Smith R., Minton R. Calculus

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15-57 SECTION 15.6

Surface Integrals 1033

, y

, f(x

, y

))

兩a兩

兩c兩

FIGURE 15.41

Portion of the tangent plane lying

above R

The basic idea for calculating a surface integral is to rewrite it as a double integral,

which is evaluated using existing techniques. To convert a given surface integral into a

double integral, we must write the integrand g(x, y, z) as a function of two variables and

write the surface area element dS in terms of the area element dA.

For the sake of simplicity, we assume that the surface is the graph of the equation

z = f (x, y), where f has continuous ﬁrst partial derivatives in some region R in the xy-

plane. Notice that for an inner partition R

, R

,...,R

of R, if we take the point (x

, y

, 0)

as the point in R

closest to the origin, then the portion of the surface S

lying above R

will

differ very little from the portion T

of the tangent plane to the surface at (x

, y

, f (x

, y

))

lying above R

. More to the point, the surface area of S

will be approximately the same as

the area of the parallelogram T

. In Figure 15.41, we have indicated the portion T

of the

tangent plane lying above R

Let the vectors u

=0, a, b and v

=c, 0, d form two adjacent sides of the paral-

lelogram T

, as indicated in Figure 15.41. Notice that since u

and v

lie in the tangent plane,

= u

× v

=ad, bc, −acisanormalvectortothetangentplane.Wesawinsection11.4

that the area of the parallelogram can be written as

S

=u

× v

=n

.

We further observe that the area of R

is given by  A

=|ac| and n

· k =−ac, so that

· k|=|ac|. We can now write

S

|ac|n



|ac|

n



· k|

 A

since ac = 0. The corresponding expression relating the surface area element dS and the

area element dA is then

dS =

n

|n · k|

dA.

In the exercises, we will ask you to derive similar formulas for the cases where the surface

S is written as a function of x and z or as a function of y and z.

If S is the surface z = f (x, y), recall from our discussion in section 13.4, that a normal

vectortoS isgivenby n =f

, f

, −1.Thisis aconvenientnormalvectorforourpurposes,

since |n · k|=1. With n=



( f

)

+ ( f

)

+ 1, we have the following result.

THEOREM 6.1 (Evaluation Theorem)

If the surface S is given by z = f (x, y) for (x, y) in the region R ⊂ R

, where f has

continuous ﬁrst partial derivatives, then



g(x, y, z)dS =



g(x, y, f (x, y))



( f

)

+ ( f

)

+ 1 dA.

PROOF

From the deﬁnition of surface integral in Deﬁnition 6.1, we have



g(x, y, z)dS = lim

||P||→0



i=1

g(x

, y

, z

)S

= lim

P→0



i=1

g(x

, y

, z

)

n



· k|

 A

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1034 CHAPTER 15

Vector Calculus 15-58

= lim

P→0



i=1

g(x

, y

, f (x

, y

))



( f

)

+ ( f

)

+ 1



)

 A



g(x, y, f (x, y))



( f

)

+ ( f

)

+ 1 dA,

as desired.

Theorem 6.1 says that we can evaluate a surface integral by evaluating a related double

integral. To convert the surface integral into a double integral, substitute z = f (x, y)in

the function g(x, y, z) and replace the surface area element dS with ndA, which for the

surface z = f (x, y)isgivenby

dS =ndA =



( f

)

+ ( f

)

+ 1 dA. (6.1)

EXAMPLE 6.1 Evaluating a Surface Integral

Evaluate



3zdS, where the surface S is the portion of the plane 2x + y + z = 2 lying

in the ﬁrst octant.

Solution On S,wehavez = 2 −2x − y, so we must evaluate



3(2 − 2x − y)dS.

Note that a normal vector to the plane 2x + y + z = 2isn =2, 1, 1, so that in this

case, the element of surface area is given by (6.1) to be

dS =ndA =

√

6dA.

