298
Chapter 14: Multiple Integrals
Summary: The previous chapter focused upon taking derivatives and limits of
multivariable functions. In this chapter, the integration of multivariable functions
is considered. This naturally leads to integrating functions with respect to two or
three variables. The integration of these types of functions over their domains is
described best using double and triple integrals. Double integrals are useful for
calculating the area of a two dimensional region or its center of mass. They are
also useful for calculating the surface area of a surface determined by a function
z = f (x, y). Triple integrals allow the mass of an object to be determined and
consequently the location of its center of mass.
In many problems, situations in two and three dimensions are not easily described
using Cartesian or rectangular coordinates. In some cases, problems may be more
easily considered if the coordinates are given as polar coordinates in two dimensions or by cylindrical coordinates or spherical coordinates in three dimensions.
Just as u-substitution played an important role in the development of integration in
one dimension, using a change of variables is important in both double and triple
integral problems. In particular, when switching from rectangular coordinates to
either polar, cylindrical or spherical coordinates, a change of variables is being
performed to switch from one set of coordinates to another.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Set up double integrals and evaluate double integrals (§14.1).
2. Write double integrals with nonconstant limits of integration (§14.2).
3. Set up double integrals over nonrectangular regions (using rectangular, §14.2,
or polar coordinates, §14.3).
4. Find the volume under a surface using double integrals (§14.1 − 14.3).
5. Switch the order of integration in double integrals (§14.2, 14.3).
6. Change between rectangular and polar coordinates in double integrals (§14.3).
7. Calculate area using double integrals (§14.2, 14.3).
8. Define surfaces parametrically (§14.4).
9. Express a parametric surface in vector form (§14.4).
10. Find tangent planes to parametric surfaces and surface area of a parametric
surface (§14.4).
11. Calculate volume under a surface using a triple integral (§14.5).
12. Determine appropriate limits of integration for triple integrals (§14.5).
13. Set up and evaluate triple integrals using both cylindrical and spherical coordinates (§14.6).
14. Convert between different coordinate systems for triple integrals (§14.6).
15. Find the Jacobian and perform general changes of variables in double and
triple integrals (§14.7).
16. Find the centroid of a lamina using double integrals (§14.8).
17. Find the centroid of a solid using triple integrals (§14.8).
14.1 Double Integrals
PURPOSE: To introduce double integrals and to explain how they
are evaluated and some of their properties.
net signed volume
region of integration
Double integrals provide a way to integrate beneath a surface such as z = f (x, y)
that is defined in terms of two variables. Single integrals found the net signed
area beneath a function. Similarly, double integrals find the net signed volume
beneath a function z = f (x, y). The limits of integration have a similar meaning
as for single integrals. In the single integral case, the limits of integration defined
the interval over which the integration would occur. Double integrals describe a
region in a plane over which the integration occurs.
To make the transition from using single integrals to double integrals requires an
understanding of two concepts. These two concepts are then repeatedly applied
throughout the remainder of this chapter (and again with triple integrals). The first
idea is that a double integral is essentially just a set of nested single integrals. That
is, a double integral can be viewed as one integral which is inside of another (this
is referred to as iterated integrals). The other concept is that the limits of integration define a two-dimensional region in the xy-plane over which the integration is
taking place.
IDEA: Two important ideas related to double and triple integrals are that (i) the
limits of integration define the domain over the integration and that (ii) multiple
integrals should be evaluated from the inside out.
evaluating double integrals
To evaluate double integrals requires that the inside integral be evaluated first.
The result of this integration becomes the integrand of the outside integral which
can then be evaluated. Each integral that is evaluated is only integrated with respect
to one variable indicated by the order of the differentials. For example,
RR
f (x, y) dxdy indicates that the inside integral is done with respect to x and then
the outer integral is done with respect to y.
IDEA: Evaluating a double integral is done in the same way as a single integral
except that two integrals will need to be evaluated.
One interesting property of double integrals is that the order
of integration may
RR
be changed. A double integral is typically represented as
f (x, y) dA where dA
can either equal dxdy (integrate with respect to x first) or dydx (integrate with
respect to y first).
order of integration
IDEA: The dA in a double integral indicates in which order the integration
should be done.
The order of integration does not affect the region, R, that is being integrated over.
However, if the order of integration is changed, care should be taken in rewriting
the limits of integration since they may change. Limits of integration are talked
about more in the next section when nonrectangular regions are considered.
CAUTION: Changing the order of integration involves changing the limits of
integration. This is not necessarily done by just switching the limits of integration.
Double integrals may be used to find the volume beneath a surface defined by
z = f (x, y). They may also be used to find the area of a region. To find the
area,
RR
the function in the integrand, f (x, y), should just be set equal to one. So R 1 dA
will calculate the area of the region R. Notice that this is written for a generic dA.
This same process will hold true with nonrectangular regions (see §14.2) and in
polar coordinates that will be introduced later (see §14.3).
Checklist of Key Ideas:
volume under a surface
infinitesimal rectangle
Riemann sum of infinitesimal rectangles
double integral of f (x, y) over a region R
net signed volume
partial integration
iterated integration and iterated integrals
order of integration
properties of double integrals
subdividing a region R into smaller subregions
calculating volume and area
14.2 Double Integrals over Nonrectangular Regions
PURPOSE: To describe how double integrals should be set up
over a nonrectangular region and how to evaluate double integrals over a nonrectangular region.
limits of integration
The main topic of discussion in this section regards the limits of integration of
double integrals. The limits of integration describe the region over which the
double integral is being evaluated. So there will be two x limits of integration and
two y limits of integration. These four things will determine what the region R
looks like. Remember that changing the limits of integration will not change the
region. On the other hand, changing the order of integration does not mean that
the limits of integration may simply be just interchanged.
CAUTION: When changing the order of integration, do not simply switch the
limits of integration. The limits of integration need to be picked carefully so
that they accurately describe the region R.
One basic idea to grasp onto is that when an integral is finished being evaluated
over a fixed region, R, the result should be a scalar value that does not involve
variables such as x or y. This does not mean that limits of integration have to
always be constant values however. On the other hand, the outer most limits of
integration should be constant values.
IDEA: Type I regions are defined by curves at the top and bottom and vertical
lines at the left and right. Type II regions are defined by curves at the left and
right and horizontal lines at the top and bottom.
type I and II regions
type I → dA = dydx
tpye II → dA = dxdy
Two types of nonrectangular regions are considered in this text: type I and type
II regions. Type I regions are bounded by curves on the top and bottom. The inner
most integration should be set RR
up between the curves. So for Type I regions, the
integral should have the order R f (x, y) dydx. The dy indicates that integration
in the y direction between the curves at the top and the bottom will occur first. For
Type II regions, the region R will be bounded on the left and right by curves and by
lines on the top and bottom. Again, the double integral should be set up so that the
inner most integral will integrate between the curves. In this case,
this is in the x
RR
direction (from left to right) so the integrals should be set up like R f (x, y) dxdy.
