Triple integral examples

Example 1

A cube has sides of length 4. Let one corner be at the origin and the adjacent corners be on the positive x, y, and z axes.

\includegraphics[width=2in]{cubeex.eps}

If the cube's density is proportional to the distance from the xy-plane, find its mass.

Solution: The density of the cube is f (x, y, z) = kz for some constant k.

If W is the cube, the mass is the triple integral

$\displaystyle \iiint_{W}^{}$kz dV = $\displaystyle \int_{0}^{4}$$\displaystyle \int_{0}^{4}$$\displaystyle \int_{0}^{4}$kz dx dy dz    
  = $\displaystyle \int_{0}^{4}$$\displaystyle \int_{0}^{4}$$\displaystyle \left(\vphantom{kxz \Big\vert _{x=0}^{x=4}}\right.$kxz$\displaystyle \Big\vert _{{x=0}}^{{x=4}}$$\displaystyle \left.\vphantom{kxz \Big\vert _{x=0}^{x=4}}\right)$dy dz    
  = $\displaystyle \int_{0}^{4}$$\displaystyle \int_{0}^{4}$4kz dy dz    
  = $\displaystyle \int_{0}^{4}$$\displaystyle \left(\vphantom{ 4kzy \Big\vert _{y=0}^{y=4}}\right.$4kzy$\displaystyle \Big\vert _{{y=0}}^{{y=4}}$$\displaystyle \left.\vphantom{ 4kzy \Big\vert _{y=0}^{y=4}}\right)$dz    
  = $\displaystyle \int_{0}^{4}$16kzdz = 8kz2$\displaystyle \Big\vert _{{z=0}}^{{z=4}}$ = 128k    

If distance is in cm and k = 1 gram per cubic cm per cm, then the mass of the cube is 128 grams.

Example 2

Evaluate the integral

$\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{x}$$\displaystyle \int_{0}^{{1+x+y}}$f (x, y, z)dz dy dx    

where f (x, y, z) = 1.

Solution:

  $\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{x}$$\displaystyle \int_{0}^{{1+x+y}}$dz dy dx    
           = $\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{x}$$\displaystyle \left(\vphantom{z\Big\vert _{z=0}^{z=1+x+y}}\right.$z$\displaystyle \Big\vert _{{z=0}}^{{z=1+x+y}}$$\displaystyle \left.\vphantom{z\Big\vert _{z=0}^{z=1+x+y}}\right)$dy dx    
           = $\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{x}$(1 + x + y)dy dx    
           = $\displaystyle \int_{0}^{1}$$\displaystyle \left(\vphantom{y + yx + \frac{y^2}{2}}\right.$y + yx + $\displaystyle {\frac{{y^2}}{{2}}}$$\displaystyle \left.\vphantom{y + yx + \frac{y^2}{2}}\right)$$\displaystyle \bigg\vert _{{y=0}}^{{y=x}}$dx    
           = $\displaystyle \int_{0}^{1}$$\displaystyle \left(\vphantom{x + x^2 + \frac{x^2}{2}}\right.$x + x2 + $\displaystyle {\frac{{x^2}}{{2}}}$$\displaystyle \left.\vphantom{x + x^2 + \frac{x^2}{2}}\right)$dx    
           = $\displaystyle \int_{0}^{1}$$\displaystyle \left(\vphantom{x + \frac{3x^2}{2} }\right.$x + $\displaystyle {\frac{{3x^2}}{{2}}}$$\displaystyle \left.\vphantom{x + \frac{3x^2}{2} }\right)$dx    
           = $\displaystyle {\frac{{x^2}}{{2}}}$ + $\displaystyle {\frac{{x^3}}{{2}}}$$\displaystyle \bigg\vert _{0}^{1}$ = $\displaystyle {\frac{{1}}{{2}}}$ + $\displaystyle {\frac{{1}}{{2}}}$ = 1    

Note: when we integrate f (x, y, z) = 1, the integral $ \iiint_{W}^{}$dV is the volume of the solid W.

Example 3a

Set up the integral of f (x, y, z) over W, the solid "ice cream cone" bounded by the cone z = $ \sqrt{{x^2+y^2}}$ and the sphere z = $ \sqrt{{1-x^2-y^2}}$.

