Lecture Note
Advanced Engineering Mathematics, 7th edition
by
Peter V. O’Neil,
Math 301 (Methods of Applied Mathematics)
Chapter 12 (Vector Integral Calculus)
12.5 Surface Integrals
Dr T. A. Apalara
IMPORTANT!! These instructor notes are made available for convenience to
reduce the amount of note-taking in class and also in case it is not possible to
attend class due to circumstances beyond control (e.g. illness). These notes are
neither intended as a replacement for class attendance and participation nor as a
substitute for using the textbook. The class presentation may differ from these
notes. Moreover, some other problems not in these notes may be discussed in class.
Math 301
Week #4
12.5 Surface Integrals
Dr T. A. Apalara
§§ 12.5 Surface Integrals
Objective: In this chapter, you will know, insha Allaah:
☛ Some facts about surfaces
☛ How to compute normal vector to a surface
☛ How to compute the tangent plane to a surface
☛ How to find the area of a smooth and piecewise smooth surface
☛ How to compute the surface integral.
Some facts about Surfaces
☛ A curve is often given by specifying coordinate functions, each of which is a function of a
single variable or parameter. For this reason we think of a curve as a one-dimensional object,
although the graph may be in 2−space or 3−space.
☛ A surface may be defined by giving coordinate or parametric functions which depend on two
independent variables, say
x = x(u, v), y = y(u, v), z = z(u, v),
with (u, v) varying over some set in the u, v plane. The locus of such points may form a
two-dimensional object in the plane or in R3
Example 1. Suppose a surface is given by the coordinate functions
x = au cos(v), y = bu sin(v), z = u,
with u and v any real numbers and a and b nonzero constants. Write z in terms of x and y and
identify the surface.
Solution
y
x
= cos v and
= sin v. So, using identity, we have
au
bu
x 2 y 2
+
= cos2 v + sin2 v = 1
au
bu
x2
y2
x2
y2
=⇒ 2 + 2 = u2 =⇒ 2 + 2 = z 2
a
b
a
b
The surface is called an elliptical cone (or just a cone) because it has elliptical cross sections parallel
to the x, y plane.
(i) Horizontal trace z = c is the ellipse
x2
y2
+
= c2
a2
b2
1
z = c.
Math 301
Week #4
12.5 Surface Integrals
Dr T. A. Apalara
(ii) Vertical traces in y = c and x = c are hyperbolas if c ̸= 0
z2 −
c2
x2
= 2,
2
a
b
y=c
and
z2 −
y2
c2
= 2,
2
b
a
x = c.
(iii) Traces in xz− and yz−planes are are pairs of lines.
z2 =
x
x2
=⇒ z = ± ,
a2
a
which are straight lines of slope ±
y=0
and
z2 =
y
y2
=⇒ z = ± ,
b2
b
x=0
1
1
and ± , respectively, through the origin.
a
b
✍
Example 2. Consider the surface having coordinate functions
x = u cos(v), y = u sin(v), z =
1 2
u sin(2v),
2
in which u and v are any real numbers. Write z in terms of x and y and identify the surface.
Solution
xy = u2 sin v cos(v) =⇒ xy = z. The surface is called a hyperbolic paraboloid.
(i) Horizontal trace z = c is the hyperbola
xy = c
z = c.
(ii) The surface intersects a plane y = ±x in a parabola z = ±x2 .
✍
Remark 1.
☛ Often a surface is defined as a level surface f (x, y, z) = k, with f a given function. For example
f (x, y, z) = (x − 1)2 + y 2 (z + 4)4 = 16
is a sphere of radius 4 and center (1, 0, −4) as its graph.
☛ We may also express a surface as a locus of points satisfying an equation z = S(x, y). Here x
and y are used as parameters, and the surface is defined by giving z as a function of x and y.
Now the graph of the surface is the locus of points (x, y, S(x, y)), as (x, y) varies over some set
of points in the x, y plane.
☛ Just as we can write a position vector to a curve, we can write a position vector
R(u, v) = x(u, v)i + y(u, v)j + z(u, v)k
for a surface.
☛ For any u and v in the parameter domain, R(u, v) can be thought of as an arrow from the
origin to the point (x(u, v), y(u, v), z(u, v)) on the surface.
