16 surface integrals - handout
DESCRIPTION
Math 55 chapter 16TRANSCRIPT
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Surface Integrals
Math 55 - Elementray Analysis III
Institute of MathematicsUniversity of the Philippines
Diliman
Math 55 Surface Integrals 1/ 16
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Surface Integrals
In the same way that a line integral is related to the arclength,surface integrals are related to surface area in a way that iff(x, y, z) = 1, the value of the surface integral of f(x, y, z) overa surface S is equal to the surface area of S.
Math 55 Surface Integrals 2/ 16
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Parametric Surfaces
Suppose f(x, y, z) is a function whose domain includes a surfaceS given by ~R(u, v) = x(u, v), y(u, v), z(u, v).
Subdivide the parameter domain D of ~R(u, v) intosubrectangles Rij with dimensions u and v. This divides Sinto corresponding patches Sij .
Let P ij be a point in each patch and Sij be the area of eachpatch.
Math 55 Surface Integrals 3/ 16
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Parametric Surfaces
Suppose f(x, y, z) is a function whose domain includes a surfaceS given by ~R(u, v) = x(u, v), y(u, v), z(u, v).Subdivide the parameter domain D of ~R(u, v) intosubrectangles Rij with dimensions u and v. This divides Sinto corresponding patches Sij .
Let P ij be a point in each patch and Sij be the area of eachpatch.
Math 55 Surface Integrals 3/ 16
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Parametric Surfaces
Suppose f(x, y, z) is a function whose domain includes a surfaceS given by ~R(u, v) = x(u, v), y(u, v), z(u, v).Subdivide the parameter domain D of ~R(u, v) intosubrectangles Rij with dimensions u and v. This divides Sinto corresponding patches Sij .
Let P ij be a point in each patch and Sij be the area of eachpatch.
Math 55 Surface Integrals 3/ 16
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Parametric Surfaces
We define the surface integral of f over the surface S asS
f(x, y, z) dS = limm,n
mi=1
nj=1
f(P ij)Sij .
We use the approximation Sij ~Ru ~Rvuv where~Ru =
xu ,
yu ,
zu
and ~Rv =
xv ,
yv ,
zv
If ~Ru and ~Rv are non-parallel and non-zero whose componentsare continuous on D, then
S
f(x, y, z) dS =
D
f(~R(u, v))~Ru ~RvdA
Math 55 Surface Integrals 4/ 16
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Parametric Surfaces
We define the surface integral of f over the surface S asS
f(x, y, z) dS = limm,n
mi=1
nj=1
f(P ij)Sij .
We use the approximation Sij ~Ru ~Rvuv where~Ru =
xu ,
yu ,
zu
and ~Rv =
xv ,
yv ,
zv
If ~Ru and ~Rv are non-parallel and non-zero whose componentsare continuous on D, then
S
f(x, y, z) dS =
D
f(~R(u, v))~Ru ~RvdA
Math 55 Surface Integrals 4/ 16
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Parametric Surfaces
We define the surface integral of f over the surface S asS
f(x, y, z) dS = limm,n
mi=1
nj=1
f(P ij)Sij .
We use the approximation Sij ~Ru ~Rvuv where~Ru =
xu ,
yu ,
zu
and ~Rv =
xv ,
yv ,
zv
If ~Ru and ~Rv are non-parallel and non-zero whose componentsare continuous on D, then
S
f(x, y, z) dS =
D
f(~R(u, v))~Ru ~RvdA
Math 55 Surface Integrals 4/ 16
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Parametric Surfaces
Example
Evaluate the surface integral
S
x2 dS where S is the sphere
x2 + y2 + z2 = 1.
Solution. We use ~R(, ) = sin cos , sin sin , cos toparametrize the sphere. Hence,
S
x2 dS =
D
sin2 cos2 ~R ~RdA
=
2pi0
pi0
sin2 cos2 sind d
=
2pi0
cos2 d
pi0
sin3 d
=4pi
3
Math 55 Surface Integrals 5/ 16
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Parametric Surfaces
Example
Evaluate the surface integral
S
x2 dS where S is the sphere
x2 + y2 + z2 = 1.