From Theorem 6.1, we then have



3(2 − 2x − y)dS =



3(2 − 2x − y)

√

6dA,

where R is the projection of the surface onto the xy-plane. A graph of the surface S is

shown in Figure 15.42a. In this case, notice that R is the triangle indicated in

Figure 15.42b. The triangle is bounded by x = 0, y = 0 and the line 2x + y = 2 (the

intersection of the plane 2x + y + z = 2 with the plane z = 0). If we integrate with

FIGURE 15.42a FIGURE 15.42b

z = 2 −2x − y The projection R of the surface S onto the xy-plane

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15-59 SECTION 15.6

Surface Integrals 1035

respect to y ﬁrst, the inside integration limits are y = 0 and y = 2 −2x, with x ranging

from 0 to 1. This gives us



3(2 − 2x − y)dS =



3(2 − 2x − y)

√

6dA



2−2x

√

6(2 − 2x − y)dy dx

= 2

√

where we leave the routine details of the integration as an exercise.



In example 6.2, we will need to rewrite the double integral using polar coordinates.

EXAMPLE 6.2 Evaluating a Surface Integral Using Polar Coordinates

Evaluate



zdS, where the surface S is the portion of the paraboloid z = 4 − x

− y

lying above the xy-plane.

Solution Substituting z = 4 − x

− y

,wehave



zdS=



(4 − x

− y

)dS.

In this case, a normal vector to the surface z = 4 − x

− y

is n =−2x, −2y, −1,so

that

dS =ndA =



+ 4y

+ 1 dA.

This gives us



(4 − x

− y

)dS =



(4 − x

− y

)



+ 4y

+ 1 dA.



FIGURE 15.43

z = 4 − x

− y

Here, the region R is enclosed by the intersection of the paraboloid with the xy-plane,

which is the circle x

+ y

= 4. (See Figure 15.43.) With a circular region of

integration and the term x

+ y

appearing (twice!) in the integrand, you had better be

thinking about polar coordinates. We have 4 − x

− y

= 4 −r



+ 4y

+ 1 =

√

+ 1 and dA = rdrdθ. For the circle x

+ y

= 4, r ranges

from 0 to 2 and θ ranges from 0 to 2π . Then, we have



(4 − x

− y

)dS =



(4 − x

− y

)



+ 4y

+ 1 dA



2π



(4 − r

)



+ 1rdrdθ

289

√

17 −

π,

where we leave the details of the ﬁnal integration to you.



Parametric Representation of Surfaces

In the remainder of this section, we study parametric representations of surface integrals.

First, we need a better understanding of surfaces that have been deﬁned parametrically.

You have already seen parametric surfaces in section 12.6. Recall that we can describe the

cone z =



+ y

in cylindrical coordinates by z = r, 0 ≤ θ ≤ 2π, which is a parametric

representation with parameters r and θ. Similarly, the equation ρ = 4, 0 ≤ θ ≤ 2π and

0 ≤ φ ≤ π ,isaparametricrepresentationofthesphere x

+ y

+ z

= 16,withparameters

θ and φ. It will be helpful to review these graphs as well as to look at some new ones.

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The general form for parametric equations representing a surface in three dimensions is

x = x(u,v), y = y(u,v)and z = z(u,v)for u

≤ u ≤ u

andv

≤ v ≤ v

.The parameters

u and v can correspond to familiar coordinates (x and y,orr and θ , for instance), or less

familiar expressions. Keep in mind that to fully describe a surface, you will need to deﬁne

two parameters.

EXAMPLE 6.3 Finding Parametric Representations of a Surface

Find a simple parametric representation for (a) the portion of the cone z =



+ y

inside the cylinder x

+ y

= 4 and (b) the portion of the sphere x

+ y

+ z

= 16

inside of the cone z =



+ y

Solution It is important to realize that both parts (a) and (b) have numerous solutions.

(In fact, every surface has an inﬁnite number of parametric representations.) The

solutions we show here are among the simplest and most useful, but they are not the

only reasonable solutions. In (a), the repeated appearance of the term x

+ y

suggests

that cylindrical coordinates (r,θ,z) might be convenient. A sketch of the surface is

shown in Figure 15.44a. Notice that the cone z =



+ y

becomes z = r in

cylindrical coordinates, where x = r cosθ and y = r sinθ. Further, the parameters r

and θ have ranges determined by the cylinder x

+ y

= 4, so that 0 ≤ r ≤ 2 and

0 ≤ θ ≤ 2π . Parametric equations for the cone are then x = r cosθ, y = r sinθ and

z = r with 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π.