☛ A surface is simple if R(u1 , v1 ) = R(u2 , v2 ) can occur only if u1 = v1 and u2 = v2 . In other
words, a simple surface is one that does not fold over and intersect itself.
2
Math 301
Week #4
12.5 Surface Integrals
Dr T. A. Apalara
Normal Vector to a Surface
We would like to define a normal vector to a surface at a point. Previously this was done for level
surfaces.
P
Let
be a surface with coordinate functions x = x(u, v), y = y(u, v), z = z(u, v). Assume that
these functionsP
are continuous with continuous first partial derivatives. Let P0 : (x(u0 , v0 ), y(u0 , v0 ), z(u0 , v0 ))
be a point on .
X
☛ If we fix v = v0 , we can define the curve
on the surface, having coordinate functions
u
x = x(u, v0 ), y = y(u, v0 ), z = z(u, v0 ).
This is a curve because
its coordinate functions are functions of the single variable u. The
X
tangent vector to
at P0 is
u
∂y
∂z
∂x
(u0 , v0 )i +
(u0 , v0 )j +
(u0 , v0 )k.
∂u
∂u
∂u
Tu0 =
☛ Similarly, if we fix u = u0 and use v as parameter, we obtain the curve
X
on the surface. This
v
curve has coordinate functions
x = x(u0 , v), y = y(u0 , v), z = z(u0 , v).
The tangent vector to
X
at P0 is
v
Tv0 =
∂y
∂z
∂x
(u0 , v0 )i +
(u0 , v0 )j +
(u0 , v0 )k.
∂v
∂v
∂v
These two curves and tangent vectors are shown below
Curves
X
u
and
X
and Tangent vectors Tu0 and Tv0 .
v
AssumingPthat neither of these tangent vectors is the zero vector, they both lie in the tangent
plane to
at P0 . Their cross product is therefore
normal (orthogonal) to this tangent plane, and
P
is the vector we define to be the normal to
at P0 :
3
Math 301
Week #4
12.5 Surface Integrals
Dr T. A. Apalara
Definition (Normal Vector to a Surface) The normal vector to the surface at P0 given by
N(P0 ) is defined as
N(P0 ) = Tu0 × Tv0
i
∂x
(u0 , v0 )
= ∂u
j
∂y
(u0 , v0 )
∂u
k
∂z
(u0 , v0 )
∂u
∂x
(u0 , v0 )
∂v
∂y
(u0 , v0 )
∂v
∂z
(u0 , v0 )
∂v
in which all the partial derivatives are evaluated at (u0 , v0 ).
Remark 2.
☛ The expression can be written as
∂y ∂z
∂z ∂y
∂z ∂x ∂x ∂z
∂x ∂y
∂y ∂x
N(P0 ) =
−
i+
−
j+
−
k.
∂u ∂v ∂u ∂v
∂u ∂v
∂u ∂v
∂u ∂v
∂u ∂v
☛ To make this vector easier to write, define the Jacobian of two functions f and g to be
∂f
∂u
∂f
∂v
∂(f, g)
∂f ∂g
∂g ∂f
=
=
−
∂(u, v)
∂u ∂v ∂u ∂v
∂g ∂g
∂u ∂v
P
Then, the normal vector to
at P0 becomes
N(P0 ) =
∂(y, z)
∂(z, x)
∂(x, y)
i+
j+
k
∂(u, v)
∂(u, v)
∂(u, v)
with all the partial derivatives evaluated at (u0 , v0 ).
☛ The last expression is easy to remember with an observation. Write x, y, z in this order. For
the i component of N, delete x, leaving y, z, (in this order) in the numerator of the Jacobian.
For the j component, delete y from x, y, z, but move to the right, getting z, x in the Jacobian.
For the k component, delete z, leaving x, y, in this order.
Example 3. Consider again the elliptical cone in Example 1, that is
x = au cos v, y = bu sin v, z = u.
!
√
a 3 b 1
Find the normal vector at P0 :
, ,
.