Sometimes an integral may be simpler to write in Type I fashion compared to
Type II fashion or vice versa. For example, consider the following double integral
written for a type II region (curves on the left and right):
y=3 x=(11−y)/2
Z
Z
y=1
f (x, y) dxdy
x=y
This may also be written in a type I fashion although it will require more integrals:
y=x
x=3 Z
Z
x=1 y=1
f (x, y) dydx +
y=3
x=4
Z Z
x=3 y=1
f (x, y) dydx +
x=5 y=11−2x
Z
Z
x=4
y=1
f (x, y) dydx
Incidentally, the region in this example is the trapezoid with corners at the points
(1, 1), (3, 3), (4, 3), and (5, 1) in the xy-plane.
IDEA: To change limits of integration, be sure that the new limits of integration
describe the same region.
CAUTION: Sometimes changing the limits of integration may require that more
than one double integral be evaluated. Also, generally the limits of integration
may not simply be interchanged.
Checklist of Key Ideas:
nonconstant limits of integration
type I nonrectangular region
type II nonrectangular region
changing order of integration with nonconstant limits of integration
finding area of a region R
14.3 Double Integrals in Polar Coordinates
PURPOSE: To explain how double integrals can be written in
either rectangular or polar coordinates and to explain how double
integrals with polar coordinates may be set up and evaluated.
In this section, double integrals using polar coordinates are described. Recall that
the relationship between rectangular coordinates and polar coordinates are give
by the following equations:
polar coordinates
see §10.2
x = r cos θ , y = r sin θ
and
r2 = x2 + y2 , tan θ = y/x
The same things that are important for double integrals in rectangular coordinates
are important for double integrals in polar coordinates. In particular, this means
that the limits of integration indicate the region that is being integrated over and
the integral should be evaluated from the inside out.
For polar coordinates, only simple polar regions will be considered. These are
regions that will resemble portions of wedges coming from a circle. The regions
will be bounded on the left and right by rays from the origin. The other two
bounds will be inner and outer curves such as r = g(θ ) and r = h(θ ). Since polar
regions of this type are the only ones that are considered here, order of integration
will
always be in the r direction first. So double polar integrals will be set up as
RR
f
R (r, θ ) rdrd θ .
simple polar regions
dA = rdrd θ
RR
Any double integral will have the format R f dA where f is a function of the
coordinate variables. The term dA, however, will vary depending upon whether
rectangular coordinates or polar coordinates are used. In rectangular coordinates
dA = dxdy or dA = dydx which represents an infinitesimal rectangle of area dA. In
polar coordinates, dA = rdrd θ . This is the area of an infinitesimal polar wedge.
A description of its derivation is given in the book. For the purposes of setting
up double integrals it may be helpful to have what dA should be represented as
written down on a notecard as problems are attempted. In this way, it will become
second nature to use the appropriate term for dA in the appropriate type of double
integral.
converting between rectangular and
polar coordinates
Conversion between rectangular and polar coordinates is a straightforward
process. Three things need to be done whenever a change is made from one coordinate system to another. The limits of integration need to be changed, the
integrand needs to be rewritten in terms of the new variables and the appropriate
form of dA needs to be used. It is important to remember that changing coordinate
systems does not alter the region R that is being integrated over.
Checklist of Key Ideas:
simple polar region
inner and outer radial boundaries
polar rectangle
infinitesimal polar rectangle and polar Riemann sums
polar double integrals
finding area using double polar integrals
converting between rectangular and polar double integrals
use of symmetry in double polar integrals
14.4 Surface Area; Parametric Surfaces
PURPOSE: To describe surfaces parametrically and to use double integrals to find the area of a parametric surface.
multivariable vector-valued functions
This section introduces two important concepts. The first is a multivariable function that is vector-valued and the second is the calculation of the area of a surface. Vector-valued functions have already been seen in the text (see Chapter 12).
For example, the vector form of a parametric equation of a line is actually a vectorvalued function. Each output of the function for a given input determines one of
the coordinates of a point on the line. Another example of a vector-valued function is the gradient function of a multivariable function. The gradient vector of a
multivariable function is the vector-value of the gradient function at a particular
point. Multivariable functions have also been considered (see Chapter 13). This
section makes use of both the vector-valued and multivariable ideas.
Multivariable functions that are vector-valued are treated in much the same way as
regular vector-valued functions. For example, the continuity and derivatives of a
vector-valued function are investigated by looking at the component functions of
the function. Since the components are each multivariable functions themselves,
they will each have appropriate partial derivatives as well. Multivariable vectorvalued functions are introduced here as a quick and simple way of representing a
surface parametrically.
The basic idea is to attach a grid of some sort to the surface. This is done by
parameterizing coordinates of any given point on the surface. For example, if
each point on the surface can be parameterized so that x, y, and z are all functions
of the parameters u and v, then the partial derivatives of the coordinates with
respect to the parameters will indicate how the coordinates will be changing along
the uv-grid. So if r(u, v) = hx(u, v), y(u, v), z(u, v)i then ∂ r/∂ u and ∂ r/∂ v will
indicate how the uv grid is traced onto the surface. The vector ∂ r/∂ v would
indicate how the surface is changing when u is held constant. In other words,
∂ r/∂ v will be tangent to the u-curves where u is constant. More importantly,
however, is the idea that a “square” on the grid can be defined by these two partial
derivative vectors. The area of one “square” would be given by the magnitude of
their cross product.
parameterizing a surface
This leads directly to the idea of how to calculate the surface area. First, parameterize the coordinates of a point on the surface in terms of the parameters u and
v. Then integrate the magnitude of the cross product of the two partial derivative
vectors, ∂ r/∂ u and ∂ r/∂ v, over the entire surface.
calculating area
IDEA: Surface area of z = f (x, y) can be found by parameterizing the coordinates as x = u, y = v, and z = f (u, v). Then the surface area would be found
by
area =
ZZ
R̃
∂r ∂r
×
∂u ∂v
dudv
where r = hu, v, f (u, v)i and R̃ is the space R from the xy-plane as it is represented in uv-space.
Tangent planes to a parametric surface can also be determined using the vectors
∂ r/∂ u and ∂ r/∂ v. Since these two vectors must be tangent to the surface at a
point, then they both must lie within any tangent plane at that point. So the cross
product of these two vectors will be normal to the surface and to any tangent plane.
This normal vector can be used to define the equation of the tangent plane. If this
vector is normalized, then it is called the principle unit normal vector.
Checklist of Key Ideas:
parameterizing a curve using one parameter
parameterizing a surface using two parameters
tangent planes
principle unit normal vector
constant parameter curves or parametric mesh lines
lines of latitude or longitude
surfaces of revolution represented parametrically
vector form of a parametric surface
vector-valued function of two variables
graph of a vector-valued function of two variables
continuity of a vector-valued function of two variables
partial derivatives of multivariable vector-valued functions
tangent plane to a parametric surface
principal unit normal vector
smooth parametric surface
surface area of parametric surfaces
surface area of z = f (x, y)
14.5 Triple Integrals
PURPOSE: To introduce triple integrals and to explain how they
may be set up and evaluated over both rectangular and nonrectangular regions.
