5.4 Triple integrals  (Page 3/8)

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

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 $u_1(y,z)\le u_2(y,z)$ for all $(y,z)$ in $D,$ then the solid region $E$ in ${ℝ}^3$ can be described as

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

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 ( [link] ), then

Then the triple integral becomes

If $D$ in the $xy$ -plane is of Type II ( [link] ), then

Then the triple integral becomes

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 $x+y+z=1.$

[link] shows the solid tetrahedron $E$ and its projection $D$ on the $xy$ -plane.

We can describe the solid region tetrahedron as

$E=\{(x,y,z)|0\le x\le 1,0\le y\le 1-x,0\le z\le 1-x-y\}.$

Hence, the triple integral is

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

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

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

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

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

Finally, evaluate

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

Putting it all together, we have

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

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

Finding a volume by evaluating a triple integral

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

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

$E=\{(x,y,z)|0\le z\le 1,\mathrm{-1}+z\le y\le 1-z,\mathrm{-1}+z\le x\le 1-z\}.$

Thus, we have

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

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

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