4 4 2
Solution
√
√
a 3
b
1
1
1
3
x=
=⇒ u cos v =
, y=
=⇒ u sin v = , z =
=⇒ u =
4
4
4
4
2
2
1
π
∴ tan v = √ =⇒ v =
6
3
xu = a cos v, xv = −au sin v, yu = b sin v, yv = bu cos v, zu = 1, zv = 0
√
√
a 3
a
b
b 3
1 π
xu =
, xv = − , yu = , yv =
, zu = 1, zv = 0
evaluated at (u, v) =
,
.
2
4
2
4
2 6
4
Math 301
Week #4
12.5 Surface Integrals
Dr T. A. Apalara
1 π
Now, we compute the Jacobian components evaluate at (u, v) =
,
2 6
√
∂(y, z)
b 3
= (yu zv − zu yv )
=−
∂(u, v) ( 1 , π )
4
(1,π)
2
6
2
6
∂(z, x)
a
= (zu xv − xu zv )
=−
∂(u, v) ( 1 , π )
4
( 21 , π6 )
2 6
ab
∂(x, y)
= (xu yv − yu xv )
= .
∂(u, v) ( 1 , π )
2
(1,π)
2
The normal vector at P0 is
6
2
6
√ b
a
ab
N(P0 ) = − 3 i − j + k
4
4
2
✍
We frequently encounter the case that a surface
is
given
by
an
equation
z
=
S(x,
y).
We
may
think
P
of u = x and v = y as the parameters for
and write the coordinate functions as
x = x,
y = y,
z = S(x, y).
Since ∂x/∂x = 1 = ∂y/∂y and ∂x/∂y = ∂y/∂x = 0, we have
0
∂(y, z)
∂(y, z)
=
=
∂S
∂(u, v)
∂(x, y)
∂x
∂S
∂(z, x)
∂(z, x)
∂x
=
=
∂(u, v)
∂(x, y)
1
∂(x, y)
∂(x, y)
1
=
=
0
∂(u, v)
∂(x, y)
1
∂S
∂y
∂S
∂y
=−
∂S
,
∂x
=−
∂S
,
∂y
0
0
= 1.
1
The normal at a point P0 : (x0 , y0 , f (x0 , y0 )) in this case is
∂S
∂S
(x0 , y0 )i −
(x0 , y0 )j + k
∂x
∂y
∂z
∂z
=−
(x0 , y0 )i −
(x0 , y0 )j + k.
∂x
∂y
N(P0 ) = −
Remark 3.
☛ We can also denote this vector as N(x0 , y0 ).
P
P
☛ Alternatively, if
is given by z = S(x, y), then
is a level surface of the function
φ(x, y, z) = z − S(x, y).
The gradient of this function is a normal vector, so compute
∂φ
∂φ
∂φ
i+
j+
k
∂x
∂y
∂z
∂S
∂S
=−
i−
j + k = N(P ).
∂x
∂y
∇φ =
5
Math 301
Week #4
12.5 Surface Integrals
x2 + y 2 . Find the normal vector at P0 : 3, 1,
p
z = x2 + y 2
Example 4. Consider the cone given by z =
Solution
We consider x = x,
y = y,
p
Dr T. A. Apalara
√
10 .
x = 3 =⇒ u = 3, y = 1 =⇒ v = 1
x
y
zx = p
, zy = p
x2 + y 2
x2 + y 2
3
1
zx = √ , zy = √
[evaluated at (x, y) = (3, 1)] .
10
10
The normal vector at P0 is
3
3
N(P0 ) = − √ i − √ j + k
10
10
✍
The Tangent Plane to a Surface
P
has a normal
If a surface
P vector N at a point P0 , then we can use N to determine the equation
of the tangent plane to
at P0 . Let (x, y, z) be any point on the tangent plane. Then the vector
(x − x0 )i + (y − y0 )j + (z − z0 )k is in the tangent plane, hence is orthogonal to N. Then
N(P0 ) · [(x − x0 )i + (y − y0 )j + (z − z0 )k] = 0.
More explicitly,
∂(y, z)
∂(z, x)
∂(x, y)
(x − x0 ) +
(y − y0 ) +
(z − z0 ) = 0.