The natural way to extend double integrals is to talk about triple integrals. Single integrals integrated along an interval in one dimension, and double integrals
are evaluated over a planar region in two dimensions so triple integrals are correspondingly evaluated over some three dimensional region.
RRR
Typically a triple integral may be written as R f (x, y, z) dV where dV represents
some infinitesimal box. In rectangular coordinates this box has a volume give by
(dx) × (dy) × (dz). Then in rectangular coordinates, dV = dxdydz. As is the case
with double integrals, this ordering of the differentials indicates the variable of
integration for each integral. Also like double integrals, the order of integration
may be changed. The same cautionary statement from double integrals about the
limits of integration applies here. Although the order of integration may change,
this does not change the region over which the integration is changing. It will
generally change the limits of integration. They may not be simply interchanged
in most cases.
CAUTION: When the order of integration is changed, care must be taken to
use the appropriate limits of integration. Generally, the limits of integration
may not simply be interchanged. Instead they will need to be redefined.
Evaluating triple integrals proceeds in the same fashion as for double integrals.
The integral is considered to be three nested integrals and the order of evaluation
should be from the inside to the outside. As mentioned above, the variable of integration for each integral is determined by the ordering of the differentials given by
dV = dxdydz. As is the case with single and double integrals, each integral is evaluated by treating all variables except the variable of integration as constants. This
means that the limits of integration for the innermost integral could be surfaces
described by the other two variables. Then the middle integral to be evaluated
may be between two curves that are defined in terms of the variable of the outer
integration.
evaluating triple integrals
It is again important to realize that the limits of integration are chosen so that they
will describe the region over which the integration will occur. The region over
which integration is occurring is independent of the order of integration. In other
words, although changing the order of integration will almost certainly alter the
limits of integration (except in some very simple cases), this does not mean that
the region of integration is changing.
Double integrals could be used to calculate the area of the region of integration by
allowing the integrand to be equal to one. When this is done with triple integrals,
the result is to find the volume of the region of integration.
Checklist of Key Ideas:
finite solid
infinitesimal subboxes
triple integral of f (x, y, z) over a region G
three dimensional region G
subdividing the region G
projection of G onto the xy-plane
simple xy-, xz- or yz-solid
nonconstant limits of integration g1 (x, y) and g2 (x, y)
calculating volume of the region G
changing order of integration
14.6 Triple Integrals in Cylindrical and Spherical
Coordinates
PURPOSE: To set up triple integrals using both cylindrical and
spherical coordinates and to explain how to convert between the
different coordinate systems.
triple integrals and volume
cylindrical coordinates
spherical coordinates
see §11.8
In two dimensions, double integrals in rectangular coordinates may be converted
to double integrals in polar coordinates. This is often done to make a calculation
more straightforward particularly if a region is easier to describe in one coordinate system than another. In three dimensions, rectangular coordinates may be
changed to either cylindrical or spherical coordinates. The impact this has on
triple integrals is similar to the effect that polar coordinates may have on double
integrals. The bottom line, however, is that regardless of what coordinate system
is used, the region G in three dimensions is not being changed and integrals are
still evaluated by calculating the inner integrals first and working outward.
IDEA: Regardless of what coordinate system is used, the region of integration
will not change. On the other hand, the limits of integration describing the
region will most likely change when different coordinate systems are used.
The biggest differences when using cylindrical and spherical coordinates are that
the limits of integration must be chosen correctly, and the integrand f (x, y, z) and
dV must be expressed in terms of the appropriate coordinate
system. That is to
RRR
say, all triple integrals will have the general form of G f dV . How G, f , and dV
are represented in a particular coordinate system is the main issue.
IDEA: Refer to §11.8 for the appropriate formulas for converting variables between coordinate systems. Here are appropriate forms for dV in cylindrical
and spherical coordinates.
cylindrical → dV = r dz dr d θ
spherical → dV = ρ 2 sin φ d ρ d φ d θ
IDEA: The order of integration may also be switched in cylindrical and spherical coordinates. In each case, the appropriate change of limits of integration need to be made. However, the factor of r in cylindrical coordinates and
ρ 2 sin φ in spherical coordinates must remain with the integrand regardless of
the order of integration.
Checklist of Key Ideas:
cylindrical wedges
Riemann sum of infinitesimal cylindrical wedges
triple integral using cylindrical coordinates
dV in cylindrical coordinates
projection of the solid G onto the xy-plane
converting between rectangular and cylindrical coordinates
changing limits of integration when changing coordinate systems
spherical wedge
Riemann sum of infinitesimal spherical wedges
triple integral using spherical coordinates
dV in spherical coordinates
converting between rectangular and spherical coordinates
14.7 Change of Variables in Multiple Integrals;
Jacobians
PURPOSE: To discuss a change of variables in either double or
triple integrals and to explain how the change of variables procedure relates to the Jacobian of a function.
Change of variables in double and triple integrals are an extension of the idea
of change of variables or u-substitution in single integrals. Recall that when a
u-substitution was performed where u = g(x), then the new differential in the
integral, du, would replace the term g′ (x)dx. This process is described in greater
detail here in this section.
Now in multiple integrals, the integrands are very often multivariable functions so
finding “g′ (x)” is not so simple. The concept that is introduced here to allow a
change of variables is called the Jacobian. It is related to this factor of g′ (x) from
the single variable case.
The first idea that is mentioned is that of transformation. A transformation T is
simply a rule for converting from one coordinate system to the another. So if
a region is represented in the uv-plane, then the transformation (x, y) = T (u, v)
would indicate how this region would be represented in the xy-plane by taking
each point in the uv-plane and transforming it to a corresponding point in the
xy-plane.
Two important questions need to be considered. First, if x = g(u, v) and y = h(u, v)
then how can u and v be expressed in terms of x and y? Second, if an infinitesimal area such as dA is described in the xy-plane, what does it look like in the
uv-plane? The first issue is found by solving the parametric equations for x and y
for expressions for u and v.
CAUTION: Generally dxdy 6= dudv. To do a change of variables in either a
double or triple integral requires the use of the Jacobian.
The infinitesimal area dA is related by the Jacobian. In general, it will not be true
that dudv = dxdy. Instead, the Jacobian will act as a conversion fact between the
two coordinate systems.
IDEA:
In
two
dimensions
dxdy = |J(u, v)|dudv, where
J(u, v) =
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
.
if
x = g(u, v)
and
y = h(u, v) then
expressing u and v
in terms of x and y
Here is an example of a simple change of variables. Consider the double integral
below which will calculate the volume under the surface z = x + y and above a
trapezoidal region with corners in the xy-plane at (1, 1), (2, 3), (5, 3), and (4, 1).
y=3 x=(y+7)/2
Z
Z
(x + y) dxdy
y=1 x=(y+1)/2
To change this into uv-space requires that x and y be parameterized in terms of u
and v. Suppose that it is given that x = (u + v)/2 and y = v. Then the Jacobian
would be
J(u, v) =
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
1/2 1/2
=
= 1/2.
