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 [latex]E[/latex] in [latex]{\mathbb{R}^3}[/latex]. The general bounded regions we will consider are of three types.

First, let [latex]D[/latex] be the bounded region that is a projection of [latex]E[/latex] onto the [latex]xy[/latex]-plane. Suppose the region [latex]E[/latex] in [latex]{\mathbb{R}^3}[/latex] has the form:

[latex]E = {{(x,y,z)}{\mid}{(x,y)} \ {\in} \ {{D},{u_1}{(x,y)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(x,y)} }[/latex].

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

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

[latex]{E} = \left \{{(x,y,z)}{\mid}{(x,z)} \ {\in} \ {{D},{u_1}{(x,z)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(x,z)} \right \}[/latex]

where [latex]D[/latex] is the projection of [latex]E[/latex] onto the [latex]xy[/latex]-plane and the triple integral is:

[latex]\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}[/latex].

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

[latex]{E} = \left \{{(x,y,z)}{\mid}{(y,z)} \ {\in} \ {{D},{u_1}{(y,z)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(y,z)} \right \}[/latex]

where [latex]D[/latex] is the projection of [latex]E[/latex] onto the [latex]yz[/latex]-plane and the triple integral is:

[latex]\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}[/latex].

Setting Up Triple Integrals with Type I and Type II Regions

Note that the region [latex]D[/latex] in any of the planes may be of Type I or Type II.

If [latex]D[/latex] in the [latex]xy[/latex]-plane is of Type I (Figure 4), then:

[latex]{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 \}[/latex].

Then the triple integral becomes:

[latex]\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}[/latex].

If [latex]D[/latex] in the [latex]xy[/latex]-plane is of Type II (Figure 5), then:

[latex]{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 \}[/latex].

Then the triple integral becomes:

[latex]\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}[/latex].

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

Show Solution

Figure 6 shows the solid tetrahedron [latex]E[/latex] and its projection [latex]D[/latex] on the [latex]xy[/latex]-plane.

We can describe the solid region tetrahedron as

[latex]{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 \}[/latex].

Hence, the triple integral is

[latex]\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}[/latex].

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

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

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

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

Finally, evaluate

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

Putting it all together, we have

[latex]\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}}[/latex].

Using Triple Integrals to Find Volume

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

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

Show Solution

In this pyramid the value of [latex]z[/latex] changes from 0 to 1 and at each height [latex]z[/latex], the cross section of the pyramid for any value of [latex]z[/latex] is the square [latex][-1+z,1-z] \ {\times} \ [-1+z,1-z][/latex]. Hence, the volume of the pyramid is [latex]\underset{E}{\displaystyle\iiint}{1}{dV}[/latex] where

[latex]{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 \}[/latex].

Thus, we have

[latex]\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}}[/latex].

Hence, the volume of the pyramid is [latex]{\frac{4}{3}}[/latex] cubic units.