Solution. We use ~R(, ) = sin cos , sin sin , cos toparametrize the sphere. Hence,
S
x2 dS =
D
sin2 cos2 ~R ~RdA
=
2pi0
pi0
sin2 cos2 sind d
=
2pi0
cos2 d
pi0
sin3 d
=4pi
3
Math 55 Surface Integrals 5/ 16
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Parametric Surfaces
Example
Evaluate the surface integral
S
x2 dS where S is the sphere
x2 + y2 + z2 = 1.
Solution. We use ~R(, ) = sin cos , sin sin , cos toparametrize the sphere. Hence,
S
x2 dS =
D
sin2 cos2 ~R ~RdA
=
2pi0
pi0
sin2 cos2 sind d
=
2pi0
cos2 d
pi0
sin3 d
=4pi
3
Math 55 Surface Integrals 5/ 16
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Parametric Surfaces
Example
Evaluate the surface integral
S
x2 dS where S is the sphere
x2 + y2 + z2 = 1.
Solution. We use ~R(, ) = sin cos , sin sin , cos toparametrize the sphere. Hence,
S
x2 dS =
D
sin2 cos2 ~R ~RdA
=
2pi0
pi0
sin2 cos2 sind d
=
2pi0
cos2 d
pi0
sin3 d
=4pi
3
Math 55 Surface Integrals 5/ 16
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Parametric Surfaces
Example
Evaluate the surface integral
S
x2 dS where S is the sphere
x2 + y2 + z2 = 1.
Solution. We use ~R(, ) = sin cos , sin sin , cos toparametrize the sphere. Hence,
S
x2 dS =
D
sin2 cos2 ~R ~RdA
=
2pi0
pi0
sin2 cos2 sind d
=
2pi0
cos2 d
pi0
sin3 d
=4pi
3
Math 55 Surface Integrals 5/ 16
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Graphs
Suppose the surface S has equation z = g(x, y).
This can beregarded as a parametric surface given by
~R(x, y) = x, y, g(x, y)
and so we have ~Rx =
1, 0, gx
and ~Ry =
0, 1, gy
.
Thus ~Rx ~Ry = gx ,gy , 1
and therefore,
S
f(x, y, z) dS =
D
f(x, y, g(x, y))
(g
x
)2+
(g
y
)2+ 1 dA
Similar formulas can be derived when the equation of the sufaceis in the form x = g(y, z) or y = g(x, z).
Math 55 Surface Integrals 6/ 16
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Graphs
Suppose the surface S has equation z = g(x, y). This can beregarded as a parametric surface given by
~R(x, y) = x, y, g(x, y)
and so we have ~Rx =
1, 0, gx
and ~Ry =
0, 1, gy
.
Thus ~Rx ~Ry = gx ,gy , 1
and therefore,
S
f(x, y, z) dS =
D
f(x, y, g(x, y))
(g
x
)2+
(g
y
)2+ 1 dA
Similar formulas can be derived when the equation of the sufaceis in the form x = g(y, z) or y = g(x, z).
Math 55 Surface Integrals 6/ 16
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Graphs
Suppose the surface S has equation z = g(x, y). This can beregarded as a parametric surface given by
~R(x, y) = x, y, g(x, y)
and so we have ~Rx =
1, 0, gx
and ~Ry =
0, 1, gy
.
Thus ~Rx ~Ry = gx ,gy , 1
and therefore,
S
f(x, y, z) dS =
D
f(x, y, g(x, y))
(g
x
)2+
(g
y
)2+ 1 dA
Similar formulas can be derived when the equation of the sufaceis in the form x = g(y, z) or y = g(x, z).
Math 55 Surface Integrals 6/ 16
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Graphs
Suppose the surface S has equation z = g(x, y). This can beregarded as a parametric surface given by
~R(x, y) = x, y, g(x, y)
and so we have ~Rx =
1, 0, gx
and ~Ry =
0, 1, gy
.
Thus ~Rx ~Ry = gx ,gy , 1
and therefore,
S
f(x, y, z) dS =
D
f(x, y, g(x, y))
(g
x
)2+
(g
y
)2+ 1 dA
Similar formulas can be derived when the equation of the sufaceis in the form x = g(y, z) or y = g(x, z).