FIGURE 15.44a FIGURE 15.44b

The cone z =



+ y

and the The portion of the sphere

cylinder x

+ y

= 4 inside the cone

The surface in part (b) is a portion of a sphere, which suggests (what else?)

spherical coordinates: x = ρ sinφ cosθ, y = ρ sinφ sin θ and z = ρ cos φ, where

= x

+ y

+ z

. The equation of the sphere x

+ y

+ z

= 16 is then ρ = 4. Using

this, a parametric representation of the sphere is x = 4sinφ cosθ, y = 4sinφ sinθ and

z = 4cos φ, where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π. To ﬁnd the portion of the sphere inside

the cone, observe that the cone can be described in spherical coordinates as φ =

Referring to Figure 15.44b, note that the portion of the sphere inside the cone is then

described by x = 4sinφ cosθ, y = 4sinφ sinθ and z = 4cos φ, where 0 ≤ θ ≤ 2π

and 0 ≤ φ ≤



Suppose that we have a parametric representation for the surface S: x = x(u,v),

y = y(u,v)and z = z(u,v),deﬁnedon therectangle R ={(u,v)|a ≤ u ≤ b andc ≤ v ≤ d}

in the uv-plane. It is often convenient to use parametric equations to evaluate the surface

integral



f (x, y, z) dS. Of course, to do this, we must substitute for x, y and z to rewrite

the integrand in terms of the parameters u and v,as

g(u,v) = f (x(u,v), y(u,v), z(u,v)).

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15-61 SECTION 15.6

Surface Integrals 1037

We must also write the surface area element dS in terms of the area element dA for the

uv-plane. Unfortunately, we can’t use (6.1) here, since this holds only for the case of a

surface written in the form z = f (x, y). Instead, we’ll need to back up just a bit.

First, notice that the position vector for points on the surface S is

r(u,v) =x(u,v), y(u,v), z(u,v). We deﬁne the vectors r

and r

(the subscripts denote

partial derivatives) by

(u,v) =x

(u,v), y

(u,v), z

(u,v)

and r

(u,v) =x

(u,v), y

(u,v), z

(u,v).

Notice that for any ﬁxed (u,v), both of the vectors r

(u,v) and r

(u,v) lie in the tangent

plane to S at the point (x(u,v), y(u,v), z(u,v)). So, unless these two vectors are parallel,

n = r

× r

is a normal vector to the surface at the point (x(u,v), y(u,v), z(u,v)). We say

that the surface S is smooth if r

and r

are continuous and r

× r

= 0, for all (u,v) ∈ R.

(This says that the surface will not have any corners.) We say that S is piecewise-smooth

if we can write S = S

∪ S

∪···∪S

, for some smooth surfaces S

, S

,...,S

a b

v

u

, v

)

FIGURE 15.45a

Partition of parameter domain

(uv-plane)

)  r

)

u

)

v

)

FIGURE 15.45b FIGURE 15.45c

Curvilinear region S

The parallelogram T

Aswehave done manytimes now,we partitiontherectangle R in the uv-plane.For each

rectangle R

in the partition, let (u

) be the closest point in R

to the origin, as indicated

in Figure 15.45a. Notice that each of the sides of R

gets mapped to a curve in xyz-space, so

that R

gets mapped to a curvilinear region S

in xyz-space, as indicated in Figure 15.45b.

Observethat if we locate their initial points at the point P

(x(u

), y(u

), z(u

)), the

vectors r

) and r

) lie tangent to two adjacent curved sides of S

. So, we can

approximate the area S

of S

by the area of the parallelogram T

whose sides are formed

by the vectors u

) and v

). (See Figure 15.45c.) As we know, the area

of the parallelogram is given by the magnitude of the cross product

u

) × v

)=r

) × r

)u

v

=r

) × r

) A

where  A

is the area of the rectangle R

. We then have that

S

≈r

) × r

) A

and it follows that the element of surface area can be written as

dS =r

× r

dA. (6.2)

Notice that this corresponds closely to (6.1), as r

× r

is a normal vector to S. Finally, we

developed (6.2) in the comparatively simple case where the parameter domain R (that is,

the domain in the uv-plane) was a rectangle. If the parameter domain is not a rectangle, you

should recognize that we can do the same thing by constructing an inner partition of the

region.Wecan nowevaluatesurfaceintegralsusingparametricequations, as in example6.4.

