Triple Integrals

Section 3.6 Triple Integrals

Motivating Questions

How are a triple Riemann sum and the corresponding triple integral of a continuous function $f = f (x, y, z)$ defined?
What are two things the triple integral of a function can tell us?

We have now learned that we define the double integral of a continuous function

f = f (x, y)

over a rectangle

R = [a, b] \times [c, d]

as a limit of a double Riemann sum, and that these ideas parallel the single-variable integral of a function

g = g (x)

on an interval

[a, b] .

Moreover, this double integral has natural interpretations and applications, and can even be considered over non-rectangular regions,

D .

For instance, given a continuous function

f

over a region

D,

the average value of

f,

f_{AVG (D)},

is given by

f_{AVG (D)} = \frac{1}{A (D)} \iint_{D} f (x, y) d A,

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where

A (D)

is the area of

D .

Likewise, if

δ (x, y)

describes a mass density function on a lamina over

D,

the mass,

M,

of the lamina is given by

M = \iint_{D} δ (x, y) d A .

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It is natural to wonder if it is possible to extend these ideas of double Riemann sums and double integrals for functions of two variables to triple Riemann sums and then triple integrals for functions of three variables. We begin investigating in Preview Activity 3.6.1.

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Preview Activity 3.6.1.

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Consider a solid piece of granite in the shape of a box

B = {(x, y, z) : 0 \leq x \leq 4, 0 \leq y \leq 6, 0 \leq z \leq 8},

whose density varies from point to point. Let

δ (x, y, z)

represent the mass density of the piece of granite at point

(x, y, z)

in kilograms per cubic meter (so we are measuring

x,

y,

and

z

in meters). Our goal is to find the mass of this solid.

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Recall that if the density was constant, we could find the mass by multiplying the density and volume; since the density varies from point to point, we will use the approach we did with two-variable lamina problems, and slice the solid into small pieces on which the density is roughly constant.

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Partition the interval

[0, 4]

into 2 subintervals of equal length, the interval

[0, 6]

into 3 subintervals of equal length, and the interval

[0, 8]

into 2 subintervals of equal length. This partitions the box

B

into sub-boxes as shown in Figure 3.6.1.

Figure 3.6.1. A partitioned three-dimensional domain.

Let $0 = x_{0} < x_{1} < x_{2} = 4$ be the endpoints of the subintervals of $[0, 4]$ after partitioning. Draw a picture of Figure 3.6.1 and label these endpoints on your drawing. Do likewise with $0 = y_{0} < y_{1} < y_{2} < y_{3} = 6$ and $0 = z_{0} < z_{1} < z_{2} = 8$ What is the length $Δ x$ of each subinterval $[x_{i - 1}, x_{i}]$ for $i$ from 1 to 2? the length of $Δ y ?$ of $Δ z ?$
The partitions of the intervals $[0, 4],$ $[0, 6]$ and $[0, 8]$ partition the box $B$ into sub-boxes. How many sub-boxes are there? What is volume $Δ V$ of each sub-box?
Let $B_{i j k}$ denote the sub-box $[x_{i - 1}, x_{i}] \times [y_{j - 1}, y_{j}] \times [z_{k - 1}, z_{k}] .$ Say that we choose a point $(x_{i j k}^{*}, y_{i j k}^{*}, z_{i j k}^{*})$ in the $i, j, k$ th sub-box for each possible combination of $i, j, k .$ What is the meaning of $δ (x_{i j k}^{*}, y_{i j k}^{*}, z_{i j k}^{*}) ?$ What physical quantity will $δ (x_{i j k}^{*}, y_{i j k}^{*}, z_{i j k}^{*}) Δ V$ approximate?
What final step(s) would it take to determine the exact mass of the piece of granite?

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Subsection 3.6.1 Triple Riemann Sums and Triple Integrals

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Through the application of a mass density distribution over a three-dimensional solid, Preview Activity 3.6.1 suggests that the generalization from double Riemann sums of functions of two variables to triple Riemann sums of functions of three variables is natural. In the same way, so is the generalization from double integrals to triple integrals. By simply adding a

z

-coordinate to our earlier work, we can define both a triple Riemann sum and the corresponding triple integral.