Math 55 Surface Integrals 6/ 16
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Graphs
Suppose the surface S has equation z = g(x, y). This can beregarded as a parametric surface given by
~R(x, y) = x, y, g(x, y)
and so we have ~Rx =
1, 0, gx
and ~Ry =
0, 1, gy
.
Thus ~Rx ~Ry = gx ,gy , 1
and therefore,
S
f(x, y, z) dS =
D
f(x, y, g(x, y))
(g
x
)2+
(g
y
)2+ 1 dA
Similar formulas can be derived when the equation of the sufaceis in the form x = g(y, z) or y = g(x, z).
Math 55 Surface Integrals 6/ 16
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Graphs
Suppose the surface S has equation z = g(x, y). This can beregarded as a parametric surface given by
~R(x, y) = x, y, g(x, y)
and so we have ~Rx =
1, 0, gx
and ~Ry =
0, 1, gy
.
Thus ~Rx ~Ry = gx ,gy , 1
and therefore,
S
f(x, y, z) dS =
D
f(x, y, g(x, y))
(g
x
)2+
(g
y
)2+ 1 dA
Similar formulas can be derived when the equation of the sufaceis in the form x = g(y, z) or y = g(x, z).
Math 55 Surface Integrals 6/ 16
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Graphs
Example
Evaluate
S
yz dS where S is the part of the plane
2x+ y + z = 2 that lies in the first octant.
Solution. The projection of the surface onto the xy-plane is theregion
D = {(x, y) : 0 y 2 2x, 0 x 1} .Therefore,
S
yz dS =
D
y(2 2x y)
(2)2 + (1)2 + 1 dA
=
6
10
22x0
2y 2xy y2 dydx
=
6
3
Math 55 Surface Integrals 7/ 16
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Graphs
Example
Evaluate
S
yz dS where S is the part of the plane
2x+ y + z = 2 that lies in the first octant.
Solution. The projection of the surface onto the xy-plane is theregion
D = {(x, y) : 0 y 2 2x, 0 x 1} .
Therefore,S
yz dS =
D
y(2 2x y)
(2)2 + (1)2 + 1 dA
=
6
10
22x0
2y 2xy y2 dydx
=
6
3
Math 55 Surface Integrals 7/ 16
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Graphs
Example
Evaluate
S
yz dS where S is the part of the plane
2x+ y + z = 2 that lies in the first octant.
Solution. The projection of the surface onto the xy-plane is theregion
D = {(x, y) : 0 y 2 2x, 0 x 1} .Therefore,
S
yz dS =
D
y(2 2x y)
(2)2 + (1)2 + 1 dA
=
6
10
22x0
2y 2xy y2 dydx
=
6
3
Math 55 Surface Integrals 7/ 16
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Graphs
Example
Evaluate
S
yz dS where S is the part of the plane
2x+ y + z = 2 that lies in the first octant.
Solution. The projection of the surface onto the xy-plane is theregion
D = {(x, y) : 0 y 2 2x, 0 x 1} .Therefore,
S
yz dS =
D
y(2 2x y)
(2)2 + (1)2 + 1 dA
=
6
10
22x0
2y 2xy y2 dydx
=
6
3
Math 55 Surface Integrals 7/ 16
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Graphs
Example
Evaluate
S
yz dS where S is the part of the plane
2x+ y + z = 2 that lies in the first octant.
Solution. The projection of the surface onto the xy-plane is theregion
D = {(x, y) : 0 y 2 2x, 0 x 1} .Therefore,
S
yz dS =
D
y(2 2x y)
(2)2 + (1)2 + 1 dA
=
6
10
22x0
2y 2xy y2 dydx
=
6
3
Math 55 Surface Integrals 7/ 16
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Examples
1 Evaluate the surface integral
S
yz dS where S is the surface
with parametric equations x = u2, y = u sin v, z = u cos v,0 u 1 and 0 v pi2 .
2 Evaluate the surface integral
S
x2z2 dS where S is the part of
the cone z2 = x2 + y2 that lies between the planes z = 1 andz = 3.
