Subsections

• Purpose
• Background
  • Mass, moments, and center of mass
  • Change to cylindrical coordinates
  
  
• Exercises

Triple Integrals

Purpose

The purpose of this lab is to acquaint you with the Maple commands for triple integrals.

Background

For last week's lab on double integrals, we calculated volume by integrating the differece between the upper function $z_{up}(x,y)$ and the lower function $z_{low}(x,y)$ over a bounded two-dimensional region $R$. We can also calculate volume using a triple integral using the lower and upper functions as the lower and upper bounds of the inner integral with respect to $z$ and $1$ as the integrand. In many cases, we may need to evaluate a triple integral where the integrand is $f(x,y,z)$ instead of 1. Although these integrals do not represent volume, the bounds on the integral are over a bounded 3-D region $D$ that can be set up the same way as a volume problem.

$$\lim_{\vert P\vert \to \infty}\sum_{i=1}^n f(x_{i}^*,y_{i}^*,z_{i}^*)\Delta V = \int\!\!\int\int_{D}f(x,y,z)dV$$

Mass, moments, and center of mass

Given a mass density function $\delta(x,y,z)$ over a 3-D region $D$, the mass of the object is defined by

$$M=\int\!\!\int\int_{D}\delta(x,y,z)dV$$

and the first moments of the solid region $D$:
About the $yz$-plane is:

$$M_{yz}=\int\!\!\int\int_{D}x \delta(x,y,z)dV$$

About the $xz$-plane is:

$$M_{xz}=\int\!\!\int\int_{D}y \delta(x,y,z)dV$$

About the $xy$-plane is:

$$M_{xy}=\int\!\!\int\int_{D}z \delta(x,y,z)dV$$

and the center of mass $(\bar{x},\bar{y},\bar{z})$ is:

$$\bar{x}=\frac{M_{yz}}{M}, \hspace{5ex} \bar{y}=\frac{M_{xz}}{M}, \hspace{5ex} \bar{z}=\frac{M_{xy}}{M}$$

The following Maple commands show an example of a triple integral to calculate the volume of a hemisphere of radius 2 after solving $x^2+y^2+z^2=4$ for $z$ as a function of $(x,y)$. If the mass density function is $\delta(x,y,z)=1$, then the mass is equivalent to the volume of the hemisphere and the $z$-coordinate for the center of mass can be calculated using the Maple commands shown below.

> with(plots):
> sphere := x^2+y^2+z^2=4
> implicitplot3d(sphere,x=-2..2,y=-2..2,z=0..2)
> M:=int(int(int(1,z=0..sqrt(4-x^2-y^2)),y=-sqrt(4-x^2)..sqrt(4-x^2)),x=-2..2)
> Mxy:=int(int(int(z,z=0..sqrt(4-x^2-y^2)),y=-sqrt(4-x^2)..sqrt(4-x^2)),x=-2..2)
> zbar:=Mxy/M

Change to cylindrical coordinates

Volumes for some surfaces calculated as a triple integral can be drastically simplified by a change of coordinates. Just as polar coordinates simplify many two-dimensional area problems, cylindrical coordinates simplify many three-dimensional volume problems. The equations relating rectangular $(x,y,z)$ coordinates to cylindrical $(r,\theta,z)$ are:

$$x = r \cos(\theta) , \hspace{2ex} y = r \sin(\theta), \hspace{2ex} z = z,$$

$$r^2 = x^2 + y^2, \hspace{2ex} tan(\theta) = \frac{y}{x}$$

So, the hemisphere of radius 2 described above would have cylindrical coordinate equation $r^2+z^2=4$ and the $z$-coordinate for the center of mass can be calculated using a triple integral in spherical coordinates using the Maple commands shown below.

$$\int\!\!\int\int_{D} 1 dV = \int\!\!\int\int_{D} r dz dr d \theta$$

> with(plots):
> sphere_cylindrical := simplify(subs({x=r*cos(theta),y=r*sin(theta),z=z},sphere))
> implicitplot3d(r^2+z^2=4,r=0..2,theta=0..2*Pi,z=0..2,coords=cylindrical)
> solve(sphere_cylindrical,r)
> M:=int(int(int(r,z=0..sqrt(4-r^2)),r=0..2),theta=0..2*Pi)
> Mxy:=int(int(int(z*r,z=0..sqrt(4-r^2)),r=0..2),theta=0..2*Pi)
> zbar:=Mxy/M

Exercises

1. For the cylinder $x^2+y^2=4$ bounded by the planes $z=0$ and $z=3$, with radius 2 and height 3,

   a

   Use the implicitplot3d command with the first argument $\left[x^2+y^2 = 4, z = 0, z = 3\right]$ and plotting ranges $-2 \leq x \leq 2$, $-2 \leq y \leq 2$, $-1 \leq z \leq 4$ to generate a surface plot of the cylinder.

   b

   Show, using a triple integral that the volume is $\pi r^2 h = 12 \pi$.

   c

   Assuming that the mass density function $\delta(x,y,z)=1$, use triple integrals to find the $z$-coordinate for the center of mass. (Hint: the volume you found above is the mass, so label as $M$ and using symmetry, you know that the center of mass should be $$(0,0,\frac{3}{2})$$.

   d

   Show that the equation for the cylinder above in cylindrical coordinates is $r=2$ and generate a cylindrical coordinate plot of the surfaces given in part a.

   e

   Find the volume/mass and the $z$-coordinate for the center of mass using triple integrals in cylindrical coordinates.

2. For the cylinder $x^2+(y-2)^2=4$ bounded by the planes $z=0$ and $z=3$, also having radius 2 and height 3,

   a

   Use the implicitplot3d command with plotting ranges $-2 \leq x \leq 2$, $0 \leq y \leq 4$, $-1 \leq z \leq 4$ to generate a surface plot of the cylinder.

   b

   Show again, using a triple integral that the volume is $\pi r^2 h = 12 \pi$.

   c

   Assuming that the mass density function $\delta(x,y,z)=1$, use triple integrals to find the $z$-coordinate for the center of mass. (Although the $x$ and $y$ coordinates for the center of mass are different, the $z$ coordinate should be the same as the cylinder in exercise 1.)

   d

   Show that the equation for the cylinder above in cylindrical coordinates is $r=4\sin(\theta)$ and generate a cylindrical coordinate plot of the surfaces given in part a.

   e

   Find the volume/mass and the $z$-coordinate for the center of mass using triple integrals in cylindrical coordinates.