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Definition 3.6.2.

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Let

f = f (x, y, z)

be a continuous function on a box

B = [a, b] \times [c, d] \times [r, s] .

The triple Riemann sum of

f

over

B

is created as follows.

Partition the interval $[a, b]$ into $m$ subintervals of equal length $Δ x = \frac{b - a}{m} .$ Let $x_{0},$ $x_{1},$ $\dots,$ $x_{m}$ be the endpoints of these subintervals, where $a = x_{0} < x_{1} < x_{2} < \dots < x_{m} = b .$ Do likewise with the interval $[c, d]$ using $n$ subintervals of equal length $Δ y = \frac{d - c}{n}$ to generate $c = y_{0} < y_{1} < y_{2} < \dots < y_{n} = d,$ and with the interval $[r, s]$ using $ℓ$ subintervals of equal length $Δ z = \frac{s - r}{ℓ}$ to have $r = z_{0} < z_{1} < z_{2} < \dots < z_{l} = s .$
Let $B_{i j k}$ be the sub-box of $B$ with opposite vertices $(x_{i - 1}, y_{j - 1}, z_{k - 1})$ and $(x_{i}, y_{j}, z_{k})$ for $i$ between $1$ and $m,$ $j$ between $1$ and $n,$ and $k$ between 1 and $ℓ .$ The volume of each $B_{i j k}$ is $Δ V = Δ x \cdot Δ y \cdot Δ z .$
Let $(x_{i j k}^{*}, y_{i j k}^{*}, z_{i j k}^{*})$ be a point in box $B_{i j k}$ for each $i,$ $j,$ and $k .$ The resulting triple Riemann sum for $f$ on $B$ is

$\sum_{i = 1}^{m} \sum_{j = 1}^{n} \sum_{k = 1}^{ℓ} f (x_{i j k}^{*}, y_{i j k}^{*}, z_{i j k}^{*}) \cdot Δ V .$

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f (x, y, z)

represents the mass density of the box

B,

then, as we saw in Preview Activity 3.6.1, the triple Riemann sum approximates the total mass of the box

B .

In order to find the exact mass of the box, we need to let the number of sub-boxes increase without bound (in other words, let

m,

n,

and

ℓ

go to infinity); in this case, the finite sum of the mass approximations becomes the actual mass of the solid

B .

More generally, we have the following definition of the triple integral.

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Definition 3.6.3.

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With following notation defined as in a triple Riemann sum, the triple integral of

f

over

B

∭_{B} f (x, y, z) d V = lim_{m, n, ℓ \to \infty} \sum_{i = 1}^{m} \sum_{j = 1}^{n} \sum_{k = 1}^{ℓ} f (x_{i j k}^{*}, y_{i j k}^{*}, z_{i j k}^{*}) \cdot Δ V .

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As we noted earlier, if

f (x, y, z)

represents the density of the solid

B

at each point

(x, y, z),

then

M = ∭_{B} f (x, y, z) d V

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is the mass of

B .

Even more importantly, for any continuous function

f

over the solid

B,

we can use a triple integral to determine the average value of

f

over

B,

f_{AVG (B)} .

We note this generalization of our work with functions of two variables along with several others in the following important boxed information. Note that each of these quantities may actually be considered over a general domain

S

R^{3},

not simply a box,

B .

The triple integral

$V (S) = ∭_{S} 1 d V$

represents the volume of the solid $S$ .
The average value of the function $f = f (x, y, x)$ over a solid domain $S$ is given by

$f_{AVG (S)} = (\frac{1}{V (S)}) ∭_{S} f (x, y, z) d V,$

where $V (S)$ is the volume of the solid $S .$
The center of mass of the solid $S$ with density $δ = δ (x, y, z)$ is $(\overset{―}{x}, \overset{―}{y}, \overset{―}{z}),$ where