0
1
Also to change the limits of integration, requires them to be written in terms
of u and v. So y = 1 = v and y = 3 = v remain unchanged. On the other hand
x = (y + 7)/2 becomes (u + v)/2 = (v + 7)/2. Solving for u we find u = 7. Similarly, the lower limit of integration becomes u = 1. Finally the surface becomes
z = x + y = (u + v)/2 + v = (u + 3v)/2. So the new integral is
Z 3Z 7
(u + 3v)
1
·
· dudv = 30.
2
2
1 1
Checklist of Key Ideas:
change of variables in a single integral
the Jacobian
transformation from the xy-plane to the uv-plane
the transformation T (u, v) from (u, v) to (x, y) coordinates
the image of (u, v) in the xy-plane
the image of S under T
the inverse transformation T −1
constant u and v curves
the Jacobian of T (u, v)
“secant vectors” and tangent vectors
change of variables for double integrals
the transformation T (u, v, w) from (u, v, w) to (x, y, z) coordinates
change of variables for triple integrals
using change of variables as a guide for transforming between rectangular,
cylindrical and spherical coordinates
14.8 Centers of Gravity Using Multiple Integrals
PURPOSE: To use double and triple integrals to find the center
of gravity for laminas and solids.
The focus of this section is to find the center of gravity (sometimes called the
center of mass) of two and three dimensional regions. The ideas all hinge upon
the units of the integrals that will be used. In finding centers of mass, the idea is
that a sum of moments divided by a sum of masses will yield a center of mass.
The term centroid simply refers to a center of gravity when the density function
δ is constant. These types of regions are called homogeneous.
The region that is being considered is essentially partitioned into much smaller
regions of size dA and dV for two and three dimensions respectively. Each of
these smaller portions has some mass given by δ (x, y)dA or δ (x, y, z)dV . In two
dimensions, the moments are defined by distance from a particular axis. So the
moment of one of these regions from the y-axis, My , would be xδ (x, y)dA and the
moment of one of these regions from the x-axis, Mx , would be yδ (x, y). If these
moments are divided by the mass then they give the corresponding x or y distance.
This is the main idea behind center of mass.
IDEA: The center of mass in two dimensions is described by the point (x̄, ȳ)
where
x̄ =
My
M
In the formulas above, My =
constant density
homogeneous region
“center of mass equals
moment divided by mass”
Mx
.
M
ȳ =
and
center of gravity
center of mass
centroid
RR
R xδ
dA, Mx =
RR
R yδ
dA and M =
Rδ
RR
dA.
It may be confusing to use the formulas above since the subscripts do not seem
to match the value being calculated. Instead, use the integrands to remember how
the calculation is done. For example, x̄ =“(xδ (x, y)dA)/(δ (x, y)dA)” which would
yield an x-value. The units that are being used should also serve to confirm which
integrals are to appear where in the formulas.
CAUTION: When calculating x̄ and ȳ, the integrals Mx , My , and M need to be
calculated separately. The integrands may not be combined.
The center of mass in three dimensions is very similar to the center of mass in
two dimensions. The only significant difference is that the moments are calculated
using distances from coordinate planes. So instead of Mx and My there are now
Mxy , Mxz and Myz . The term Mxy is the moment with respect to the xy-plane.
Again the subscripts may be confusing. The key here is the distance involved. For
Mxy , the distance to the xy-plane from a point is z (the distance will always be the
variable not found in the subscript).
IDEA: The center of mass in three dimensions (x̄, ȳ, z̄) is found by
x̄ =
Myz
,
M
whereRRR Mxy =
M = G δ dV .
ȳ =
G zδ
RRR
Mxz
,
M
dV ,
and
Mxz =
RRR
z̄ =
G yδ
Mxy
M
dV ,
Myz =
G xδ
RRR
dV ,
and
center of mass in 3-space
Theorem of Pappus
The Theorem of Pappus provides an interesting application of the center of mass
of a two dimensional region. If this region R with a center of mass located at (x̄, ȳ)
is rotated around a particular axis to generate a solid, then the volume will be
the area of the region times the circumference of the circle traveled by the center
of mass during the revolution. When rotated about the x-axis, for example, the
distance traveled by the center of mass would be 2π ȳ.
Checklist of Key Ideas:
lamina (planar region)
homogeneous or nonhomogeneous
density function δ (x, y) or δ (x, y, z)
mass of a lamina or solid
moment about the line x = a or y = c
lever arm
equilibrium of a system of masses
first moment about the x-, y-, or z-axis
moment about the xy-, xz-, or yz-planes
centroid of a planar region (a lamina) R
centroid of a solid region G
Theorem of Pappus
Chapter 14 Sample Tests
Section 14.1
1.
Z 1Z 3
0
(x + 4) dx dy =
0
33
2
27
(b)
2
9
(c)
2
11
(d)
2
(a)
2.
Z 1Z 3
0
4, −1 ≤ y ≤ 2} is
(x + 4) dy dx =
x2 sin y dx dy.
1
0
dx dy =
10. Find the volume of the solid bounded by z = 10 − 4x − 2y and
the rectangle R = [0, 2] × [0, 1].
(a) 1
(b) 10
(b) 6
(c) 9
(c) 0
(d) 2
ex dx dy =
0
(b) e2 − 2
(c) 2e
(d) e2
Z πZ π
0
x2 sin y dA; R = {(x, y) : 0 ≤ x ≤
(c) 20
(a) e2 − 1
5.
Z 3 Z 2R
5xy3 dx dy.
(d) 25
(a) 5
0
ZZ
(b) 12
(d) 36
4.
−2 −1
(a) 10
0
Z 1Z 2
5xy3 dA; R = {(x, y) : −2 ≤ x ≤
9. Find the volume of the solid bounded by z = −2x − 2y and
the rectangle R = [0, 1] × [0, 3].
33
2
27
(b)
2
21
(c)
2
11
(d)
2
0
2, 1 ≤ y ≤ 3} is
ZZ
Z 4 ZR 2
8. Answer true or false.
(a)
3.
(b) 2
35
(c)
3
(d) 7
7. Answer true or false.
0
Z 2Z 3
(a) 4
π /2
cos x dx dy =
Z 1 Z 2
x3 y dy dx.
Z 1 Z 2
sin x cos y dy dx.
Z 1 Z 2
e3xy dy dx.
−1 −1
(a) π
14. Answer true or false. The volume of the solid bounded by
z = sin x cos y and R = {(x, y) : −1 ≤ x ≤ 1, −1 ≤ y ≤ 2} is
(b) −π
π
(c)
2
−π
(d)
2
6. Evaluate
11. Answer true or false. The average value of the function
f (x, y) = sin x cos y over the rectangle [0, π ] × [0, 2π ] is
Z Z
1 π 2π
sin x cos y dy dx.