3 Evaluate
S
z dS where S is the surface whose sides S1 are
given by the cylinder x2 + y2 = 1, whose bottom S2 is the diskx2 + y2 1 in the xy-plane, and whose top S3 is the part of theplane z = 1 + x that lies above S2.
4 Evaluate
S
z dS where S is the surface x = y + 2z2,
0 y, z 1.
Math 55 Surface Integrals 8/ 16
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Examples
1 Evaluate the surface integral
S
yz dS where S is the surface
with parametric equations x = u2, y = u sin v, z = u cos v,0 u 1 and 0 v pi2 .
2 Evaluate the surface integral
S
x2z2 dS where S is the part of
the cone z2 = x2 + y2 that lies between the planes z = 1 andz = 3.
3 Evaluate
S
z dS where S is the surface whose sides S1 are
given by the cylinder x2 + y2 = 1, whose bottom S2 is the diskx2 + y2 1 in the xy-plane, and whose top S3 is the part of theplane z = 1 + x that lies above S2.
4 Evaluate
S
z dS where S is the surface x = y + 2z2,
0 y, z 1.
Math 55 Surface Integrals 8/ 16
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Examples
1 Evaluate the surface integral
S
yz dS where S is the surface
with parametric equations x = u2, y = u sin v, z = u cos v,0 u 1 and 0 v pi2 .
2 Evaluate the surface integral
S
x2z2 dS where S is the part of
the cone z2 = x2 + y2 that lies between the planes z = 1 andz = 3.
3 Evaluate
S
z dS where S is the surface whose sides S1 are
given by the cylinder x2 + y2 = 1, whose bottom S2 is the diskx2 + y2 1 in the xy-plane, and whose top S3 is the part of theplane z = 1 + x that lies above S2.
4 Evaluate
S
z dS where S is the surface x = y + 2z2,
0 y, z 1.
Math 55 Surface Integrals 8/ 16
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Examples
1 Evaluate the surface integral
S
yz dS where S is the surface
with parametric equations x = u2, y = u sin v, z = u cos v,0 u 1 and 0 v pi2 .
2 Evaluate the surface integral
S
x2z2 dS where S is the part of
the cone z2 = x2 + y2 that lies between the planes z = 1 andz = 3.
3 Evaluate
S
z dS where S is the surface whose sides S1 are
given by the cylinder x2 + y2 = 1, whose bottom S2 is the diskx2 + y2 1 in the xy-plane, and whose top S3 is the part of theplane z = 1 + x that lies above S2.
4 Evaluate
S
z dS where S is the surface x = y + 2z2,
0 y, z 1.Math 55 Surface Integrals 8/ 16
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Oriented Surfaces
A Mobius strip only has one side and is not orientable.
For an orientable surface, there are two possible orientation:
Math 55 Surface Integrals 9/ 16
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Oriented Surfaces
A Mobius strip only has one side and is not orientable.
For an orientable surface, there are two possible orientation:
Math 55 Surface Integrals 9/ 16
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Oriented Surfaces
1 If a surface S is given by the vector function ~R(u, v), thenS is automatically supplied with the orientation of the unit
normal vector ~N =~Ru ~Rv~Ru ~Rv
.
2 If a surface S is given by the equation z = g(x, y), then Shas a natural (upward) orientation given by the unit
normal vector ~N =gx,gy, 1
1 + (gx)2 + (gy)2.
3 If S a closed surface (the boundary of a region E), thepositive orientation is the one for which the normal vectorspoint outward from E
Math 55 Surface Integrals 10/ 16
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Oriented Surfaces
1 If a surface S is given by the vector function ~R(u, v), thenS is automatically supplied with the orientation of the unit
normal vector ~N =~Ru ~Rv~Ru ~Rv
.
2 If a surface S is given by the equation z = g(x, y), then Shas a natural (upward) orientation given by the unit
normal vector ~N =gx,gy, 1
1 + (gx)2 + (gy)2.
3 If S a closed surface (the boundary of a region E), thepositive orientation is the one for which the normal vectorspoint outward from E
Math 55 Surface Integrals 10/ 16
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Oriented Surfaces
1 If a surface S is given by the vector function ~R(u, v), thenS is automatically supplied with the orientation of the unit
normal vector ~N =~Ru ~Rv~Ru ~Rv
.