$\begin{aligned} \overset{―}{x} & = \frac{∭_{S} x δ (x, y, z) d V}{M}, \\ \overset{―}{y} & = \frac{∭_{S} y δ (x, y, z) d V}{M}, \\ \overset{―}{z} & = \frac{∭_{S} z δ (x, y, z) d V}{M}, \end{aligned}$

and $M = ∭_{S} δ (x, y, z) d V$ is the mass of the solid $S .$

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In the Cartesian coordinate system, the volume element

d V

d z d y d x,

and, as a consequence, a triple integral of a function

f

over a box

B = [a, b] \times [c, d] \times [r, s]

in Cartesian coordinates can be evaluated as an iterated integral of the form

∭_{B} f (x, y, z) d V = \int_{a}^{b} \int_{c}^{d} \int_{r}^{s} f (x, y, z) d z d y d x .

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If we want to evaluate a triple integral as an iterated integral over a solid

S

that is not a box, then we need to describe the solid in terms of variable limits.

Checkpoint 3.6.4.

Set up and evaluate the triple integral of

f (x, y, z) = x + 2 y - 3 z

over the box

B = [- 1, 3] \times [0, 4] \times [- 1, 2] .

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Example 3.6.5.

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Let

S

be the solid cone bounded by

z = \sqrt{x^{2} + y^{2}}

and

z = 3 .

A picture of

S

is shown at right in Figure 3.6.6. Our goal in what follows is to set up an iterated integral of the form

\begin{matrix} (3.6.1) & \int_{x = ?}^{x = ?} \int_{y = ?}^{y = ?} \int_{z = ?}^{z = ?} δ (x, y, z) d z d y d x \end{matrix}

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to represent the mass of

S

in the setting where

δ (x, y, z)

tells us the density of

S

at the point

(x, y, z) .

Our particular task is to find the limits on each of the three integrals.

Figure 3.6.6. Left: The cone. Right: Its projection.

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If we think about slicing up the solid, we can consider slicing the domain of the solid’s projection onto the $x y$ -plane (just as we would slice a two-dimensional region in $R^{2}$ ), and then slice in the $z$ -direction as well. The projection of the solid onto the $x y$ -plane is shown at left in Figure 3.6.6. If we decide to first slice the domain of the solid’s projection perpendicular to the $x$ -axis, over what range of constant $x$ -values would we have to slice?
If we continue with slicing the domain, what are the limits on $y$ on a typical slice? How do these depend on $x ?$ What, therefore, are the limits on the middle integral?
Finally, now that we have thought about slicing up the two-dimensional domain that is the projection of the cone, what are the limits on $z$ in the innermost integral? Note that over any point $(x, y)$ in the plane, a vertical slice in the $z$ direction will involve a range of values from the cone itself to its flat top. In particular, observe that at least one of these limits is not constant but depends on $x$ and $y .$
In conclusion, write an iterated integral of the form (3.6.1) that represents the mass of the cone $S .$

Solution.

The shadow of the figure on the left on the $x$ -axis is the interval $[- 3, 3] .$ So this is our range of $x$ -values.
For a fixed $x$ between $- 3$ and $3,$ the possible $y$ -values range from $- \sqrt{9 - x^{2}}$ (the lower semicircle) to $\sqrt{9 - x^{2}}$ (the upper semicircle). So the middle integral in the triple integral should look like $\int_{y = - \sqrt{9 - x^{2}}}^{\sqrt{9 - x^{2}}} .$
Over a fixed point $(x, y)$ in the shadow of the conic region, the possible $z$ -values range from the cone up to the plane $z = 3 .$ The equation of the cone is $z = \sqrt{x^{2} + y^{2}},$ so the innermost integral should look like $\int_{z = \sqrt{x^{2} + y^{2}}}^{3} .$
$\int_{x = - 3}^{3} \int_{y = - \sqrt{9 - x^{2}}}^{\sqrt{9 - x^{2}}} \int_{z = \sqrt{x^{2} + y^{2}}}^{3} δ (x, y, z) d z d y d x$

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Example 3.6.7.

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Set up the integral of a function

f (x, y, z)

over the region bounded by the surface

z = 4 - x^{2}

and by the planes

y = 1,

y = 2,

and the

x y

-plane.