∂(u, v) (u0 ,v0 )
∂(u, v) (u0 ,v0 )
∂(u, v) (u0 ,v0 )
P
This is the equation of the tangent plane to
at P0 .
P
Remark 4. In the special case that
is given by z = S(x, y), then the normal vector at P0 is
N(P0 ) = −
∂S
∂S
i−
j + k,
∂x
∂y
so the equation of the tangent plane becomes
−
∂S
∂x
(x − x0 ) −
(x0 ,y0 )
∂S
∂y
(y − y0 ) + (z − z0 ) = 0.
(x0 ,y0 )
Example 5. For the elliptical cone considered in 3, find the tangent plane at
Solution
!
√
a 3 b 1
, ,
.
4 4 2
We found in 3 that the normal vector at P0 is
√ b
a
ab
N(P0 ) = − 3 i − j + k.
4
4
2
So, the tangent plane to
P
at this point has equation
√ !
√ b
a 3
a
b
ab
1
− 3
x−
−
y−
+
z−
= 0.
4
4
4
4
2
2
✍
6
Math 301
Week #4
Example 6. Let
P
12.5 Surface Integrals
be the hemisphere given by z =
Dr T. A. Apalara
p
9 − x2 − y 2 .
(a) Find the normal vector at P0 : (2, −2, 1) .
(b) Find the tangent plane at (2, −2, 1) .
Solution
(a) Let’s use the other method. Let S(x, y) =
p
9 − x2 − y 2 .
φ(x, y, z) = z − S(x, y) =⇒ φ(x, y, z) = z −
x
y
φx = p
, φy = p
,
2
2
9−x −y
9 − x2 − y 2
φx = 2, φy = −2,
p
9 − x2 − y 2
φz = 1
[evaluated at (x, y) = (2, −2)] .
φz = 1,
The gradient of this function is
∇φ =2i − 2j + k.
So, the normal vector at P0 is
N(P0 ) = 2i − 2j + k.
(b) The tangent plane at (2, −2, 1) is 2(x − 2) − 2(y + 2) + z − 1 = 0 =⇒ 2x − 2y + z = 9.
✍
Smooth and Piecewise Smooth Surfaces
Recall that a curve is smooth if it has a continuous tangent. Similarly, a surface is smooth if it
has a continuous normal vector. A surface is piecewise smooth if it consists of a finite number of
smooth surfaces. For example, a sphere is smooth, and the surface of a cube is piecewise smooth.
A cube consists of six square pieces, which are smooth, but does not have a normal vector along
any of its edges.
Area of a Smooth Surfaces
P
In calculus, it is shown that the area of a smooth surface
given by z = S(x, y) is the integral
s
2 2
X ZZ
∂S
∂S
area of
1+
=
+
dA.
(1)
∂x
∂y
D
where D is the set of points in the x, y−plane for which f is defined. This may also be written
s
2 2
X ZZ
∂z
∂z
area of
=
1+
+
dxdy.
∂x
∂y
D
Note that the length of the normal vector is given as
s
2 2
∂S
∂S
∥N(x, y)∥ = 1 +
+
.
∂x
∂y
7
Math 301
Week #4
12.5 Surface Integrals
Dr T. A. Apalara
So, Equation (1) is the integral of the length of the normal vector
X ZZ
area of
=
∥N(x, y)∥ dxdy.
D
This is analogous to the formula for the length of a curve as the integral of the length of the tangent
vector.
P
Area of a smooth Surface: If
is given by coordinate functions x = x(u, v), y = y(u, v) and
z = z(u, v), with (u, v) varying over some set D in the u, v plane, then
X ZZ
area of
=
∥N(u, v)∥dudv,
(2)
D
the integral of the length of the normal vector, which is given by
N(P0 ) =
∂(y, z)
∂(z, x)
∂(x, y)
i+
j+
k.
∂(u, v)
∂(u, v)
∂(u, v)
p
Example 7. Compute the area of the hemisphere given by z = S(x, y) = 9 − x2 − y 2 .
p
Solution Note that the domain of the hemisphere S(x, y) = 9 − x2 − y 2 is
D = {(x, y) | 9 − x2 − y 2 ≥ 0} = {(x, y) | x2 + y 2 ≤ 9}
which is the disk with center (0, 0) and radius 3. In other words, D consists of all points on or
inside the circle of radius 3 about the origin in the x, y−plane.
x
y
∂z
∂z
= −p
= −p
,
.