2 If a surface S is given by the equation z = g(x, y), then Shas a natural (upward) orientation given by the unit
normal vector ~N =gx,gy, 1
1 + (gx)2 + (gy)2.
3 If S a closed surface (the boundary of a region E), thepositive orientation is the one for which the normal vectorspoint outward from E
Math 55 Surface Integrals 10/ 16
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Surface Integrals of Vector Fields
Suppose S is an orientable surface with normal vector ~N andfluid flows through S with density (x, y, z) and velocity~v(x, y, z). The flux or rate of flow (mass per unit time) per unitarea is ~v
If we divide S into smaller patches Sij (as before) then we can
approximate the flux along Sij in the direction of ~N as
(~v ~N)A(Sij).By increasing the number of subdivisions we get the surfaceintegral of the function ~v ~N over S, i.e.
S
~v ~N dS
Math 55 Surface Integrals 11/ 16
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Surface Integrals of Vector Fields
Suppose S is an orientable surface with normal vector ~N andfluid flows through S with density (x, y, z) and velocity~v(x, y, z). The flux or rate of flow (mass per unit time) per unitarea is ~v
If we divide S into smaller patches Sij (as before) then we can
approximate the flux along Sij in the direction of ~N as
(~v ~N)A(Sij).
By increasing the number of subdivisions we get the surfaceintegral of the function ~v ~N over S, i.e.
S
~v ~N dS
Math 55 Surface Integrals 11/ 16
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Surface Integrals of Vector Fields
Suppose S is an orientable surface with normal vector ~N andfluid flows through S with density (x, y, z) and velocity~v(x, y, z). The flux or rate of flow (mass per unit time) per unitarea is ~v
If we divide S into smaller patches Sij (as before) then we can
approximate the flux along Sij in the direction of ~N as
(~v ~N)A(Sij).By increasing the number of subdivisions we get the surfaceintegral of the function ~v ~N over S, i.e.
S
~v ~N dS
Math 55 Surface Integrals 11/ 16
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Surface Integrals of Vector Fields
By letting ~F = ~v, the previous integral can be written asS
~F ~N dS and thus the following definition.
Definition
If ~F is a continuous vector field defined on an oriented surface Swith unit normal vector ~N , then the surface integral of ~Fover S is
S
~F d~S =S
~F ~N dS.
This integral is also called the flux of ~F across S.
Math 55 Surface Integrals 12/ 16
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Surface Integrals of Vector Fields
If the surface S is given by ~R(u, v), then ~N =~Ru ~Rv~Ru ~Rv
.
Therefore,S
~F d~S =S
~F ~Ru ~Rv~Ru ~Rv
dS
=
D
(~F
~Ru ~Rv~Ru ~Rv
)~Ru ~Rv dA
=
D
~F (~R(u, v)) (~Ru ~Rv) dA
where D is the parameter domain.
Math 55 Surface Integrals 13/ 16
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Surface Integrals of Vector Fields
If the surface S is given by ~R(u, v), then ~N =~Ru ~Rv~Ru ~Rv
.
Therefore,S
~F d~S =S
~F ~Ru ~Rv~Ru ~Rv
dS
=
D
(~F
~Ru ~Rv~Ru ~Rv
)~Ru ~Rv dA
=
D
~F (~R(u, v)) (~Ru ~Rv) dA
where D is the parameter domain.
Math 55 Surface Integrals 13/ 16
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Surface Integrals of Vector Fields
If the surface S is given by ~R(u, v), then ~N =~Ru ~Rv~Ru ~Rv
.
Therefore,S
~F d~S =S
~F ~Ru ~Rv~Ru ~Rv
dS
=
D
(~F
~Ru ~Rv~Ru ~Rv
)~Ru ~Rv dA
=
D
~F (~R(u, v)) (~Ru ~Rv) dA
where D is the parameter domain.
Math 55 Surface Integrals 13/ 16
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Surface Integrals of Vector Fields
If the surface S is given by ~R(u, v), then ~N =~Ru ~Rv~Ru ~Rv
.