Solution.

The surface

z = 4 - x^{2}

intersects the

x y

-plane in the lines

x = - 2

and

x = 2 .

(Remember that the

x y

-plane is defined by the equation

z = 0 .

Set

z = 0

in the equation

z = 4 - x^{2}

and solve for

x

to get

\pm 2 .

) So the shadow of our region in the

x y

-plane is the rectangle

[- 2, 2] \times [1, 2] .

Noting that the region is bounded below by the

x y

-plane and above by the surface

z = 4 - x^{2},

we are ready to set up our integral:

\int_{x = - 2}^{2} \int_{y = 1}^{2} \int_{z = 0}^{4 - x^{2}} f (x, y, z) d z d y d x

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Note well: When setting up iterated integrals, the limits on a given variable can be only in terms of the remaining variables. In addition, there are multiple different ways we can choose to set up such an integral. For example, two possibilities for iterated integrals that represent a triple integral

∭_{S} f (x, y, z) d V

over a solid

S

are

$\int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} \int_{h_{1} (x, y)}^{h_{2} (x, y)} f (x, y, z) d z d y d x$
$\int_{r}^{s} \int_{p_{1} (z)}^{p_{2} (z)} \int_{q_{1} (x, z)}^{q_{2} (x, z)} f (x, y, z) d y d x d z$

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where

g_{1},

g_{2},

h_{1},

h_{2},

p_{1},

p_{2},

q_{1},

and

q_{2}

are functions of the indicated variables. There are four other options beyond the two stated here, since the variables

x,

y,

and

z

can (theoretically) be arranged in any order. Of course, in many circumstances, an insightful choice of variable order will make it easier to set up an iterated integral, just as was the case when we worked with double integrals.

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Example 3.6.8.

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Find the mass of the tetrahedron in the first octant bounded by the coordinate planes and the plane

x + 2 y + 3 z = 6

if the density at point

(x, y, z)

is given by

δ (x, y, z) = x + y + z .

A picture of the solid tetrahedron is shown at left in Figure 3.6.9.

Figure 3.6.9. Left: The tetrahedron. Right: Its projection.

Solution.

We find the mass,

M,

of the tetrahedron by the triple integral

M = ∭_{S} δ (x, y, z) d V,

where

S

is the solid tetrahedron described above. In this example, we choose to integrate with respect to

z

first for the innermost integral. The top of the tetrahedron is given by the equation

x + 2 y + 3 z = 6;

solving for

z

then yields

z = \frac{1}{3} (6 - x - 2 y) .

The bottom of the tetrahedron is the

x y

-plane, so the limits on

z

in the iterated integral will be

0 \leq z \leq \frac{1}{3} (6 - x - 2 y) .

To find the bounds on

x

and

y

we project the tetrahedron onto the

x y

-plane; this corresponds to setting

z = 0

in the equation

z = \frac{1}{3} (6 - x - 2 y) .

The resulting relation between

x

and

y

x + 2 y = 6.

The right image in Figure 3.6.9 shows the projection of the tetrahedron onto the

x y

-plane.

If we choose to integrate with respect to

y

for the middle integral in the iterated integral, then the lower limit on

y

is the

x

-axis and the upper limit is the hypotenuse of the triangle. Note that the hypotenuse joins the points

(6, 0)

and

(0, 3)

and so has equation

y = 3 - \frac{1}{2} x .

Thus, the bounds on

y

are

0 \leq y \leq 3 - \frac{1}{2} x .

