Surface Integrals

Math 55 - Elementray Analysis III

Institute of MathematicsUniversity of the Philippines

Diliman

Math 55 Surface Integrals 1/ 16

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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