\left[\frac{x}{2}\right]}_{x=0}^{3}=\frac{3}{2}-0=1.5\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

There are in all six ways ( $3!$ ) to order the three integrations; each order gives the same answer of $1.5$ .

##### Exercise

Find the volume of the solid shown in the diagram below. Check that when the order of integration is changed, the volume remains unaltered.

$32$