Finally, the

x

values run from 0 to 6, so the iterated integral that gives the mass of the tetrahedron is

\begin{matrix} (3.6.2) & M = \int_{0}^{6} \int_{0}^{3 - (1 / 2) x} \int_{0}^{(1 / 3) (6 - x - 2 y)} x + y + z d z d y d x . \end{matrix}

Evaluating the triple integral gives us

\begin{aligned} M & = \int_{0}^{6} \int_{0}^{3 - (1 / 2) x} \int_{0}^{(1 / 3) (6 - x - 2 y)} x + y + z d z d y d x \\ = \int_{0}^{6} \int_{0}^{3 - (1 / 2) x} [x z + y z + \frac{z}{2}] |_{0}^{(1 / 3) (6 - x - 2 y)} d y d x \\ = \int_{0}^{6} \int_{0}^{3 - (1 / 2) x} \frac{4}{3} x - \frac{5}{18} x^{2} - \frac{}{9} x y + \frac{2}{3} y - \frac{4}{9} y^{2} + 2 d y d x \\ = \int_{0}^{6} [\frac{4}{3} x y - \frac{5}{18} x^{2} y - \frac{7}{18} x y^{2} + \frac{1}{3} y^{2} - \frac{4}{27} y^{3} + 2 y] |_{0}^{3 - (1 / 2) x} d x \\ = \int_{0}^{6} 5 + \frac{1}{2} x - \frac{7}{12} x^{2} + \frac{13}{216} x^{3} d x \\ = [5 x + \frac{1}{4} x^{2} - \frac{7}{36} x^{3} + \frac{13}{864} x^{4}] |_{0}^{6} \\ = \frac{33}{2} . \end{aligned}

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Setting up limits on iterated integrals can require considerable geometric intuition. It is important to not only create carefully labeled figures, but also to think about how we wish to slice the solid. Further, note that when we say “we will integrate first with respect to

x,

” by “first” we are referring to the innermost integral in the iterated integral. The next activity explores several different ways we might set up the integral in the preceding example.

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Example 3.6.10.

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Let

T

be the region inside the tetrahedron with vertices

(0, 0, 0),

(5, 5, 0),

(0, 5, 0),

and

(0, 5, 6) .

Write the integral

∭_{T} f (x, y, z) d V

as an iterated (triple) integral in at least three different ways, using different orders of integration.

Solution.

\begin{aligned} ∭_{T} f (x, y, z) d V & = \int_{x = 0}^{5} \int_{y = x}^{5} \int_{z = 0}^{\frac{1}{5} (6 y - 6 x)} f (x, y, z) d z d y d x \\ = \int_{x = 0}^{5} \int_{z = 0}^{\frac{1}{5} (30 - 6 x)} \int_{y = \frac{1}{6} (6 x + 5 z)}^{5} f (x, y, z) d y d z d x \\ = \int_{y = 0}^{5} \int_{x = 0}^{y} \int_{z = 0}^{\frac{1}{5} (6 y - 6 x)} f (x, y, z) d z d x d y \\ = \int_{z = 0}^{6} \int_{y = \frac{5}{6} z}^{5} \int_{x = 0}^{\frac{1}{6} (6 y - 5 z)} f (x, y, z) d x d y d z \end{aligned}

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Now that we have begun to understand how to set up iterated triple integrals, we can apply them to determine important quantities, such as those found in the next activity.

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Activity 3.6.2.

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A solid

S

is bounded below by the square

z = 0,

- 1 \leq x \leq 1,

- 1 \leq y \leq 1

and above by the surface

z = 2 - x^{2} - y^{2} .

A picture of the solid is shown in Figure 3.6.11.

Figure 3.6.11. The solid bounded by the surface $z = 2 - x^{2} - y^{2} .$

First, set up an iterated double integral to find the volume of the solid $S$ as a double integral of a solid under a surface. Then set up an iterated triple integral that gives the volume of the solid $S .$ You do not need to evaluate either integral. Compare the two approaches.
Set up (but do not evaluate) iterated integral expressions that will tell us the center of mass of $S,$ if the density at point $(x, y, z)$ is $δ (x, y, z) = x^{2} + 1 .$
Set up (but do not evaluate) an iterated integral to find the average density on $S$ using the density function from part (b).
Use technology appropriately to evaluate the iterated integrals you determined in (a), (b), and (c); does the location you determined for the center of mass make sense?

