Section 17.7

- Surface Integrals

THE SURFACE INTEGRAL

Suppose f is a function of three variables

whose domain includes the surface S. We divide S

into patches Sij with area ?Sij. We evaluate f

at a point in each patch, multiply it by

the area ?Sij, and form the sum Then we take the

limit as the patch size approaches 0 and define

the surface integral of f over the surface S as

Suppose the surface S is a graph of a function of

two variables g(x, y), where (x, y) are in some

region D. Divide D into smaller rectangles Rij

of equal size. The patch Sij lies directly above

the rectangle Rij and the point in Sij is of

the form . We use the

approximation If f is continuous on S and g

has continuous derivatives, then the definition

on the previous slide becomes.

SUMMARY

If z g(x, y), then

NOTE Similar formulas apply when it is more

convenient to project S onto the yz-plane or

xz-plane.

EXAMPLES

1. Evaluate the surface integral where S is the

first octant portion of the plane 2x y 2z

6. 2. Evaluate the surface integral where S is

the first octant portion of the cylinder y2 z2

9 between x 0 and x 4.

MASS AND CENTER OF MASS

Suppose a thin sheet has the shape of the surface

S and that that ?(x, y, z) is the density

function for the sheet. Then the mass of the

sheet is Its center of mass is

EXAMPLE

A cone-shaped surface lamina S is given by At

each point on S, the density is proportional to

the distance between the point and the z-axis.

Find the mass m of the lamina.

ORIENTED SURFACES

We only want to consider two-sided, orientable

surfaces, We start with a surface that has

tangent planes at every point (x, y, z) on S.

There are two unit normal vectors n1 and n2

-n1. If it is possible to choose a unit normal

vector n at every point (x, y, z) so that n

varies continuously over S, then S is called an

oriented surface and the given choice of n

provides S with an orientation. Thus, there are

two possible orientations for any orientable

surface which correspond to the choice for normal

vectors.

For a surface z g(x, y) given as the graph of

g, we see that the induced orientation is given

by the unit normal vector Since the

k-component is positive, this gives the upward

orientation of the surface.

CLOSED SURFACES AND ORIENTATION

- A closed surface is a surface the is the boundary

of a (closed) solid region E. - The positive orientation of a closed surface is

the one for which the normal vectors point

outward from the region E.

FLUX INTEGRAL

Suppose that S is an oriented surface with unit

normal vector n, and imagine a fluid with density

?(x, y, z) and velocity field v(x, y, z) flowing

through S. Then the rate of flow (mass per unit

time) per unit area is ?v. If we divide S into

small patches Sij, then Sij is nearly planar and

so we can approximate the mass of fluid crossing

Sij in the direction of the normal n per unit

time by the quantity (?v n)A(Sij) where ?, v,

and n are evaluated at some point on Sij. By

summing these quantities and taking the limit we

get the surface integral of the function ?v n

over S and this is interpreted physically as

the rate of flow through S.

FLUX INTEGRAL (CONTINUED)

If we write F ?v, then F is also a vector

field on and the integral on the previous slide

becomes A surface integral of this form occurs

frequently in physics, even when F is not ?v, and

is called the surface integral (or flux integral)

of F over S.

SURFACE INTEGRAL OF A VECTOR FIELD

If F is a continuous vector field defined on an

oriented surface S with unit normal vector n,

then the surface integral of F over S is This

integral is also called the flux of F across S.

In words, this definition says that the surface

integral of a vector field over S is equal to the

surface integral of its normal component over S.

EXAMPLE

Let S be the portion of paraboloid z g (x, y)

4 - x2 - y2 lying above the xy-plane. Find the

upward flux of F(x, y, z) xi yj zk across

the surface S.

In the case of a surface S given by a graph z

g(x, y), we can find n by noting that S is also

the level surface f (x, y, z) z - g(x, y) 0.

We know that the gradient is

normal to this surface at (x, y, z) and so a unit

normal vector is Using the formula for the

surface integral from Slide 4 to evaluate the

flux integral with F(x, y, z) P(x, y, z)i

Q(x, y, z)j R(x, y, z)k we get the following

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EXAMPLE

Evaluate the flux for the vector field F(x, y, z)

xi yj zk across S which is the first

octant portion of the sphere x2 y2 z2 16,

taking n to be the upward normal.