π 0 0
12. Answer true or false. The average value of the function f (x, y) = x2 y3 over the rectangle [0, 5] × [0, 2] is
Z Z
1 5 2 2 3
x y dy dx.
10 0 0
13. Answer true or false. The volume of the solid bounded by
z = x3 y and R = {(x, y) : −1 ≤ x ≤ 1, −1 ≤ y ≤ 2} is
−1 −1
ZZ
R
2xy2 dA; R = {(x, y) : −1 ≤ x ≤ 2, 1 ≤ y ≤ 2}.
15. Answer true or false. The volume of the solid bounded by
z = e3xy and R = {(x, y) : −1 ≤ x ≤ 1, −1 ≤ y ≤ 2} is
−1 −1
Section 14.2
(c)
28
15
Z 2Z x
(d)
25 23
+
10
3
1.
0
xy dy dx =
0
(a) 1
8. Find the area of the region of the xy-plane that is enclosed by
y = x and y = x2 , for 0 ≤ x ≤ 2.
(b) 2
1
3
2
(b)
3
(c) 1
1
(d)
6
(c) 3
(a)
(d) 4
2.
Z π /2 Z cos x
0
dy dx =
0
(a) 1
(b) 0
9. Find the area of the the region in the xy-plane that is enclosed
by y = x and y = x2 , for 1 ≤ x ≤ 3.
(c) −1
(d) π
3.
Z 1Z x
0
(a) 4
14
(b)
3
16
(c)
3
(d) 6
ey dy dx =
0
(a) e − 2
(b) e + 1
(c) −e
(d) e
4.
10.
Z 1Z xp
1
x2 + 1 dy dx =
0
0
(b)
(c)
(d)
(b) 0.042
(c) 0.419
(d) 0.423
11. Answer true or false.
12.
ZZ
x4 dA, where R is the re-
R
gion bounded by y = x + 4, y = 2x, and x = 16, is
16
x4 dy dx.
x+4
6. Answer true or false.
ZZ
xy dA, where R is the region
R
bounded by y = x, y = 0, and x = 4, is
Z 4Z x
0
7.
Z 2 Z x2
0
y dy dx =
x
25
10
26
(b)
15
(a)
Z 1 Z x2
0
5. Answer true or false.
Z 28 Z 2x
(xy − 4) dy dx ≈
x
(a) 0.412
2 3/2
(2 + 1)
3
23/2 + 1
3
2 3/2
(2 − 1)
3
23/2 − 1
3
(a)
Z 2 Z x2
0
xy dy dx.
Z 0 Z x
−1 0
2
dy dx = − .
3
x
dy dx =
(a) −1
(b) 1
1
(c)
2
(d) −
1
2
13. ZAnswer
true or false.
Z 2
2
x
f (x, y) dy dx =
0
0
14. ZAnswer
true or false.
Z
x2
3
0
0
f (x, y) dy dx =
Z 4 Z √y
0
Z 3 Z y2
0
0
15. ZAnswer
true or false.
Z 2Z
2 Z ln x
f (x, y) dy dx =
1
1
1
f (x, y) dx dy.
0
1
f (x, y) dx dy.
ey
f (x, y) dx dy.
Section 14.3
1.
0
Z π /2 Z 0
r cos θ dr d θ =
sin θ
0
(a) −
1
6
π
(c)
2
8. Answer true or false.
9. Answer true or false.
1
6
2.
π
2
Z π /2 Z sin θ
r3 dr d θ ≈
−π /2 0
(a) −0.33
(b) −0.29
0
0
r2 dr d θ =
(a) 1
(b) −1
(c) 0
(d) 2
4.
Z π Z cos 2θ
0
0
(c) 0
(d) 2
Z π /6 Z sin 3θ
0
dr d θ =
(a) 3
1
(b)
3
(c) −3
1
(d) −
3
6. Answer true or false.
Z 1 Z cos θ
0
0
0
0
Z 1Z x
0
dr d θ .
7. Answer true or false.
Z πZ 1
dy dx =
0
Z πZ 2
0
0
r dr d θ .
r dr d θ .
(a)
(b) −1
0
−1 0
0
25π
3
25π
(b)
6
250π
(c)
3
125π
(d)
3
13. Find the area enclosed by the three-petaled rose r = 2 sin 3θ .
π
(a)
4
π
(b)
2
π
(c)
8
(d) π
dr d θ =
(a) 1
5.
Z 0 Z √4−x2
(a)
(d) 0.33
3.
0
Z π /2 Z 1
25π
3
25π
(b)
6
250π
(c)
3
125π
(d)
3
12. Find the volume of the solid formed by the left hemisphere
x2 + y2 + z2 = 25.
(c) 0.29
Z π Z cos θ
dy dx =
10. Find the volume of the solid formed by the left hemisphere
r2 + z2 = 4.
8π
(a)
3
16π
(b)
3
32π
(c)
3
2π
(d)
3
11. Find the volume of the solid formed by the left hemisphere
r2 + z2 = 25.
(b)
(d) −
Z 1 Z √1−x2
Z 1 Z √1−x2
0
3
rer dr d θ .
(x2 + y2 ) dy dx =
0
0
14. Find the volume of the region enclosed by x2 + y2 = 4 and
x2 + z2 = 4 that is above the xy−plane.
8π
3
16π
(b)
3
(c) 8π
(a)
ex
2
+y2
dy dx =
(d) 4π
8. Answer true or false. To find the portion of the surface
z = x + y2 that lies above the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3,
15. Find the region inside the circle r = 25 sin θ .
Z 2Z 3
evaluate
(a) 25
0
9. Answer true or false.
To find the portion of the
surface z = 3x2 + 4y2 that lies above the rectangle
Z 3Z 2q
0 ≤ x ≤ 2, 1 ≤ y ≤ 3, evaluate
6x2 + 8y2 dx dy.
(b) 5
(c) 25π
(d) 5π
1
Section 14.4
1. The surface expressed parametrically by x = r cos θ , y =
r sin θ , z = 9 − r2 is
(a) a sphere.
(c) a paraboloid.
(d) a cone.
2. The surface√expressed parametrically by x = r cos θ , y =
r sin θ , z = 9 − r2 is
(a) a sphere.
2
1
11. Answer true or false. To find the portion of the surface
z = 5xy + 3 that lies above the rectangle 1 ≤ x ≤ 3, 2 ≤ y ≤ 5,
Z 5Z 3q
evaluate
25x2 + 25y2 + 1 dx dy.
0
(d) a cone.
3. Answer true or false. A parametric representation of the surface z + x2 + y2 = 5 in terms of the parameters u = x and v = y
is x = u, y = v, z = 5 − u2 − v2 .
4. Answer true or false.
The parametric equations for
x2 + y2 = 16 from the plane z = 1 to the plane z = 2 are
x = 4 cos v, y = 4 sin v, z = u with 0 ≤ v ≤ 2π and 1 ≤ u ≤ 2.