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Subsection 3.6.2 Summary

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Let $f = f (x, y, z)$ be a continuous function on a box $B = [a, b] \times [c, d] \times [r, s] .$ The triple integral of $f$ over $B$ is defined as

$∭_{B} f (x, y, z) d V = lim_{Δ V \to 0} \sum_{i = 1}^{m} \sum_{j = 1}^{n} \sum_{k = 1}^{l} f (x_{i j k}^{*}, y_{i j k}^{*}, z_{i j k}^{*}) \cdot Δ V,$

where the triple Riemann sum is defined in the usual way. The definition of the triple integral naturally extends to non-rectangular solid regions $S .$
The triple integral $∭_{S} f (x, y, z) d V$ can tell us
- -.
  the volume of the solid $S$ if $f (x, y, z) = 1,$
- -.
  the mass of the solid $S$ if $f$ represents the density of $S$ at the point $(x, y, z) .$
Moreover,

$f_{AVG (S)} = \frac{1}{V (S)} ∭_{S} f (x, y, z) d V,$

is the average value of $f$ over $S .$

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Exercises 3.6.3 Exercises

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1.

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Find the triple integral of the function

f (x, y, z) = x^{4} \cos (y + z)

over the cube

4 \leq x \leq 7,

0 \leq y \leq π,

0 \leq z \leq π .

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2.

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Evaluate the triple integral

\int \int \int_{E} x y z d V

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where E is the solid:

0 \leq z \leq 7, 0 \leq y \leq z, 0 \leq x \leq y .

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3.

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Find the mass of the rectangular prism

0 \leq x \leq 4, 0 \leq y \leq 4, 0 \leq z \leq 3,

with density function

ρ (x, y, z) = x .

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4.

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Find the average value of the function

f (x, y, z) = y e^{- x y}

over the rectangular prism

0 \leq x \leq 5,

0 \leq y \leq 4,

0 \leq z \leq 1

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5.

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Find the volume of the solid bounded by the planes x = 0, y = 0, z = 0, and x + y + z = 1.

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6.

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Find the mass of the solid bounded by the

x y

-plane,

y z

-plane,

x z

-plane, and the plane

(x / 2) + (y / 4) + (z / 8) = 1,

if the density of the solid is given by

δ (x, y, z) = x + 4 y .

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mass =

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7.

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The moment of inertia of a solid body about an axis in 3-space relates the angular acceleration about this axis to torque (force twisting the body). The moments of inertia about the coordinate axes of a body of constant density and mass

m

occupying a region

W

of volume

V

are defined to be

I_{x} = \frac{m}{V} \int_{W} (y^{2} + z^{2}) d V I_{y} = \frac{m}{V} \int_{W} (x^{2} + z^{2}) d V I_{z} = \frac{m}{V} \int_{W} (x^{2} + y^{2}) d V

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Use these definitions to find the moment of inertia about the

z

-axis of the rectangular solid of mass

12

given by

0 \leq x \leq 2,

0 \leq y \leq 3,

0 \leq z \leq 2 .

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I_{x} =

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I_{y} =

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I_{z} =

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8.

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Express the integral

∭_{E} f (x, y, z) d V

as an iterated integral in six different ways, where E is the solid bounded by

z = 0, x = 0, z = y - 3 x

and

y = 6 .

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\int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} \int_{h_{1} (x, y)}^{h_{2} (x, y)} f (x, y, z) d z d y d x

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a =

b =

🔗

g_{1} (x) =

g_{2} (x) =

🔗

h_{1} (x, y) =

h_{2} (x, y) =

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\int_{a}^{b} \int_{g_{1} (y)}^{g_{2} (y)} \int_{h_{1} (x, y)}^{h_{2} (x, y)} f (x, y, z) d z d x d y

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a =

b =

🔗

g_{1} (y) =

g_{2} (y) =

🔗

h_{1} (x, y) =

h_{2} (x, y) =

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\int_{a}^{b} \int_{g_{1} (z)}^{g_{2} (z)} \int_{h_{1} (y, z)}^{h_{2} (y, z)} f (x, y, z) d x d y d z