Therefore,S
~F d~S =S
~F ~Ru ~Rv~Ru ~Rv
dS
=
D
(~F
~Ru ~Rv~Ru ~Rv
)~Ru ~Rv dA
=
D
~F (~R(u, v)) (~Ru ~Rv) dA
where D is the parameter domain.
Math 55 Surface Integrals 13/ 16
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Surface Integrals of Vector Fields
Let ~F (x, y, z) = P (x, y, z), Q(x, y, z), R(x, y, z). If S has theequation z = g(x, y), we regard x and y as parameters.
Therefore,S
~F d~S =D
~F (~R(u, v)) (~Ru ~Rv) dA
=
D
P,Q,R gx,g
y, 1
dA
=
D
(P g
xQg
y+R
)dA
Similar formulas can be obtained if S is given by x = g(y, z) ory = g(x, z).
Math 55 Surface Integrals 14/ 16
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Surface Integrals of Vector Fields
Let ~F (x, y, z) = P (x, y, z), Q(x, y, z), R(x, y, z). If S has theequation z = g(x, y), we regard x and y as parameters.
Therefore,S
~F d~S =D
~F (~R(u, v)) (~Ru ~Rv) dA
=
D
P,Q,R gx,g
y, 1
dA
=
D
(P g
xQg
y+R
)dA
Similar formulas can be obtained if S is given by x = g(y, z) ory = g(x, z).
Math 55 Surface Integrals 14/ 16
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Surface Integrals of Vector Fields
Let ~F (x, y, z) = P (x, y, z), Q(x, y, z), R(x, y, z). If S has theequation z = g(x, y), we regard x and y as parameters.
Therefore,S
~F d~S =D
~F (~R(u, v)) (~Ru ~Rv) dA
=
D
P,Q,R gx,g
y, 1
dA
=
D
(P g
xQg
y+R
)dA
Similar formulas can be obtained if S is given by x = g(y, z) ory = g(x, z).
Math 55 Surface Integrals 14/ 16
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Surface Integrals of Vector Fields
Let ~F (x, y, z) = P (x, y, z), Q(x, y, z), R(x, y, z). If S has theequation z = g(x, y), we regard x and y as parameters.
Therefore,S
~F d~S =D
~F (~R(u, v)) (~Ru ~Rv) dA
=
D
P,Q,R gx,g
y, 1
dA
=
D
(P g
xQg
y+R
)dA
Similar formulas can be obtained if S is given by x = g(y, z) ory = g(x, z).
Math 55 Surface Integrals 14/ 16
-
Surface Integrals of Vector Fields
Let ~F (x, y, z) = P (x, y, z), Q(x, y, z), R(x, y, z). If S has theequation z = g(x, y), we regard x and y as parameters.
Therefore,S
~F d~S =D
~F (~R(u, v)) (~Ru ~Rv) dA
=
D
P,Q,R gx,g
y, 1
dA
=
D
(P g
xQg
y+R
)dA
Similar formulas can be obtained if S is given by x = g(y, z) ory = g(x, z).
Math 55 Surface Integrals 14/ 16
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Examples
1 Find the flux of the vector field ~F (x, y, z) = z, y, x acrossthe unit sphere x2 + y2 + z2 = 1
2 Evaluate
S
~F d~S where ~F (x, y, z) = y, x, z and S is the
boundary of the solid region E enclosed by the paraboloidz = 1 x2 y2 and the plane z = 0.
Math 55 Surface Integrals 15/ 16
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Examples
1 Find the flux of the vector field ~F (x, y, z) = z, y, x acrossthe unit sphere x2 + y2 + z2 = 1
2 Evaluate
S
~F d~S where ~F (x, y, z) = y, x, z and S is the
boundary of the solid region E enclosed by the paraboloidz = 1 x2 y2 and the plane z = 0.
Math 55 Surface Integrals 15/ 16
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References
1 Stewart, J., Calculus, Early Transcendentals, 6 ed., ThomsonBrooks/Cole, 2008
2 Dawkins, P., Calculus 3, online notes available athttp://tutorial.math.lamar.edu/
Math 55 Surface Integrals 16/ 16