Triple Integral Definition and Applications

Section13.1Triple Integral Definition and Applications

Objectives

After completing this section you will:

Be comfortable setting up triple integrals for a solid region (domain)
Be able to switch the order of integration for a triple integral

Consider the iterated integral

\int_{- 3}^{3} \int_{0}^{\sqrt{9 - y^{2}}} \int_{0}^{9 - x^{2} - y^{2}} 1 d z d x d y .

$\begin{equation*} \ds \int_{-3}^3 \int_0^{\sqrt{9-y^2}}\int_0^{9-x^2-y^2} 1dzdxdy. \end{equation*}$

This is an integral of the form $\iiint_D dV\text{,}$ which means along some solid region $D$ in the plane, we are adding up little bits of volume. This integral will give the volume of a very familiar solid region in space. Not every triple integral will correspond to an easily drawn or visualized (even with a computer) space. However, this one does.

Exercise13.1.1

Before we compute this integral, let's visualize it. The parenthesis have been added to help you parse this first example.

\int_{- 3}^{3} (\int_{0}^{\sqrt{9 - y^{2}}} (\int_{0}^{9 - x^{2} - y^{2}} 1 d z) d x) d y .

$\begin{equation*} \ds \int_{-3}^3 (\int_0^{\sqrt{9-y^2}} (\int_0^{9-x^2-y^2} 1dz) dx ) dy. \end{equation*}$

(a)

State the bounds for each variable as inequalities.

(b)

Sketch the region $D$ in space.

(c)

Compute the innermost integral, and compare the resulting/remaining double integral to the first exercise in the double integral unit.

(d)

Now evaluate the remaining integrals (though you might want to change coordinate systems before doing so).

When working with double integrals, there were two different ways to set up the bounds for our integrals, as $dA=dxdy=dydx\text{.}$ When working with triple integrals, there are six different ways to order the differentials, and therefore set up bounds for our integrals.

d V = d x d y d z = d x d z d y = d y d x d z = d y d z d x = d z d x d y = d z d y d x

$\begin{equation*} dV=dxdydz = dxdzdy = dydxdz=dydzdx=dzdxdy=dzdydx \end{equation*}$

Consider again the iterated integral

\int_{- 3}^{3} \int_{0}^{\sqrt{9 - y^{2}}} \int_{0}^{9 - x^{2} - y^{2}} d z d x d y = \int_{0}^{9} \int_{0}^{\sqrt{9 - z}} \int_{- \sqrt{9 - x^{2} - z}}^{\sqrt{9 - x^{2} - z}} d y d x d z

$\begin{equation*} \ds \int_{-3}^3 \int_0^{\sqrt{9-y^2}}\int_0^{9-x^2-y^2} dzdxdy = \int_0^9\int_0^{\sqrt{9-z}}\int_{-\sqrt{9-x^2-z}}^{\sqrt{9-x^2-z}} dydxdz \end{equation*}$

These two integrals are equal, but required switching the order of the bounds. There are 4 other iterated integrals that are equal to these integral which can also be found by switching the order of the bounds.

Exercise13.1.2

Set up the equivalent integrals using the differential orderings: $dydzdx$ and $dxdzdy\text{.}$ We'll look at the remaining 2 in class (though you're welcome to finish them and present them with your work).

Exercise13.1.3

Consider the iterated integral

\int_{- 1}^{1} \int_{0}^{1 - x^{2}} \int_{0}^{y} d z d y d x .

$\begin{equation*} \int_{-1}^1\int_0^{1-x^2}\int_0^{y} dzdydx. \end{equation*}$

The bounds for this integral describe a region in space which satisfies the 3 inequalities $-1\leq x\leq 1\text{,}$ $0\leq y\leq 1-x^2\text{,}$ and $0\leq z\leq y\text{.}$

(a)

Draw the solid domain $D$ in space described by the bounds of the iterated integral.

(b)

There are 5 other iterated integrals equivalent to this one.

(i)

Set up the integral that use the differential ordering: $dydxdz$

(ii)

Set up the integral that use the differential ordering: $dxdzdy\text{.}$

We'll create the other 3 in class (though you are welcome to include them as part of your presentation)

Subsection13.1.1Using the triple integral

Exercise13.1.4

In each exercise below, you'll be given enough information to determine a solid domain $D$ in space.

Draw the solid $D$ and
Set up an iterated integral (pick any order you want) that would give the volume of $D\text{.}$

Note: You don't need to evaluate the integral, rather you just need to set them up.

(a)

The region $D$ under the surface $z=y^2\text{,}$ above the $xy$ -plane, and bounded by the planes $y=-1\text{,}$ $y=1\text{,}$ $x=0\text{,}$ and $x=4\text{.}$

(b)

The region $D$ in the first octant that is bounded by the coordinate planes, the plane $y+z=2\text{,}$ and the surface $x=4-y^2\text{.}$

(c)

The pyramid $D$ in the first octant that is below the planes $\ds\frac{x}{3}+\frac{z}{2}=1$ and $\ds\frac{y}{5}+\frac{z}{2}=1\text{.}$

Hint

Don't let $z$ be the inside bound.

(d)

The region $D$ that is inside both right circular cylinders $x^2+z^2=1$ and $y^2+z^2=1\text{.}$

We can find average value, centroids, centers of mass, moments of inertia, and radii of gyration exactly as before. We just now need to integrate using three integrals, and replace $ds\text{,}$ $dA$ or $d\sigma\text{,}$ with $dV\text{.}$

Exercise13.1.5

Consider the triangular wedge $D$ that is in the first octant, bounded by the planes $\ds\frac{y}{7}+\frac{z}{5}=1$ and $x=12\text{.}$ In the $yz$ plane, the wedge forms a triangle that passes through the points $(0,0,0)\text{,}$ $(0,7,0)\text{,}$ and $(0,0,5)\text{.}$