\includegraphics[width=2in]{icecream.eps}

Solution: The "ice cream cone" is between these two surfaces:

$\displaystyle \sqrt{{x^2+y^2}}$$\displaystyle \le$z$\displaystyle \le$$\displaystyle \sqrt{{1-x^2-y^2}}$.    

This gives the range of z as a function of x and y.

Now we need to find the maximal range of x and y. Inside the ice cream cone, the maximal range of x and y occurs where the two surfaces meet, i.e., where the "ice cream" (the sphere) meets the cone. From the figure, you can see that the surfaces meet in a circle, and the range of x and y is the disk that is the interior of that circle.

The surfaces meet when $ \sqrt{{x^2+y^2}}$ = $ \sqrt{{1-x^2-y^2}}$, which means x2 + y2 = 1 - x2 - y2 or

x2 + y2 = $\displaystyle {\frac{{1}}{{2}}}$.    

We have shown that in the "ice cream cone"

x2 + y2$\displaystyle \le$$\displaystyle {\frac{{1}}{{2}}}$.    

We write this as

- $\displaystyle \sqrt{{1/2 - x^2}}$$\displaystyle \le$y$\displaystyle \le$$\displaystyle \sqrt{{1/2 - x^2}}$    

This gives the range of y as a function of x.

Finally, we need the maximal range of x alone. Given that x2 + y2$ \le$1/2, the maximal range occurs when y = 0 so that x2$ \le$1/2. The maximal range of x is

-1/$\displaystyle \sqrt{{2}}$$\displaystyle \le$x$\displaystyle \le$1/$\displaystyle \sqrt{{2}}$.    

We have determined all the limits on the interated integral. The integral is

$\displaystyle \iiint_{W}^{}$f dV = $\displaystyle \int_{{-1/\sqrt{2}}}^{{1/\sqrt{2}}}$$\displaystyle \int_{{-\sqrt{1/2-x^2}}}^{{\sqrt{1/2-x^2}}}$$\displaystyle \int_{{\sqrt{x^2+y^2}}}^{{\sqrt{1-x^2-y^2}}}$f (x, y, z)dz dy dx    

Example 3b

Find the volume of the ice cream cone.

Solution: Simply set f = 1 in Example 3a.

Volume is

$\displaystyle \iiint_{W}^{}$dV = $\displaystyle \int_{{-1/\sqrt{2}}}^{{1/\sqrt{2}}}$$\displaystyle \int_{{-\sqrt{1/2-x^2}}}^{{\sqrt{1/2-x^2}}}$$\displaystyle \int_{{\sqrt{x^2+y^2}}}^{{\sqrt{1-x^2-y^2}}}$dz dy dx    

We won't attempt to evaluate this integral in rectangular coordinates.

Example 4

Find volume of the tetrahedron bounded by the coordinate planes and the plane through (2, 0, 0), (0, 3, 0), and (0, 0, 1).

\includegraphics[width=2in]{tetra.eps}

Solution:

The first step is to find the equation for the plane. You can follow the procedure in forming plane example 2 to calculate that the plane is

3x + 2y + 6z = 6. (1)

In the tetrahedron, the total range of z is

0$\displaystyle \le$z$\displaystyle \le$1    

Now we need to find the maximal total range of x for each value of z. For any value of z, x has the largest range when y = 0. So look at the cross section of the tetrahedron in the xz-plane (i.e., where y = 0).

\includegraphics[width=2in]{tetrasub.eps}

The lower limit of x is zero, and the upper limit corresponds to the plane given by equation (1) when y = 0. Plugging y = 0 into equation (1), we obtain 3x + 6z = 6, or x = 2(1 - z). Hence, for a given z, the range of x is

0$\displaystyle \le$x$\displaystyle \le$2(1 - z).    

Last, we need to find the range of y for a given value of z and x. The lowest y gets is zero. The upper value of y is given by equation (1) for the upper plane, which we can solve for y to write as y = 3(1 - x/2 - z). Hence, for a given z and x, the range of y is

0$\displaystyle \le$y$\displaystyle \le$3$\displaystyle \left(\vphantom{1 - \frac{x}{2} - z}\right.$1 - $\displaystyle {\frac{{x}}{{2}}}$ - z$\displaystyle \left.\vphantom{1 - \frac{x}{2} - z}\right)$.    