5. Answer true or false. Parametric equations for x2 + z2 = 25
from y = 0 to y = 1 are x = 5 cos u, y = v, z = 5 sin u with
0 ≤ u ≤ 2π and 0 ≤ v ≤ 1.
6. The cylindrical parameterization of z = xex
2
+y2
Z 4Z 2
0
1
15. Answer true or false. To find the portion of the surface
z = x3 + y3 that lies above the rectangle 0 ≤ x ≤ 1, 0 ≤ y ≤ 3,
Z 3Z 1q
evaluate
9x4 + 9y4 + 1 dx dy.
0
0
Section 14.5
1.
Z 1Z 2Z 3
0
0
x2 yz dx dy dz =
0
(a) 18
(a) x = r cos θ , y = r sin θ , z = er .
(b) 9
(b) x = r sin θ , y = r cos θ , z = er .
(c) 27
z = rer cos θ .
(d) 6
2.
Z 2 Z π /2 Z π /2
0
7. The equation of the tangent plane to x = u, y = v, z = u + v2
where u = 0 and v = 1 is
0
sin x sin y dx dy dz =
0
(a) −2
(b) 2
(a) x + 2(y − 2) + z − 1 = 0.
(c) 1
(b) x + 2(y − 2) − z + 1 = 0.
(d) x + 2y − 2 − z + 1 = 0.
2x dx dy.
0
14. Answer true or false. To find the portion of the surface
z = x2 − 2y that lies above the rectangle 1 ≤ x ≤ 2, 0 ≤ y ≤ 1,
Z 1Z 2p
4x2 − 1 dx dy.
evaluate
is
(d) x = r sin θ , y = r cos θ , z = rer cos θ .
0
13. Answer true or false. To find the portion of the surface
z = x2 − y that lies above the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 4,
0
(c) a paraboloid.
(c) x + 2y − 2 + z + 1 = 0.
1
12. Answer true or false.
To find the portion of the
surface z = x2 − x + y2 that lies above the rectangle
Z 3Z 4q
0 ≤ x ≤ 4, 0 ≤ y ≤ 3, evaluate
4x2 + 4y2 dx dy.
evaluate
(b) an ellipsoid.
0
10. Answer true or false.
To find the portion of the surface z = x2 + 3y2 + 4 that lies
above the rectangle 1 ≤ x ≤ 2, 2 ≤ y ≤ 4, evaluate
Z 4Z 2q
4x2 + 36y2 + 1 dx dy.
2
(b) an ellipsoid.
(c) x = r cos θ , y = r sin θ ,
(x + y2 ) dy dx.
0
(d) −1
3.
Z 2 Z z2 Z y
0
0
0
y dx dy dz =
(a) 4
128
(b)
21
(c) 8
8
(d)
5
4.
9.
0
0
(c) 2
(d) 4
3 dz dy dx =
0
10.
(d) 5
11.
x dx dy dz =
12. Answer true or false.
z3 dx dy dz =
13. Answer true or false.
0
14. Answer true or false.
1
3
0
0
0
Z 5Z zZ y
0
0
1.
Z π /2 Z π /2 Z 1
0
0
(b) 4
(c) 0
(d) 1
2.
0
dx dy dz = 25.
Z 4 Z z 2 Z y3
0
dx dy dz = 45 .
0
(a) 2
(a)
0
8
3
(c) 6.095
8
3
(c) 2π 3
(d) 5
(d) π 3
(b) 6.082
0
Z π /2 Z π /2 Z 2
0
z3 y dx dy dz ≈
0
ρ 4 sin φ cos θ d ρ d φ d θ =
(a) 1
1
(b)
5
(c) −1
1
(d) −
5
sin y dx dy dz =
0
Z 2Z zZ y
0
dz dy dx = x2 .
Section 14.6
(a) 2
0
Z 5 Z x Z x+y
0
Z 5 Z π /2 Z cosy
0
sin x dx dy dz =
(d) 5
2
3
4
(d)
5
8.
−z2 0
(a) 8
14
(b)
3
(c) 6
0
(c)
1
Z 2 Z 0 Z 3π
1
5
1
(a)
3
7.
dx dy dz =
(c) 6
Z 1 Z zZ 2
(b) −
0
(b) 7
1
7
2
(b)
7
1
(c)
5
(d) 0
−1 0
0
(a) 8
(a)
6.
Z 2 Z z2 Z 3
1
Z 1 Z zZ y
−1 0
yz dx dy dz =
0
(b) 0
(a) 1
1
(b)
12
1
(c)
3
3
(d)
20
5.
0
(a) 0.29
Z 1 Z x Z y3
0
Z 1 Z z Z √y2 +5
(b) −
0
ρ 2 sin φ cos θ d ρ d φ d θ =
3. Answer true
√ or false.
Z 2π Z π Z
0
4.
0
36−r2
0
Z 2π Z π Z 2
0
0
−2
2r dz dr d θ =
√
12. Answer true or false.
16π 2 − 24π
3
0
13.
(a) 0
(a) 6π
(b) 4
(b) 9π
−π /2 1
14.
sin φ d ρ d θ d φ =
δ (r, θ , z) dr d θ dz, where δ (r, θ , z) = r, is the
−2 0
0
0
0
δ (r, θ , z) dr d θ dz, where δ (r, θ , z) = rz, is the
following.
(a) 3π
9π
(b)
2
3
(c)
2
3π
(d)
2
(c) 4π
3π
(d)
2
Z 2 Z π /2 Z 2π
Z 1 Z 2π Z 3
0
(b) 3π
ρ 3 sin φ cos θ d φ d θ d ρ =
15.
(a) 16
Z 1 Z 2π Z 3
0
0
0
δ (r, θ , z) dr d θ dz, where δ (r, θ , z) = z2 , is the
following.
(b) 0
(a) 6π
27π
(b)
2
(c) 3
(c) 8
(d) 4
Z 2π Z 6 Z 2
3
sin θ d ρ d φ d θ = 0
(d) 3π
(a) π
0
0
0
(c) 3
Z π Z π /2 Z 3
0
7.
0
0
following.
(d) 6
6.
Z 1 Z 2π Z 3
0
sin φ cos θ ρ d ρ d θ d φ =
(c) −4
5.
Z 2π Z 1−sin2 θ Z 1
1
dz dr d θ =
(d) 2π
(a) 6π
(b) 2π
Section 14.7
(c) 4π
(d) π
8.
Find the center of gravity of the sphere x2 +y2 +z2
δ (x, y, z) = 6x2 y2 z2 .
= 4 where
(a) (2, 2, 2)
1. Find
∂ (x, y)
, if x = u + 2v and y = 3u + v.
∂ (u, v)
(a) 5
(b) −5
(b) (4, 4, 4)
(c) 7
(c) (1, 1, 1)
(d) −7
(d) (0, 0, 0)
9. Answer true orpfalse. The center of
p gravity of the solid enclosed by z = x2 + y2 and z = − x2 + y2 , if the density is
δ (x, y, z) = x2 + y2 + z2 , is at the origin.