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1038 CHAPTER 15

Vector Calculus 15-62

EXAMPLE 6.4 Evaluating a Surface Integral Using Spherical Coordinates

Evaluate



(3x

+ 3y

+ 3z

)dS, where S is the sphere x

+ y

+ z

= 4.

Solution Since the surface is a sphere and the integrand contains the term

+ y

+ z

, spherical coordinates are indicated. Notice that the sphere is described by

ρ = 2 and on the surface of the sphere, the integrand becomes 3(x

+ y

+ z

) = 12.

Further, we can describe the sphere ρ = 2 with the parametric equations

x = 2sinφ cos θ, y = 2sinφ sinθ and z = 2cos φ, for 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π.

This says that the sphere is traced out by the endpoint of the vector-valued function

r(φ, θ) =2sin φ cosθ,2sinφ sinθ,2cosφ.

We then have the partial derivatives

=−2sinφ sinθ,2 sinφ cos θ,0

and

=2cos φ cosθ,2 cosφ sin θ,−2 sin φ.

We leave it as an exercise to show that a normal vector to the surface is given by

n = r

× r

=−4sin

φ cos θ,−4 sin

φ sin θ,−4 sinφ cos φ,

so that n=4|sin φ|. Equation (6.2) now gives us dS = 4|sinφ|dA, so that



(3x

+ 3y

+ 3z

)dS =



(12)(4) |sinφ|dA



2π



48sin φ dφ dθ

= 192π,

where we replaced |sin φ| by sin φ by using the fact that for 0 ≤ φ ≤ π, sinφ ≥ 0.



NOTES

For the sphere x

= R

the surface area element in

spherical coordinates is

dS = R

sinφ dφ dθ.

If you did the calculation of dS in example 6.4, you may not think that parametric

equations lead to simple solutions. (That’s why we didn’t show all of the details!) However,

recall that for changing a triple integral from rectangular to spherical coordinates, you re-

place dx dydz by ρ

sinφ dρ dφ dθ. In example 6.4, we have ρ

= 4 and dS = 4sinφ dA,

which looks more familiar. This shortcut is valuable, since surface integrals over spheres

are reasonably common.

We close the section with several illustrations of how surface integrals are used. The

ﬁrst is familiar: observe that the surface integral of the function f (x, y, z) = 1 over the

surface S is simply the surface area of S. That is,



1 dS = Surfacearea of S.

The proof of this follows directly from the deﬁnition of the surface integral and is left as an

exercise.

EXAMPLE 6.5 Using a Surface Integral to Compute Surface Area

Compute the surface area of the portion of the hyperboloid x

+ y

− z

= 4 between

z = 0 and z = 2.

Solution We need to evaluate



1dS. Notice that we can write the hyperboloid

parametrically as x = 2cosu cosh v, y = 2sin u cosh v and z = 2 sinh v.(Youcan

derive parametric equations in the following way. To get a circular cross section of

radius 2 in the xy-plane, start with x = 2cosu and y = 2sinu. To get a hyperbola in the

xz- or yz-plane, multiply x and y by coshv and set z = sinhv.) We have 0 ≤ u ≤ 2π to

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15-63 SECTION 15.6

Surface Integrals 1039

get the circular cross sections and 0 ≤ v ≤ sinh

−1

1(≈ 0.88). The hyperboloid is traced

out by the endpoint of the vector-valued function

r(u,v) =2cos u cosh v, 2sinu coshv, 2 sinhv,

so that

=−2sinu coshv, 2 cosu cosh v, 0

and

=2cos u sinhv, 2sinu sinhv, 2 cosh v.

This gives us the normal vector

n = r

× r

=4cos u cosh

v, 4sinu cosh

v, −4coshv sinhv,

where n=4cosh v



cosh

v + sinh

v.Wenowhave



1dS =



4cosh v



cosh

v + sinh

v dA



sinh

−1



2π

4cosh v



cosh

v + sinh

v du dv

≈ 31.95,

where we evaluated the ﬁnal integral numerically.