To find the volume, we integrate the function 1 over this region:

  $\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{{2(1-z)}}$$\displaystyle \int_{0}^{{3(1- x/2 - z)}}$dy dx dz    
           = $\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{{2(1-z)}}$3$\displaystyle \left(\vphantom{1 - \frac{x}{2} - z}\right.$1 - $\displaystyle {\frac{{x}}{{2}}}$ - z$\displaystyle \left.\vphantom{1 - \frac{x}{2} - z}\right)$dx dz    
           = $\displaystyle \int_{0}^{1}$3$\displaystyle \biggl[$x - $\displaystyle {\frac{{x^2}}{{4}}}$ - zx$\displaystyle \biggr]_{{x=0}}^{{x=2(1-z)}}$dz    
           = $\displaystyle \int_{0}^{1}$3$\displaystyle \left(\vphantom{2(1-z) - (1-z)^2 - 2z(1-z)}\right.$2(1 - z) - (1 - z)2 - 2z(1 - z)$\displaystyle \left.\vphantom{2(1-z) - (1-z)^2 - 2z(1-z)}\right)$dz    
           = $\displaystyle \int_{0}^{1}$3(1 - 2z + z2)dz    
           = 3$\displaystyle \left[\vphantom{ z - z^2 + \frac{z^3}{3} }\right.$z - z2 + $\displaystyle {\frac{{z^3}}{{3}}}$$\displaystyle \left.\vphantom{ z - z^2 + \frac{z^3}{3} }\right]_{0}^{1}$    
           = 3$\displaystyle \left(\vphantom{1 - 1 +\frac{1}{3}}\right.$1 - 1 + $\displaystyle {\frac{{1}}{{3}}}$$\displaystyle \left.\vphantom{1 - 1 +\frac{1}{3}}\right)$ = 3$\displaystyle \left(\vphantom{\frac{1}{3}}\right.$$\displaystyle {\frac{{1}}{{3}}}$$\displaystyle \left.\vphantom{\frac{1}{3}}\right)$ = 1    

Example 5

Change the order of x and y in the integral we derived above,

$\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{{2(1-z)}}$$\displaystyle \int_{0}^{{3(1- x/2 - z)}}$dy dx dz,    

so that the order will be dx dy dz.

Solution: If y will be middle integral, we need limits of y in terms of z (independent of x).

For given z, how large can y range? From the above limits, we know

0$\displaystyle \le$y$\displaystyle \le$3$\displaystyle \left(\vphantom{1 - \frac{x}{2} - z}\right.$1 - $\displaystyle {\frac{{x}}{{2}}}$ - z$\displaystyle \left.\vphantom{1 - \frac{x}{2} - z}\right)$.    

The range is largest when x = 0, so

0$\displaystyle \le$y$\displaystyle \le$3$\displaystyle \left(\vphantom{1 - z}\right.$1 - z$\displaystyle \left.\vphantom{1 - z}\right)$    

Then, given z and y, we need to know the range of the x. The following relationship must still be true.

y$\displaystyle \le$3$\displaystyle \left(\vphantom{1 - \frac{x}{2} - z}\right.$1 - $\displaystyle {\frac{{x}}{{2}}}$ - z$\displaystyle \left.\vphantom{1 - \frac{x}{2} - z}\right)$    

Rewrite in terms of x:

$\displaystyle {\frac{{3x}}{{2}}}$$\displaystyle \le$3 - 3z - y    

or

x$\displaystyle \le$2$\displaystyle \left(\vphantom{1 - z - \frac{y}{3}}\right.$1 - z - $\displaystyle {\frac{{y}}{{3}}}$$\displaystyle \left.\vphantom{1 - z - \frac{y}{3}}\right)$    

Since we also know x$ \ge$ 0, the new limits of integration are

$\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{{3(1-z)}}$$\displaystyle \int_{0}^{{2(1-z-y/3)}}$dx dy dz.    


Duane Nykamp
nykamp@math.umn.edu
2006-02-25