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a =

b =

🔗

g_{1} (z) =

g_{2} (z) =

🔗

h_{1} (y, z) =

h_{2} (y, z) =

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\int_{a}^{b} \int_{g_{1} (y)}^{g_{2} (y)} \int_{h_{1} (y, z)}^{h_{2} (y, z)} f (x, y, z) d x d z d y

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a =

b =

🔗

g_{1} (y) =

g_{2} (y) =

🔗

h_{1} (y, z) =

h_{2} (y, z) =

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\int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} \int_{h_{1} (x, z)}^{h_{2} (x, z)} f (x, y, z) d y d z d x

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a =

b =

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g_{1} (x) =

g_{2} (x) =

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h_{1} (x, z) =

h_{2} (x, z) =

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\int_{a}^{b} \int_{g_{1} (z)}^{g_{2} (z)} \int_{h_{1} (x, z)}^{h_{2} (x, z)} f (x, y, z) d y d x d z

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a =

b =

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g_{1} (z) =

g_{2} (z) =

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h_{1} (x, z) =

h_{2} (x, z) =

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9.

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Calculate the volume under the elliptic paraboloid

z = 2 x^{2} + 6 y^{2}

and over the rectangle

R = [- 1, 1] \times [- 2, 2] .

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10.

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The motion of a solid object can be analyzed by thinking of the mass as concentrated at a single point, the center of mass. If the object has density

ρ (x, y, z)

at the point

(x, y, z)

and occupies a region

W,

then the coordinates

(\overset{―}{x}, \overset{―}{y}, \overset{―}{z})

of the center of mass are given by

\overset{―}{x} = \frac{1}{m} \int_{W} x ρ d V \overset{―}{y} = \frac{1}{m} \int_{W} y ρ d V \overset{―}{z} = \frac{1}{m} \int_{W} z ρ d V,

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Assume

x,

y,

z

are in cm. Let

C

be a solid cone with both height and radius 3 and contained between the surfaces

z = \sqrt{x^{2} + y^{2}}

and

z = 3 .

C

has constant mass density of 3 g/cm

^{3},

find the

z

-coordinate of

C

’s center of mass.

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\overset{―}{z} =

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(Include units
¹
/webwork2_files/helpFiles/Units.html
.)

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11.

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Without calculation, decide if each of the integrals below are positive, negative, or zero. Let W be the solid bounded by

z = \sqrt{x^{2} + y^{2}}

and

z = 2 .

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$\underset{W}{∭} (z - \sqrt{x^{2} + y^{2}}) d V$
$\underset{W}{∭} e^{- x y z} d V$
$\underset{W}{∭} (z - 2) d V$

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12.

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Set up a triple integral to find the mass of the solid tetrahedron bounded by the xy-plane, the yz-plane, the xz-plane, and the plane

x / 6 + y / 5 + z / 30 = 1,

if the density function is given by

δ (x, y, z) = x + y .

Write an iterated integral in the form below to find the mass of the solid.

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\underset{R}{∭} f (x, y, z) d V = \int_{A}^{B} \int_{C}^{D} \int_{E}^{F}

d z d y d x

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with limits of integration

A =

B =

C =

D =

E =

F =

13.

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Consider the solid

S

that is bounded by the parabolic cylinder

y = x^{2}

and the planes

z = 0

and

z = 1 - y

as shown in Figure 3.6.12.

Figure 3.6.12. The solid bounded by $y = x^{2}$ and the planes $z = 0$ and $z = 1 - y .$

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Assume the density of

S

is given by

δ (x, y, z) = z

Set up (but do not evaluate) an iterated integral that represents the mass of $S .$ Integrate first with respect to $z,$ then $y,$ then $x .$ A picture of the projection of $S$ onto the $x y$ -plane is shown at left in Figure 3.6.13.
Set up (but do not evaluate) an iterated integral that represents the mass of $S .$ In this case, integrate first with respect to $y,$ then $z,$ then $x .$ A picture of the projection of $S$ onto the $x z$ -plane is shown at center in Figure 3.6.13.
Set up (but do not evaluate) an iterated integral that represents the mass of $S .$ For this integral, integrate first with respect to $x,$ then $y,$ then $z .$ A picture of the projection of $S$ onto the $y z$ -plane is shown at right in Figure 3.6.13.
Which of these three orders of integration is the most natural to you? Why?

