5.4 Triple integrals (Page 3/8)

Calculus volume 3 Page 3 / 8

E = {(x, y, z) | (x, z) \in D, u_{1} (x, z) \leq y \leq u_{2} (x, z)}

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

\underset{E}{∭} f (x, y, z) d V = \iint_{D} [\int_{u_{1} (x, z)}^{u_{2} (x, z)} f (x, y, z) d y] d A .

Finally, if $D$ is a general bounded region in the $y z$ -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 $ℝ^{3}$ can be described as

E = {(x, y, z) | (y, z) \in D, u_{1} (y, z) \leq x \leq u_{2} (y, z)}

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

\underset{E}{∭} f (x, y, z) d V = \iint_{D} [\int_{u_{1} (y, z)}^{u_{2} (y, z)} f (x, y, z) d x] d A .

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 $x y$ -plane is of Type I ( [link] ), then

E = {(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)} .

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). — A box $E$ where the projection $D$ in the $x y$ -plane is of Type I.

Then the triple integral becomes

\underset{E}{∭} f (x, y, z) d V = \int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} \int_{u_{1} (x, y)}^{u_{2} (x, y)} f (x, y, z) d z d y d x .

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

E = {(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)} .

Then the triple integral becomes

\underset{E}{∭} f (x, y, z) d V = \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) d z d x d y .

Evaluating a triple integral over a general bounded region

Evaluate the triple integral of the function $f (x, y, z) = 5 x - 3 y$ 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 $x y$ -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. — The solid $E$ has a projection $D$ on the $x y$ -plane of Type I.

We can describe the solid region tetrahedron as

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

Hence, the triple integral is

\underset{E}{∭} f (x, y, z) d V = \int_{x = 0}^{x = 1} \int_{y = 0}^{y = 1 - x} \int_{z = 0}^{z = 1 - x - y} (5 x - 3 y) d z d y d x .

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

\int_{z = 0}^{z = 1 - x - y} (5 x - 3 y) d z = (5 x - 3 y) (1 - x - y) .

Now evaluate the integral $\int_{y = 0}^{y = 1 - x} (5 x - 3 y) (1 - x - y) d y,$ obtaining

\int_{y = 0}^{y = 1 - x} (5 x - 3 y) (1 - x - y) d y = \frac{1}{2} {(x - 1)}^{2} (6 x - 1) .

Finally, evaluate

\int_{x = 0}^{x = 1} \frac{1}{2} {(x - 1)}^{2} (6 x - 1) d x = \frac{1}{12} .

Putting it all together, we have

\underset{E}{∭} f (x, y, z) d V = \int_{x = 0}^{x = 1} \int_{y = 0}^{y = 1 - x} \int_{z = 0}^{z = 1 - x - y} (5 x - 3 y) d z d y d x = \frac{1}{12} .

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Just as we used the double integral $\iint_{D} 1 d A$ to find the area of a general bounded region $D,$ we can use $\underset{E}{∭} 1 d V$ 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 $x y$ -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). — Finding the volume of a pyramid with a square base.

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 $\underset{E}{∭} 1 d V$ where

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

Thus, we have

\underset{E}{∭} 1 d V = \int_{z = 0}^{z = 1} \int_{y = 1 + z}^{y = 1 - z} \int_{x = 1 + z}^{x = 1 - z} 1 d x d y d z = \int_{z = 0}^{z = 1} \int_{y = 1 + z}^{y = 1 - z} (2 - 2 z) d y d z = \int_{z = 0}^{z = 1} {(2 - 2 z)}^{2} d z = \frac{4}{3} .

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

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Consider the solid sphere $E = {(x, y, z) | x^{2} + y^{2} + z^{2} = 9} .$ Write the triple integral $\underset{E}{∭} f (x, y, z) d V$ 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.

$\underset{E}{∭} 1 d V = 8 \int_{x = −3}^{x = 3} \int_{y = − \sqrt{9 - x^{2}}}^{y = \sqrt{9 - x^{2}}} \int_{z = − \sqrt{9 - x^{2} - y^{2}}}^{z = \sqrt{9 - x^{2} - y^{2}}} 1 d z d y d x = 36 π .$

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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