∂x
9 − x2 − y 2 ∂y
9 − x2 − y 2
Then
area of
X
ZZ
s
=
1+
∂z
∂x
2
+
∂z
∂y
2
dxdy
D
ZZ s
1+
=
x2
y2
+
dxdy = 3
2
2
9−x −y
9 − x2 − y 2
ZZ
D
D
1
p
dxdy.
9 − x2 − y 2
This is an improper double integral which we can evaluate easily by converting it to polar coordinates. Let x = r cos θ, y = r sin θ. Since D is the disk of radius 3 about the origin, we have
0 ≤ r ≤ 3 and 0 ≤ θ ≤ 2π. Then
ZZ
Z 2π Z 3
Z 3
X
1
1
1
p
√
√
dxdy = 3
rdrdθ = 2π
rdrdθ
area of
=3
2
2
2
9−r
9 − r2
9−x −y
0
0
0
D
3
= − 6π
p
9 − r2 dθ
[substitution method u = 9 − r2 ]
0
=18π.
✍
This is the area of a hemisphere of radius 3.
8
Math 301
Week #4
12.5 Surface Integrals
Dr T. A. Apalara
Remark 5.
☞ Area of a hemisphere of radius R is 2πR2 . Area of a whole sphere of radius R is 4πR2 .
☞ (Integrals in Polar Coordinates). Suppose that a function f (r, θ) is defined over a region
R that is bounded by the rays θ = α and θ = β and by the continuous curves r = g1 (θ) and
r = g2 (θ), then
ZZ
Z θ=β Z r=g2 (θ)
f (r, θ)r dr dθ.
f (r, θ) dA =
θ=α
R
r=g1 (θ)
We are now ready to define the integral of a function over a surface.
Surface Integral
P
Surface Integral: Let
be a smooth surface having coordinate functions x = x(u,
Pv), y =
y(u, v), z = z(u, v) for (u, v) in some set DZ of
the
u,
v
plane.
Let
f
be
continuous
on
. Then
Z
P
the surface integral of f over
is denoted
f (x, y, z)dσ, and is defined by
P
ZZ
ZZ
f (x(u, v), y(u, v), z(u, v)) ∥N(u, v)∥dudv.
f (x, y, z) dσ =
P
D
Remark 6.
X
X
P
☛ If
is a piecewise smooth surface having smooth components
,··· ,
, with each compo1
n
nent either disjoint from the others, or intersecting another component in a set of zero area (for
example, along a curve), then
ZZ
ZZ
ZZ
f (x, y, z) dσ =
f (x, y, z) dσ + · · · +
f (x, y, z) dσ.
P
X
X
n
1
For example, we would integrate over the surface of a cube by summing the surface integrals
over the six faces. Two such faces either do not intersect, or intersect each other along a line
segment having zero area. The intersection condition is to prevent the selection of surface
components that overlap each other in significant ways.
P
☛ If
is described by z = S(x, y), then
s
2 2
ZZ
ZZ
∂S
∂S
f (x, y, z) dσ =
f (x, y, S(x, y)) 1 +
+
dxdy.
∂x
∂y
P
D
ZZ
Example 8. Compute the surface integral
xyzdσ over the part of the surface
P
x = u cos v, y = u sin v, z =
corresponding to (u, v) in D : 1 ≤ u ≤ 2, 0 ≤ v ≤ π.
9
1 2
u sin 2v
2
Math 301
Solution
Week #4
12.5 Surface Integrals
Dr T. A. Apalara
First we need the normal vector. So
xu = cos v, xv = −u sin v, yu = sin v, yv = u cos v, zu = u sin 2v, zv = u2 cos 2v
The components of N(u, v) are the Jacobians:
∂(y, z)
= (yu zv − zu yv ) = u2 [sin v cos 2v − cos v sin 2v] = u2 sin(v − 2v) = −u2 sin v
∂(u, v)
∂(z, x)
= (zu xv − xu zv ) = −u2 [sin v sin 2v + cos v cos 2v] = −u2 cos(v − 2v) = −u2 cos v
∂(u, v)
∂(x, y)
= (xu yv − yu xv ) = u cos2 v + sin2 v = u.