10. Answer true or false. The center of gravity of the solid enclosed by x2 + y2 = 1 and 2y2 + z2 = 1 is at the origin if
δ (x, y, z) is constant.
11. Answer true or false. The center of gravity of the solid enclosed by x2 + y2 = 1 and 3y2 + z2 = 1 is at the origin if
δ (x, y, z) = x2 + 1.
2. Find
∂ (x, y)
, if x = u2 and y = u − v.
∂ (u, v)
(a) 2u + 1
(b) 2u
(c) −2u
(d) −2u − 1
3. Find the Jacobian if x = eu and y = ev .
(a) 0
(b) euv
Section 14.8
(c) eu−v
(d) eu+v
4. Find the Jacobian if u = ex and v = yex .
ln v − v
(a)
u
ln v + v
(b)
u
1
(c) 2
u
1
(d) − 2
u
5. Find the Jacobian if x = 3u + w, y = vw, and z = u2 v.
3u2 v + 2uvw
3u2 v + 2uw2
6(u2 v + uw2 )
−6(u2 v + uw2 )
(a)
(b)
(c)
(d)
If x = uv, y = u + 2v, then
∂ (x, y)
=
∂ (u, v)
∂ (x, y)
=
∂ (u, v)
2u
1
v
1
u
.
2
1
.
2v
9. Answer true or false.
If x = u + 2v + 3w, y = 7 + uv + 4v, z = eu + vw, then
1 2v 3
∂ (x, y, z)
= 7
4 u .
∂ (u, v, w)
eu w v
10. Answer true or false. If x = eu , y = ev , z = ew , then
eu 0
0
∂ (x, y, z)
= 0 ev 0 .
∂ (u, v, w)
0 0 ew
11. Answer true or false. If u = x + 2y and v = 3x + v,
Z 1Z 1
0
0
(b) 18 kg
(c) 8 kg
(d) 10 kg
(a) 2
7. Answer true or false.
If x = u2 , y = v2 , then
(a) 14 kg
2. A lamina with density δ (x, y) = xy is bounded by x = 2,
x = 0, y = x, and y = 0. Find its mass.
y
6. Find the Jacobian if u = x, v = , and w = x + z.
x
(a) −u
(b) −uvw
(c) u
(d) uvw
8. Answer true or false.
1. A uniform beam 10 m in length is supported at its center by a
fulcrum. A mass of 10 kg is placed at the left end, a mass of
4 kg is placed on the beam 4 m from the left end, and a third
mass is placed 2 m from the right end. What mass should the
third mass be to achieve equilibrium?
ex+2y e3x+y dx dy =
Z 3Z 4
0
eu ev du dv.
0
12. Answer true or false.
If u = 4x + y and v = x2 y,
Z 2Z 2
Z 10 Z 4
4x + y
u
dx dy =
dv du.
2
x y
1 1
5
1 v
13. Answer true or false. If u = x + y and v = 2x − y,
Z 2Z 2
Z 4Z 2
(x + y)2
u2
dy dx =
− dv du.
v
1 1 2x − y
2 1
14. Answer true or false. If x = uvw, y = eu , and z = 2u, the
Jacobian is 0.
15. Answer true or false. If x = uvw, y = u − v2 w, and z = u, the
Jacobian has no dependence on v nor on w.
(b) 4
(c) 1
(d) 8
3. A lamina with density δ (x, y) = xy is bounded by x = 2,
x = 0, y = x, and y = 0. Find its center of mass.
8 16
(a)
,
5 5
8 8
(b)
,
5 5
16 16
(c)
,
5 5
16 16
(d)
,
15 15
4. A lamina with density δ (x, y) = x2 +2y2 is bounded by x = y,
x = 0, y = 0, and y = 2. Find its mass.
(a) 4
28
3
20
(c)
5
(d) 2
(b)
5. A lamina with density δ (x, y) = x2 +2y2 is bounded by x = y,
x = 0, y = 0, and y = 2. Find its center of mass.
1 1
(a)
,
3 3
6 8
(b)
,
7 5
(c)
(d)
8 6
,
5 7
224
,8
15
6. A lamina with density δ (x, y) = x2 +2y2 is bounded by x = y,
x = 0, y = 0, and y = 2. Find its moment of inertia about the
x−axis.
224
9
(b) 8
224
(c)
15
416
(d)
45
(a)
7. A lamina with density δ (x, y) = x2 +2y2 is bounded by x = y,
x = 0, y = 0, and y = 2. Find its moment of inertia about the
y−axis.
224
9
(b) 8
224
(c)
15
416
(d)
45
11. Answer true or false. The centroid given by z =
0.
13. The centroid of a rectangular solid in the first octant with vertices (0, 0, 0), (0, 0, 2), and (2, 2, 2) is the following.
1 1 1
(a)
, ,
2 2 2
(b) (0, 1, 2)
(c) (1, 1, 1)
2 2 2
(d)
, ,
3 3 3
14. The centroid of the solid given by (x − 2)2 + y2 + (z + 3)2 = 9
is the following.
(a) (2, 0, 3)
(b) (−2, 0, 3)
8. A lamina with density δ (x, y) = xy is bounded by x = 2,
x = 0, y = x, and y = 0. Find its moment of inertia about the
x−axis.
(a)
(b)
(c)
(d)
9. A lamina with density δ (x, y) = xy is bounded by x = 2,
x = 0, y = x, and y = 0. Find its moment of inertia about the
y−axis.
16
5
32
(b)
5
8
(c)
3
16
(d)
3
(c) (0, 0, 0)
(d) (2, 0, −3)
15. The centroid of the solid given by
1 is the following.
(x−3)2
4
+
(y−5)2
16
+
(a) (−3, −5, 2)
(b) (0, 0, 0)
(c) (2, 4, 3)
(d) (3, 5, −2)
Chapter 14 Test
1.
Z 4Z 2
0
(a)
10. Answer true or false. The moment of inertia about x = a,
where a is the x−coordinate of the center of mass, is 0.
x2 + y2 is
12. The centroid of a rectangular solid in the first octant with vertices (0, 0, 0), (0, 0, 1), and (1, 1, 1) is the following.
1 1 1
(a)
, ,
2 2 2
(b) (1, 1, 1)
1 1 1
(c)
, ,
3 3 3
1 1 1
, ,
(d)
4 4 4
(a)
16
5
32
5
8
3
16
3
p
dx dy =
0
(a) 6
(b) 8
(c) 0
(d) 20
2.
ZZ
R
2yx2 dA; R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2} =
(a)
4
3
(z+2)2
9
=
8
3
(c) 0
1
(d)
3
3. Answer true or false. The volume of the solid bounded by
z = x7 y3 and R = {(x, y) : −1 ≤ x ≤ 1, −1 ≤ y ≤ 2} is
(b)
Z 1 Z 2
x7 y3 dy dx.