Figure 3.6.13. Projections of $S$ onto the $x y,$ $x z,$ and $y z$ -planes.

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14.

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This problem asks you to investigate the average value of some different quantities.

Set up, but do not evaluate, an iterated integral expression whose value is the average sum of all real numbers $x,$ $y,$ and $z$ that have the following property: $y$ is between 0 and 2, $x$ is greater than or equal to 0 but cannot exceed $2 y,$ and $z$ is greater than or equal to 0 but cannot exceed $x + y .$
Set up, but do not evaluate, an integral expression whose value represents the average value of $f (x, y, z) = x + y + z$ over the solid region in the first octant bounded by the surface $z = 4 - x - y^{2}$ and the coordinate planes $x = 0,$ $y = 0,$ $z = 0 .$
How are the quantities in (a) and (b) similar? How are they different?

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15.

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Consider the solid that lies between the paraboloids

z = g (x, y) = x^{2} + y^{2}

and

z = f (x, y) = 8 - 3 x^{2} - 3 y^{2} .

By eliminating the variable $z,$ determine the curve of intersection between the two paraboloids, and sketch this curve in the $x y$ -plane.
Set up, but do not evaluate, an iterated integral expression whose value determines the mass of the solid, integrating first with respect to $z,$ then $y,$ then $x .$ Assume the the solid’s density is given by $δ (x, y, z) = \frac{1}{x^{2} + y^{2} + z^{2} + 1} .$
Set up, but do not evaluate, iterated integral expressions whose values determine the mass of the solid using all possible remaining orders of integration. Use $δ (x, y, z) = \frac{1}{x^{2} + y^{2} + z^{2} + 1}$ as the density of the solid.
Set up, but do not evaluate, iterated integral expressions whose values determine the center of mass of the solid. Again, assume the the solid’s density is given by $δ (x, y, z) = \frac{1}{x^{2} + y^{2} + z^{2} + 1} .$
Which coordinates of the center of mass can you determine without evaluating any integral expression? Why?

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16.

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In each of the following problems, your task is to

(i).
sketch, by hand, the region over which you integrate
(ii).
set up iterated integral expressions which, when evaluated, will determine the value sought
(iii).
use appropriate technology to evaluate each iterated integral expression you develop

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Note well: in some problems you may be able to use a double rather than a triple integral, and polar coordinates may be helpful in some cases.

Consider the solid created by the region enclosed by the circular paraboloid $z = 4 - x^{2} - y^{2}$ over the region $R$ in the $x y$ -plane enclosed by $y = - x$ and the circle $x^{2} + y^{2} = 4$ in the first, second, and fourth quadrants. Determine the solid’s volume.
Consider the solid region that lies beneath the circular paraboloid $z = 9 - x^{2} - y^{2}$ over the triangular region between $y = x,$ $y = 2 x,$ and $y = 1 .$ Assuming that the solid has its density at point $(x, y, z)$ given by $δ (x, y, z) = x y z + 1,$ measured in grams per cubic cm, determine the center of mass of the solid.
In a certain room in a house, the walls can be thought of as being formed by the lines $y = 0,$ $y = 12 + x / 4,$ $x = 0,$ and $x = 12,$ where length is measured in feet. In addition, the ceiling of the room is vaulted and is determined by the plane $z = 16 - x / 6 - y / 3 .$ A heater is stationed in the corner of the room at $(0, 0, 0)$ and causes the temperature in the room at a particular time to be given by

$T (x, y, z) = \frac{80}{1 + \frac{x^{2}}{1000} + \frac{y^{2}}{1000} + \frac{z^{2}}{1000}}$

What is the average temperature in the room?
Consider the solid enclosed by the cylinder $x^{2} + y^{2} = 9$ and the planes $y + z = 5$ and $z = 1 .$ Assuming that the solid’s density is given by $δ (x, y, z) = \sqrt{x^{2} + y^{2}},$ find the mass and center of mass of the solid.

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