HISTORICAL

NOTES

August Ferdinand M¨obius

(1790–1868) German

astronomer and mathematician

who gave one of the earliest

descriptions of the one-sided

surface that bears his name.

obius’ doctoral thesis was in

astronomy, but his astronomy

teachers included the great

mathematician Carl Friedrich

Gauss. M

obius did research in

both fields. His mathematics

publications, primarily in

geometry and topology, were

exceedingly clear and well

developed.

Our next example of a surface integral requires some preliminary discussion. First, we

say that a surface S is orientable (or two-sided) if it is possible to deﬁne a unit normal

vector n at each point (x, y, z) not on the boundary of the surface and if n is a continuous

function of (x, y, z). In this case, S has two identiﬁable sides (a top and a bottom or an inside

and an outside). Once we choose a consistent direction for all normal vectors to point, we

call the surface oriented. For instance, a sphere is a two-sided surface; the two sides of

the surface are the inside and the outside. Notice that you can’t get from the inside to the

outside without passing through the sphere. The positive orientation for the sphere (or

for any other closed surface) is to choose outward normal vectors (normal vectors pointing

away from the interior).

All of the surfaces we have seen so far in this course are two-sided, but it’s not difﬁcult

to construct a one-sided surface. Perhaps the most famous example of a one-sided surface

is the M¨obius strip, named after the German mathematician A. F. M¨obius. You can easily

construct a M¨obius strip by taking a long rectangular strip of paper, giving it a half-twist

and then taping the short edges together, as illustrated in Figures 15.46a through 15.46c.

Notice that if you started painting the strip, you would eventually return to your starting

point, having painted both “sides” of the strip, but without having crossed any edges. This

says that the M¨obius strip has no inside and no outside and is therefore not orientable.

FIGURE 15.46a FIGURE 15.46b FIGURE 15.46c

A long, thin strip Make one half-twist A M¨obius strip

One reason we need to be able to orient a surface is to compute the ﬂux of a vector

ﬁeld. It’s easiest to visualize the ﬂux for a vector ﬁeld representing the velocity ﬁeld for a

ﬂuid in motion. In this context, the ﬂux measures the net ﬂow rate of the ﬂuid across the

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surface in the direction of the speciﬁed normal vectors. (Notice that for this to make sense,

the surface must have two identiﬁable sides. That is, the surface must be orientable.) The

orientation of the surface lets us distinguish one direction from the other. In general, we

have the following deﬁnition.

DEFINITION 6.2

Let F(x, y, z) be a continuous vector ﬁeld deﬁned on an oriented surface S with unit

normal vector n. The surface integral of F over S (or the ﬂux of F over S)isgiven



F · n dS.

Think carefully about the role of the unit normal vector in Deﬁnition 6.2. Notice that

since n is a unit vector, the integrand F ·n gives (at any given point on S) the component

of F in the direction of n. So, if F represents the velocity ﬁeld for a ﬂuid in motion, F ·n

corresponds to the component of the velocity that moves the ﬂuid across the surface (from

one side to the other). Also, note that F · n can be positive or negative, depending on which

normal vector we have chosen. (Keep in mind that at each point on a surface, there are two

unit normal vectors, one pointing toward each side of the surface.) You should recognize

that this is why we need to have an oriented surface.

EXAMPLE 6.6 Computing the Flux of a Vector Field

Compute the ﬂux of the vector ﬁeld F(x, y, z) =x, y, 0 over the portion of the

paraboloid z = x

+ y

below z = 4 (oriented with upward-pointing normal vectors).

Solution First, observe that at any given point, the normal vectors for the paraboloid

z = x

+ y

are ±2x, 2y, −1. For the normal vector to point upward, we need a

positive z-component. In this case,

m =−2x, 2y, −1=−2x, −2y, 1

is such a normal vector. A unit vector pointing in the same direction as m is then

n =

−2x, −2y, 1



+ 4y

+ 1

Before computing F · n, we use the normal vector m to write the surface area increment

dS in terms of dA. From (6.1), we have

dS =mdA =



+ 4y

+ 1 dA.

