Triple Integrals

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15.6 Triple Integrals

Triple Integrals

We have defined single integrals for functions of one

variable and double integrals for functions of two variables, so we can define triple integrals for functions of three

variables.

Let’s first deal with the simplest case where f is defined on a rectangular box:

B = {(x, y, z) | a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s}

The first step is to divide B into sub-boxes. We do this by dividing the interval [a, b] into l subintervals [x_{i – 1}, x_i] of equal width ∆x, dividing [c, d] into m subintervals of width

∆y, and dividing [r, s] into n subintervals of width ∆z.

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Triple Integrals

The planes through the endpoints of these subintervals parallel to the

coordinate planes divide the box B into lmn sub-boxes

B_ijk = [x_{i –1}, x_i] × [y_{j –1}, y_j] × [z_{k – 1}, z_k] which are shown in Figure 1.

Each sub-box has volume

∆V = ∆x ∆y ∆z.

Figure 1

Triple Integrals

Then we form the triple Riemann sum

where the sample point is in B_ijk.

By analogy with the definition of a double integral, we define the triple integral as the limit of the triple Riemann sums in (2).

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Triple Integrals

Again, the triple integral always exists if f is continuous. We can choose the sample point to be any point in the

sub-box, but if we choose it to be the point (x_i, y_j, z_k) we get a simpler-looking expression for the triple integral:

Triple Integrals

Just as for double integrals, the practical method for

evaluating triple integrals is to express them as iterated integrals as follows.

The iterated integral on the right side of Fubini’s Theorem means that we integrate first with respect to x (keeping

y and z fixed), then we integrate with respect to y (keeping z fixed), and finally we integrate with respect to z.

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Triple Integrals

There are five other possible orders in which we can integrate, all of which give the same value.

For instance, if we integrate with respect to y, then z, and then x, we have

Example 1

Evaluate the triple integral

∫∫∫

B = {(x, y, z) | 0 ≤ x ≤ 1, –1 ≤ y ≤ 2, 0 ≤ z ≤ 3}

Solution:

We could use any of the six possible orders of integration.

If we choose to integrate with respect to x, then y, and then z, we obtain

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Example 1 – Solution

Triple Integrals

Now we define the triple integral over a general bounded region E in three-dimensional space (a solid) by much the same procedure that we used for double integrals.

We enclose E in a box B of the type given by Equation 1.

Then we define F so that it agrees with f on E but is 0 for points in B that are outside E.

By definition,

This integral exists if f is continuous and the boundary of E is “reasonably smooth.”

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Triple Integrals

The triple integral has essentially the same properties as the double integral.

We restrict our attention to continuous functions f and to certain simple types of regions.

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Triple Integrals

A solid region E is said to be of type 1 if it lies between the graphs of two continuous functions of x and y, that is,

E = {(x, y, z) | (x, y) ∈ D, u₁(x, y) ≤ z ≤ u₂(x, y)}

where D is the projection of E onto the xy-plane as shown in Figure 2.

A type 1 solid region

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Triple Integrals

Notice that the upper boundary of the solid E is the surface with equation z = u₂(x, y), while the lower boundary is the surface z = u₁(x, y).

By the same sort of argument, it can be shown that if E is a type 1 region given by Equation 5, then

The meaning of the inner integral on the right side of Equation 6 is that x and y are held fixed, and therefore u₁(x, y) and u₂(x, y) are regarded as constants, while f(x, y, z) is integrated with respect to z.

Triple Integrals

In particular, if the projection D of E onto the xy-plane is a type I plane region (as in Figure 3).

Figure 3

A type 1 solid region where the projection D is a type I plane region

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Triple Integrals

Then

E = {(x, y, z) | a ≤ x ≤ b, g₁(x) ≤ y ≤ g₂(x), u₁(x, y) ≤ z ≤ u₂(x, y)}

and Equation 6 becomes

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Triple Integrals

If, on the other hand, D is a type II plane region (as in Figure 4), then

E = {(x, y, z) | c ≤ y ≤ d, h₁(y) ≤ x ≤ h₂(y), u₁(x, y) ≤ z ≤ u₂(x, y)}

and Equation 6 becomes

A type 1 solid region with a type II projection

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A solid region E is of type 2 if it is of the form

E = {(x, y, z) | (y, z) ∈ D, u₁(y, z) ≤ x ≤ u₂(y, z)}

where, this time, D is the projection of E onto the yz-plane (see Figure 7).

The back surface is x = u₁(y, z), the front surface is x = u₂(y, z), and we have

Figure 7

A type 2 region

Triple Integrals

Triple Integrals

Finally, a type 3 region is of the form

E = {(x, y, z) | (x, z) ∈ D, u₁(x, z) ≤ y ≤ u₂(x, z)}

where D is the projection of E onto the xz-plane, y = u₁(x, z) is the left surface, and y = u₂(x, z) is the right surface

(see Figure 8).

A type 3 region

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Triple Integrals

For this type of region we have

In each of Equations 10 and 11 there may be two possible expressions for the integral depending on whether D is a type I or type II plane region (and corresponding to

Equations 7 and 8).

Applications of Triple Integrals

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Applications of Triple Integrals

We know that if f(x) ≥ 0, then the single integral

represents the area under the curve y = f(x) from a to b, and if f(x, y) ≥ 0, then the double integral

∫∫

represents the volume under the surface z = f(x, y) and above D.

The corresponding interpretation of a triple integral

∫∫∫

four-dimensional object and, of course, that is very difficult to visualize. (Remember that E is just the domain of the function f; the graph of f lies in four-dimensional space.)

Applications of Triple Integrals

Nonetheless, the triple integral

∫∫∫

interpreted in different ways in different physical situations, depending on the physical interpretations of x, y, z and

f(x, y, z).

Let’s begin with the special case where f(x, y, z) = 1 for all points in E. Then the triple integral does represent the

volume of E:

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Applications of Triple Integrals

For example, you can see this in the case of a type 1 region by putting f(x, y, z) = 1 in Formula 6:

and we know this represents the volume that lies between the surfaces z = u₁(x, y) and z = u₂(x, y).

Example 5

Use a triple integral to find the volume of the tetrahedron T bounded by the planes x + 2y + z = 2, x = 2y, x = 0, and z = 0.

Solution:

The tetrahedron T and its projection D onto the xy-plane are shown in Figures 14 and 15.

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Example 5 – Solution

The lower boundary of T is the plane z = 0 and the upper boundary is the plane x + 2y + z = 2, that is, z = 2 – x – 2y.

Therefore we have

(Notice that it is not necessary to use triple integrals to

compute volumes. They simply give an alternative method for setting up the calculation.)

cont’d

Applications of Triple Integrals

All the applications of double integrals can be immediately extended to triple integrals.

For example, if the density function of a solid object that occupies the region E is ρ(x, y, z), in units of mass per unit volume, at any given point (x, y, z), then its mass is

and its moments about the three coordinate planes are

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Applications of Triple Integrals

The center of mass is located at the point where

If the density is constant, the center of mass of the solid is called the centroid of E.

The moments of inertia about the three coordinate axes are

The total electric charge on a solid object occupying a region E and having charge density σ(x, y, z) is

If we have three continuous random variables X, Y, and Z, their joint density function is a function of three variables such that the probability that (X, Y, Z) lies in E is

Applications of Triple Integrals

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In particular,

The joint density function satisfies

Applications of Triple Integrals