∂(u, v)
Then
∥N(u, v)∥2 = u4 sin2 v + u2 cos2 v + u2 = u4 + u2 = u2 (1 + u2 ).
Therefore,
p
∥N(u, v)∥ = u 1 + u2 .
The surface integral is
p
ZZ
ZZ
1
xyzdσ =
[u cos v] [u sin v] u2 sin 2v u 1 + u2 dA
2
P
D
1
=
2
Z
=
π
Z
2
Z
cos v sin v sin(2v)u5
0
1
πZ 2
sin2 (2v)u5
0
1
p
1 + u2 dudv
Z
Z
p
1 π
1 2 4p
1 + u2 dudv =
[1 − cos(4v)] dv ·
u
1 + u2 2udu
| {z }
2 0
2 1
t=1+u2
1
sin(4v)
= v−
4
4
π
5
Z
(t − 1)2 t1/2 dt =
2
0
π
4
11 √
100 √
21 −
2 .
21
1051
✍
ZZ
Example 9. Compute the surface integral
zdσ over the part of the plane x + y + z = 4 lying
P
above the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.
Solution
ZZ
With z = S(x, y) = 4 − x − y, we have
zdσ =
P
s
ZZ
z
1+
∂S
∂x
2
+
∂S
∂y
2
ZZ
dxdy =
q
z
D
2
2
1 + (−1) + (−1) dxdy
D
√ Z
= 3
0
1
Z
0
2
√ Z
(4 − x − y) dxdy = 3
1
0
x2
4x −
− xy
2
END
10
2
0
√ Z
dy = 3
1
√
(6 − 2y) dy = 5 3.
0
✍
Math 301
Week #4
12.5 Surface Integrals
Dr T. A. Apalara
More examples for your practice
ZZ
Example. Evaluate
h(x, y, z)dσ, where
P
is the part of cylinder z = 1 + y 2 for which −1 ≤ x ≤ 1, 0 ≤ y ≤ 1.
P
(i) h(x, y, z) = xyz and
P
(ii) h(x, y, z) = xyz and
is the part of paraboloid z = x2 + y 2 for which 0 ≤ z ≤ 4.
p
P
(iii) h(x, y, z) = z and
is the part of cone z = x2 + y 2 in the first octant and between the
panes z = 2 and z = 4
Solution
(i) Here z = S(x, y) = 1 + y 2 =⇒ Sx = 0 and Sy = 2y, then
ZZ
ZZ
q
2
2
h (x, y, z) dσ =
h (x, y, S(x, y)) 1 + (Sx ) + (Sy ) dxdy
P
D
ZZ
ZZ
xyzdσ =
P
xy
p
1 + y2
xy
p
y2
q
2
1 + 02 + (2y) dxdy
D
ZZ
=
1+
p
1+
4y 2
Z
=
Z
xy
y
p
1 + y2
p
1 + 4y 2 dy
Z
p
p
1 + y 2 1 + 4y 2 dxdy
−1
0
1
1
dxdy =
D
Z
1
1
x dx = 0.
| {z }
−1
0
=0
(ii) Here z = S(x, y) = x2 + y 2 =⇒ Sx = 2x and Sy = 2y, then
ZZ
ZZ
q
2
2
h (x, y, z) dσ =
h (x, y, S(x, y)) 1 + (Sx ) + (Sy ) dxdy
P
D
ZZ
ZZ
xyzdσ =
P
q
2
2
xy(x2 + y 2 ) 1 + (2x) + (2y) dxdy
D
ZZ
=
p
xy(x2 + y 2 ) 1 + 4x2 + 4y 2 dxdy.
D
When 0 ≤ z ≤ 4 we have that the point (x, y) lies in the disk D, centered at (0, 0) with radius
2. So, using the polar coordinates, we have
ZZ
ZZ
xyzdσ =
P
p
xy(x2 + y 2 ) 1 + 4x2 + 4y 2 dxdy
D
Z
2π
Z
=
0
Z
=
0
|
2
(r cos θ)(r sin θ)r2
p
1 + 4r2 rdrdθ =
0
Z
0
2π
Z
cos θ sin θ dθ
{z
}
2
r5
p
1 + 4r2 dr = 0.