Putting this all together gives us



F · n dS =



x, y, 0·

−2x, −2y, 1



+ 4y

+ 1



+ 4y

+ 1 dA



x, y, 0·−2x, −2y, 1dA



(−2x

− 2y

)dA,

z  4

FIGURE 15.47

z = x

+ y

where the region R is the projection of the portion of the paraboloid under consideration

onto the xy-plane. Note how the square roots arising from the calculation of n and dS

canceled out one another. Look at the graph in Figure 15.47 and recognize that this

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15-65 SECTION 15.6

Surface Integrals 1041

projection is bounded by the circle x

+ y

= 4. You should quickly realize that the

double integral should be set up in polar coordinates. We have



F · n dS =



(−2x

− 2y

)dA



2π



(−2r

)rdrdθ =−16π.



Flux integrals enable engineers and physicists to compute the ﬂow of a variety of

quantities in three dimensions. If F represents the velocity ﬁeld of a ﬂuid, then the ﬂux

gives the net amount of ﬂuid crossing the surface S. In example 6.7, we compute heat ﬂow

across a surface. If T (x, y, z) givesthe temperature at (x, y, z), then the net heat ﬂow across

the surface S is the ﬂux of F =−k∇T , where the constant k is called the heat conductivity

of the material.

EXAMPLE 6.7 Computing the Heat Flow Out of a Sphere

For T (x, y, z) = 30 −

and k = 2, compute the heat ﬂow out of the region bounded

by x

+ y

+ z

= 9.

Solution We compute the ﬂux of −2∇(30 −

) =−20, 0, −

z=0, 0,

z.

Since we want the ﬂow out of the sphere, we need to ﬁnd an outward unit normal to

the sphere. An outward normal is ∇(x

+ y

+ z

) =2x, 2y, 2z,sowetake

n =

2x, 2y, 2z



+ 4y

+ 4z

. On the sphere, we have 4x

+ 4y

+ 4z

= 36, so that the

denominator simpliﬁes to

√

36 = 6. This gives us n =

x, y, z and F ·n =

Since the surface is a sphere, we will use spherical coordinates for the surface integral.

On the sphere, z = ρ cosφ = 3cosφ, so that F ·n =

(3cos φ)

cos

φ. Further,

dS = ρ

sinφ dA = 9 sinφ dA. The ﬂux is then



F · n ds =





cos



9sin φ dA =



2π



6cos

φ sinφ dφ dθ = 8π.

Since the ﬂux is positive, the heat is ﬂowing out of the sphere. Finally, observe that

since the temperature decreases as |z| increases, this result should make sense.



BEYOND FORMULAS

Surface integrals complete the

set of integrals introduced in

this book. (There are many

more types of integrals used in

more advanced mathematics

courses and applications areas.)

Although the different types of

integrals sometimes require

different methods of evaluation,

the underlying concepts are the

same. In each case, the integral

is a limit of approximating sums

and in some cases can be

evaluated directly with some

form of an antiderivative.

EXERCISES 15.6

WRITING EXERCISES

1. For deﬁnition 6.1, we deﬁned the partition of a surface and

took the limit as the norm of the partition tends to 0. Explain

why it would not be sufﬁcient to have the number of segments

in the partition tend to ∞.

2. In example 6.2, you could alternatively start with cylindrical

coordinates and use a parametric representation as we did

in example 6.4. Discuss which method you think would be

simpler.

3. Explain in words why



1dS equals the surface area of S.

4. For example 6.6, sketch a graph showing the surface S and

several normal vectors to the surface. Also, show several vec-

tors in the graph of the vector ﬁeld F. Explain why the ﬂux is

negative.

In exercises 1–8, ﬁnd a parametric representation of the

surface.

1. z = 3x + 4y 2. x

+ y

+ z

= 4

3. x

+ y

− z

= 1 4. x

− y

+ z

= 4

5. The portion of x

+ y

= 4 from z = 0toz = 2

6. The portion of y

+ z

= 9 from x =−1tox = 1

7. The portion of z = 4 − x

− y

above the xy-plane

8. The portion of z = x

+ y

below z = 4

............................................................

