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.
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
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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
f (x, y, z)dz dy dx |
Solution:
dz dy dx |
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= ![]() |
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Note: when we integrate
f (x, y, z) = 1, the integral
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 =
and the sphere
z =
.
Solution: The "ice cream cone" is between these two surfaces:
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
=
,
which means
x2 + y2 = 1 - x2 - y2 or
| x2 + y2 = |
| x2 + y2 |
| - |
Finally, we need the maximal range of x alone. Given that
x2 + y2
1/2, the maximal range occurs when y = 0 so that
x2
1/2. The
maximal range of x is
| -1/ |
We have determined all the limits on the interated integral. The integral is
![]() ![]() 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
![]() ![]() dz dy dx |
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).
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 |
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).
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 |
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 |
To find the volume, we integrate the function 1 over this region:
![]() dy dx dz |
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= 3 |
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= dz |
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| = 3 |
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| = 3 |
Example 5
Change the order of x and y in the integral we derived above,
![]() dy dx 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 |
| 0 |
Then, given z and y, we need to know the range of the x. The following relationship must still be true.
| y |
| x |
Since we also know x
0, the new limits of integration are
![]() dx dy dz. |