0
=0
11
2π
Z
0
2
r5 cos θ sin θ
p
1 + 4r2 drdθ
Math 301
Week #4
12.5 Surface Integrals
Dr T. A. Apalara
p
y
x
and Sy = p
, then
x2 + y 2 =⇒ Sx = p
2
2
2
x +y
x + y2
ZZ
ZZ
q
2
2
h (x, y, z) dσ =
h (x, y, S(x, y)) 1 + (Sx ) + (Sy ) dxdy
(iii) Here z = S(x, y) =
P
D
ZZ
zdσ =
P
ZZ p
v
u
u
2
2
x + y t1 +
D
=
ZZ p
x
!2
+
p
x2 + y 2
!2
y
p
x2 + y 2
dxdy
√
x2 + y 2 2 dxdy.
D
When 2 ≤ z ≤ 4 =⇒ 2 ≤
ZZ
P
p
x2 + y 2 ≤ 4. So, using the polar coordinates, we have
√ ZZ p
x2 + y 2 dxdy
zdσ = 2
D
π
√ Z 2Z
= 2
0
2
4
r rdrdθ =
√
π
2·
2
Z
4
2
r dr =
2
√
π 1
2 · · r3
2 3
4
2
√
28π 2
=
.
3
✍
12
12.6 Applications of Surface Integrals
Then
395
√ 2 7
√
z dσ = 3
− x d x = 5 3. 2
0
PROBLEMS
SECTION 12.5
evaluate
5. f (x, y, z) = z, is the part of the cone z = x 2 + y 2
in the first octant and between the planes z = 2 and
z = 4.
1. f (x, y, z) = x, is the part of the plane x + 4y + z =
10 in the first octant.
6. f (x, y, z) = x yz, is the part of the plane z = x + y
with (x, y) in the square with vertices (0, 0), (1, 0),
(0, 1) and (1, 1).
In
each of Problems
f (x, y, z)dσ .
1
through
10,
is the part of the plane z = x for
2. f (x, y, z) = y 2 ,
0 ≤ x ≤ 2, 0 ≤ y ≤ 4.
7. f (x, y, z) = y, is the part of the cylinder z = x 2 for
0 ≤ x ≤ 2, 0 ≤ y ≤ 3.
3. f (x, y, z) = 1,
is the part of the paraboloid
z = x 2 + y 2 lying between the planes z = 2
and z = 7.
is the part of the paraboloid z =
8. f (x, y, z) = x 2 ,
4 − x 2 − y 2 lying above the x, y - plane.
4. f (x, y, z) = x + y,
is the part of the plane
4x + 8y + 10z = 25 lying above the triangle
in the x, y - plane having vertices (0, 0), (1, 0)
and (1, 1).
10. f (x, y, z) = x yz,
is the part of the cylinder z =
1 + y 2 for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.
12.6
9. f (x, y, z) = z, is the part of the plane z = x − y for
0 ≤ x ≤ 1 and 0 ≤ y ≤ 5.
Applications of Surface Integrals
12.6.1
If
Surface Area
is a piecewise smooth surface, then
dσ =
N(u, v) du dv = area of
.
D
This assumes a bounded surface having finite area. Clearly we do not need surface integrals to
compute areas of surfaces. However, we mention this result because it is in the same spirit as
other familiar mensuration formulas:
ds = length of C,
C
d A = area of D,
D
d V = volume of M.
M
12.6.2
Mass and Center of Mass of a Shell
Imagine a shell of negligible thickness in the shape of a piecewise smooth surface . Let
δ(x, y, z) be the density of the material of the shell at point (x, y, z). We want to compute the
mass of the shell.
Let
have coordinate functions x(u, v), y(u, v), z(u, v) for (u, v) in D. Form a grid of
lines over D, as in Figure 12.15, by drawing vertical lines u units apart and horizontal lines
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
October 14, 2010
14:53
THM/NEIL
Page-395
27410_12_ch12_p367-424