−1 −1
4. Answer true or false. The average value of the function f (x, y) = x3 y4 over the rectangle [0, 2] × [0, 3] is
Z Z
1 2 3 3 4
x y dy dx.
6 0 0
5.
Z π /2 Z cos x
0
0
dy dx =
(a) 1
(b) 0
(c) −1
(d) π
4π
3
8π
(b)
3
2π
(c)
3
16π
(d)
3
11. Find the area enclosed by one petal of the three-petaled rose
r = 2 sin 3θ .
π
(a)
12
π
(b)
6
π
(c)
24
π
(d)
3
12. The surface expressed
parametrically by x = r cos θ ,
√
y = r sin θ , and z = 16 − r2 is a(n)
(a)
(a) sphere.
6. Answer true or false.
ZZ
(b) ellipsoid.
sin x dA, where R is the
(c) paraboloid.
R
region bounded by y = x + 4, y = x, and x = 16 is
Z 28 Z 2x
16
7.
(a) x = r cos θ , y = r sin θ , z = er .
y dy dx =
(b) x = r sin θ , y = r cos θ , z = er .
0
35
(c) x = r cos θ , y = r sin θ , z = er cos θ .
10
35
(b)
5
35
(c)
−9
10
35
(d)
+9
10
(d) x = r sin θ , y = r cos θ , z = er cos θ .
(a)
8.
(d) cone.
13. The cylindrical parameterization of z = (x2 + y2 )ex is
x+4
Z 3 Z x2
0
sin x dy dx.
Z 0
Z sin θ
−π /2 0
(a) −
14. The equation of the tangent plane to x = u, y = v, z = u + v2
where u = 1 and v = 1 is
(a) x − 1 + 2(y − 2) + z − 2 = 0.
(b) x − 1 − 2(y − 1) − z + 2 = 0.
(c) x − 1 + 2y − 2 + z + 2 = 0.
(d) x − 1 + 2y − 2 − z + 2 = 0.
r cos θ dr d θ =
15. Answer true or false.
To find the portion of the
surface z = 3x2 + 4y2 that lies above the rectangle
Z 3Z 2q
0 ≤ x ≤ 2, 1 ≤ y ≤ 3, evaluate
36x2 + 64y2 + 1 dx dy.
1
6
1
1
6
π
(c)
2
(b)
π
(d) −
2
9. Answer true or false.
16.
Z 3 Z π /2 Z π /2
1
0
sin x sin y dx dy dz =
0
(a) −2
(b) 2
Z 3 Z √9−x2
−3 0
dx dy =
Z πZ 3
0
10. Find the volume of the solid left hemisphere if
(c) 1
(d) −1
r dr d θ
0
r2 + z2
= 1.
17.
Z 5 Z π /2 Z sin y
1
0
0
0
cos y dx dy dz =
(a) 2
(b) 8π
(b) 4
(c) 0
(c) 0
(d) 2π
(d) 1
18. A lamina with density δ (x, y) = xy is bounded by x = 0,
x = y, y = 0, and y = 2. Find its mass.
(b) (6, 6, 6)
(b) 4
(c) (1, 1, 1)
(c) 1
(d) (0, 0, 0)
19. A lamina with density δ (x, y) = xy is bounded by x = 0,
x = y, y = 0, and y = 2. Find its center of mass.
8 16
(a)
,
5 5
8 8
,
(b)
5 5
16 16
(c)
,
5 5
16 16
,
(d)
15 15
20. The centroid of a rectangular solid in the first octant with vertices (0, 0, 0), (0, 1, 1), and (1, 0, 0) is
(a) (0, 1, 2).
1 1 1
(b)
, ,
.
2 2 2
(c) (1, 1, 1).
2 2 2
(d)
, ,
.
3 3 3
Z 2π Z π /2 Z 2
0
0
0
ρ 2 sin φ d ρ d φ d θ =
16π
(a)
3
16π
(b) −
3
3
(c) 4π
Z π /2 Z π Z 8
0
−π 4
(a) 4π
24. Answer true or false. The center of gravity of the solid enclosed by x2 + y2 = 1 and y2 + z2 = 1 is at the origin if
δ (x, y, z) = x + 1.
25. Answer true or false.
Z 2π Z 1−cos2 θ Z 1
0
26. Find
0
0
cos θ d ρ d φ d θ = 0
∂ (x, y)
, if x = 5u + 2v and y = 7u + v.
∂ (u, v)
(a) −9
(b) 9
(c) −19
(d) 19
27. Find the Jacobian if u = xy and v = 2x.
2u
v2
2u 1
(b) − 2 −
v
v
2u 1
(c) − 2 +
v
v
1
(d)
v
(a) −
28. Find the Jacobian if x = 4u + w, y = vw, and z = u2 v.
(a) 4u2 v + 2uvw
(b) 4u2 v + 2uw2
(c) 8(u2 v + uw2 )
(d) −8(u2 v + uw2 )
(d) −4π 3
22.
(a) (3, 3, 3)
(a) 2
(d) 8
21.
23. Find the center of gravity of the sphere x2 +y2 +z2 = 6 where
δ (x, y, z) = 6x2 y2 z2 .
sin φ d ρ d θ d φ =
29. Answer true or false.
Z 2Z 2
x+y 2
1
Z 4Z 2
2
1
−
u2
v2
1
2x − y
dx dy =
dv du, where u = x + y and v = 2x − y.
Chapter 14: Answers to Sample Tests
Section 14.1
1. a
9. b
2. b
10. b
3. b
11. false
4. a
12. true
5. b
13. true
6. d
14. true
7. false
15. true
8. true
2. a
10. b
3. a
11. false
4. d
12. d
5. false
13. false
6. true
14. false
7. c
15. false
8. c
2. c
10. b
3. c
11. c
4. c
12. c
5. b
13. d
6. false
14. b
7. false
15. c
8. true
2. a
10. true
3. true
11. true
4. true
12. false
5. true
13. false
6. c
14. false
7. d
15. true
8. false
2. b
10. b
3. b
11. b
4. d
12. false
5. d
13. false
6. d
14. false
7. a
8. c
2. a
10. true
3. false
11. false
4. a
12. true
5. c
13. b
6. b
14. b
7. a
15. d
8. d
2. c
10. true
3. d
11. false
4. c
12. false
5. a
13. false
6. c
14. true
7. true
15. false
8. false
2. a
10. false
3. a
11. false
4. b
12. a
5. b
13. c
6. a
14. d
7. d
15. d
8. c
2. a
10. c
18. a
26. a
3. true
11. d
19. d
27. d
4. true
12. a
20. b
28. a
5. a
13. c
21. a
29. false
6. false
14. d
22. b
7. a
15. true
23. d
8. b
16. b
24. false
Section 14.2
1. b
9. b
Section 14.3
1. a
9. false
Section 14.4
1. c
9. false
Section 14.5
1. b
9. a
Section 14.6
1. b
9. true
Section 14.7
1. b
9. false
Section 14.8
1. b
9. d
Chapter 14 Test
1. b
9. true
17. a
25. true