In exercises 9–18, sketch a graph of the parametric surface.

9. x = u, y = v, z = u

+ 2v

10. x = u, y = v, z = 4 − u

− v

11. x = u cosv, y = u sin v, z = u

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1042 CHAPTER 15

Vector Calculus 15-66

12. x = u cosv, y = u sin v, z = u

13. x = 2sinu cosv, y = 2sinu sinv, z = 2cosu

14. x = 2cosv, y = 2sinv, z = u

15. x = u, y = sinu cosv, z = sinu sinv

16. x = cosu cosv, y = u, z = cos u sinv

17. x = (4 +2cos v) cos u, y = (4 + 2 cos v) sin u, z = 2sinv

18. x = (−2 +

√

16 − v

)cosu, y = (−2 +

√

16 − v

)sinu,

z = v, −

√

12 ≤ v ≤

√

............................................................

19. Match the parametric equations a–c with the surfaces 1–3

a. x = u cosv, y = u sin v, z = v

b. x = v, y = u cos v, z = u sinv

c. x = u, y = u cos v, z = u sinv

100

SURFACE 1 SURFACE 2

SURFACE 3

20. In example 6.4, show that

× r

=−4sin

φ cos θ,−4sin

φ sin θ,−4sinφ cosφ

and then show that n=4|sinφ|.

............................................................

In exercises 21–32, ﬁnd the surface area of the given surface.

21. The portion of the cone z =



+ y

below the plane

z = 4

22. The portion of the paraboloid z = x

+ y

below the plane

z = 4

23. The portion of the plane 3x + y + 2z = 6 inside the cylinder

+ y

= 4

24. The portion of the plane x + 2y + z = 4 above the region

bounded by y = x

and y = 1

25. The portion of the cone z =



+ y

above the triangle with

vertices (0, 0), (1, 0) and (1, 1)

26. The portion of the paraboloid z = x

+ y

inside the cylinder

+ y

= 4

27. The portion of the hemisphere z =



4 − x

− y

above the

plane z = 1

28. The lemon (−2 +

√

16 − v

)cosu, (−2 +

√

16 − v

)sinu,

v, −

√

12 ≤ v ≤

√

29. The torus (c +a cosv)cos u, (c +a cosv) sin u, a sin v,

c > a > 0

30. Thesurface(−a +b cosv) cos u, (−a +b cosv) sin u, b sin v,

b > a > 0

31. The portion of x

+ y

− z

= 1 with 0 ≤ z ≤ 1 (Hint: Re-

quires trigonometric substitution)

32. The portion of the parabolic cylinder y = 4 − x

with y ≥ 0

and between z = 0 and z = 2 (Hint: Requires trigonometric

substitution)

............................................................

In exercises 33–42, set up a double integral that equals



g(x, y, z) dS and evaluate the surface integral



g(x, y, z) dS.

33.



xzdS, S is the portion of the plane z = 2x + 3y above the

rectangle 1 ≤ x ≤ 2, 1 ≤ y ≤ 3

34.



(z − y

)dS, S is the portion of the paraboloid z = x

+ y

below z = 4

35.



+ y

+ z

)

3/2

dS, S is the lower hemisphere

z =−



9 − x

− y

36.





+ y

+ z

dS, S is the sphere x

+ y

+ z

= 9

37.



+ y

− z) dS, S is the portion of the paraboloid

z = 4 − x

− y

between z = 1 and z = 2

38.



zdS, S is the hemisphere z =−



9 − x

− y

39.



dS, S is the portion of the cone z

= x

+ y

between

z =−4 and z = 4

40.



√

dS, S is the portion of the hemisphere

z =



4 − x

− y

above the cone z =



+ y

41.



xdS, S is the portion of x

+ y

− z

= 1 between z = 0

and z = 1

42.





+ y

+ z

dS, S is the portion of x =−



4 − y

− z

between y = 0 and y = x

............................................................

In exercises 43–54, evaluate the ﬂux integral



F ·n dS.

43. F =x, y, z, S is the portion of z = 4 − x

− y

above the

xy-plane (n upward)

44. F =y, −x, 1, S is the portion of z = x

+ y

below z = 4

(n downward)