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.

PIC

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

∫∫ ∫W kzdV = ∫ 04 ∫ 04 ∫ 04kzdxdydz
= ∫ 04 ∫ 04( ) kxz |xx==40dydz
= ∫ 04 ∫ 044kzdydz
= ∫ 04( ) 4kzy|yy==40dz
= ∫ 0416kzdz = 2||z=4 8kz z=0 = 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

∫ 01 ∫ 0x ∫ 01+x+yf(x,y,z)dzdydx
where f(x,y,z) = 1.

Solution:

∫ 01 ∫ 0x ∫ 01+x+ydzdydx
= ∫ 01 ∫ 0x( ) z|zz==10+x+ydydx
= ∫ 01 ∫ 0x(1 + x + y)dydx
= ∫ 01( 2) ||y=x y + yx + y-- | 2 |y=0 dx
= ∫ 01( x2 ) x + x2 + --- 2dx
= ∫ 01( 3x2 ) x + ---- 2dx
= x2 --- 2 + | x3|1 --|| 2 0 = 1 -- 2 + 1 -- 2 = 1

Note: when we integrate f(x,y,z) = 1, the integral ∫∫∫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 = ∘x2--+--y2- and the sphere z = ∘1----x2---y2.

PIC

Solution: The “ice cream cone” is between these two surfaces:

-------- ∘ x2 + y2 ≤ z ≤ ----------- ∘ 1 - x2 - y2.
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 ∘ -------- x2 + y2 = ∘ ----------- 1 - x2 - y2, which means x2 +y2 = 1-x2 -y2 or

x2 + y2 = 1 -- 2.
We have shown that in the “ice cream cone”
x2 + y2 ≤1 -- 2.
We write this as
-∘ --------- 1∕2 - x2 ≤ y ≤∘ --------- 1∕2 - x2
This gives the range of y as a function of x.

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∕√ -- 2 ≤ x ≤ 1∕√ -- 2.

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

√- √ ------ √ ------- ∫∫ ∫ ∫ 1∕ 2∫ 1∕2-x2 ∫ 1-x2-y2 f dV = √- √ ------ √ ----- f(x, y,z)dzdydx W -1∕ 2 - 1∕2- x2 x2+y2

Example 3b

Find the volume of the ice cream cone.

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

Volume is

√ ------ √ ------- ∫∫ ∫ ∫ 1∕√2-∫ 1∕2-x2∫ 1- x2- y2 dV = √------ √ ----- dzdydx W -1∕√2- - 1∕2-x2 x2+y2
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).

PIC

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 ≤ z ≤ 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).

PIC

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 ≤ x ≤ 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 ≤ y ≤ 3( ) 1 - x-- z 2.

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

∫ 01 ∫ 02(1-z) ∫ 03(1-x∕2-z)dydxdz
= ∫ 01 ∫ 02(1-z)3( x ) 1 - --- z 2dxdz
= ∫ 013[ ] x2 x=2(1- z) x - ---- zx 4 x=0 dz
= ∫ 013( 2 ) 2 (1 - z) - (1 - z) - 2z(1 - z)dz
= ∫ 013(1 - 2z + z2)dz
= 3[ z3 ]1 z - z2 + --- 3 0
= 3( ) 1- 1 - 1 + 3 = 3( ) 1- 3 = 1

Example 5

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

∫ 01 ∫ 02(1-z) ∫ 03(1-x∕2-z)dydxdz,
so that the order will be dxdydz.

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 ≤ y ≤ 3( x ) 1 - --- z 2.
The range is largest when x = 0, so
0 ≤ y ≤ 3(1 - z)

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

y ≤ 3( x ) 1 - --- z 2
Rewrite in terms of x:
3x- 2 ≤ 3 - 3z - y
or
x ≤ 2( ) 1 - z - y- 3

Since we also know x ≥ 0, the new limits of integration are

∫ 01 ∫ 03(1-z) ∫ 02(1-z-y∕3)dxdydz.