Evaluate the surface integrals.

1. , where is the portion of the cone between the planes and

2. where is the portion of the cylinder between the planes , and above the xy-plane.

3. where is the portion of the plane in the first octant between and

4. Evaluate over the surface of the cube defined by the inequalities [Hint: Integrate over each face separately.]

5. Evaluate over the portion of the cone below the plane

Find the mass of the given lamina assuming the density to be a constant .

6. The lamina that is the portion of the paraboloid above the -plane.

7. The lamina that is the portion of the circular cylinder that lies directly above the rectangle in the xy-plane.

8. Find the mass of the lamina that is the portion of the surface between the planes and if the density is

9. Find the centroid of the lamina that is the portion of the paraboloid below the plane

10. Evaluate the surface integral where is the portion of the cone for which

11. Evaluate the surface integral , where is the portion of the paraboloid for which

The vector differential operator
is defined by setting

This is no ordinary vector. Its ``components'' are differentiation symbols. As
the term ``operator'' suggests,
is to be thought of as something that ``operates'' on things. What sort of
things? Scalar fields and vector fields.

Suppose that
is a differentiable scalar field. Then
operates on
as follows:

This is just the *gradient* of
,
with which we are already familiar.

How does
operate on vector fields? In two ways. If
is a differentiable vector field, then by definition

and

The first ``product,''
,
defined in imitation of the ordinary dot product, is called the
**divergence** of
:

The second ``product,''
,
defined in imitation of the ordinary cross product, is called the
**curl** of
:

Suppose we know the divergence of a field and also the curl. What does that tell us? For definitive answers we must wait for the divergence theorem and Stokes's theorem, but, in a preliminary way, we can give some rough answers right now. View as the velocity field of some fluid. The divergence of at a point gives us an indication of whether the fluid tends to accumulate near (negative divergence) or tends to move away from (positive divergence). The curl at measures the rotational tendency of the fluid.

**Example 1.** Set

Divergence:

Curl:

because the partial derivatives that appear in the expanded determinant are
all zero.

The field of Example 1

can be viewed as the velocity field of a fluid in radial motion--toward the
origin if
,
away from the origin if
.
Consider a point
a spherical neighborhood of that point, and a cone emanating from the origin
that is tangent to the boundary of the neighborhood.

Note two things: all the fluid in the cone stays in the cone, and the speed of the fluid is proportional to its distance from the origin. Therefore, if the divergence is negative, then is negative, the motion is toward the origin, and the neighborhood gains fluid because the fluid coming in is moving more quickly than the fluid going out. (Also the entry area is greater than the exit area.) If, however, the divergence is positive, then is positive, the motion is away from the origin, and the neighborhood loses fluid because the fluid coming in is moving more slowly than the fluid going out. (Also the entry area is smaller than the exit area.)

Since the motion is radial, the fluid has no rotational tendency whatsoever, and we would expect the curl to be identically zero. It is.

**Example 2.** Set

Divergence:

Curl:

The field of Example 2

is
the velocity field of uniform counterclockwise rotation about the
-axis
with angular speed
.
We can see this by noting that
is perpendicular to

and the speed at each point is
where
is the radius of rotation:

How is the curl,
,
related to the rotation? The angular velocity vector is the vector
.
In this case then the curl of
is twice the angular velocity vector.

With this rotation no neighborhood gains any fluid and no neighborhood loses any fluid. As we saw, the divergence is identically zero.

For vectors we have
Is it true that
?
We make it true by defining

**THEOREM** (THE CURL OF A GRADIENT IS ZERO)

If
is a scalar field with continuous second partials, then

Proof

by the equality of the mixed partials.

For vectors we have The analogous operator formula,, is also valid.

**THEOREM** (THE DIVERGENCE OF A CURL IS ZERO)

If the components of the vector field
have continuous second partials, then

Proof. Again the key is the equality of the mixed partials:

since for each component the mixed partials cancel.

The next two identities are product rules. Here
is a scalar field and
is a vector field.

We know from Example 1 that
and
at all points of space. Now we can show that, if
is an integer, then, for all

Proof. Recall that
We have

Since
it
follows that

From the operator
we can construct other operators. The most important of these is the Laplacean
The Laplacean (named after the French mathematician Pierre- Simon Laplace)
operates on scalar fields according to the following rule:

**Example 3.** If
then

**Example 4.** If
then

**Example 5.** To calculate
we could write

and proceed from there. The calculations are straightforward but lengthy. We
will do it in a different way.

Recall that

Using these relations, we have

Green's theorem enables us to express a double integral over a Jordan region
with a piecewise-smooth boundary
as a line integral over

In vector terms this relation can be written

Here
is the outer unit normal and the integral on the right is taken with respect
to arc length.

Proof. Set
Then

All we have to show then is that

For
traversed counterclockwise,
where
is the unit tangent vector. Thus

Since
we have
.
Therefore

Green's theorem expressed has a higher dimensional analog that is known as the
**divergence theorem**.

**THEOREM** THE DIVERGENCE THEOREM

Let
be a solid bounded by a closed surface
which, if not smooth, is piecewise smooth. If the vector field
is continuously differentiable throughout
,
then

where
is the outer unit normal.

Proof. We will carry out the proof under the assumption that
is smooth and that any line parallel to a coordinate axis intersects
at most twice. Our first step is to express the outer unit normal
in terms of its direction cosines:

Then

The idea of the proof is to show that

All three equations can be verified in much the same manner. We will carry out
the details only for the third equation.

Let
be the projection of
onto the
-plane.
If
then, by assumption, the vertical line through
intersects
in at most two points, an upper point
and a lower point
.
(If the vertical line intersects
at only one point
,
we set
.)
As
ranges over
,
the upper point
describes a surface

and the lower point describes a surface

By our
assumptions,
and
are continuously differentiable,
,
and the solid
is the set of all points
with

Now let
be the angle between the positive
-axis
and the upper unit normal. On
the outer unit normal
is the upper unit normal. Thus on

On
the outer unit normal
is the lower unit normal. In this case

Thus,

and

It follows that

This confirms the third equation. The second equation can be confirmed by
projection onto the
-plane;
the first equation can be confirmed by projection onto the
-plane.

Choose a point
and surround it by a closed ball
,
of radius
According to the divergence theorem

Thus

(average divergence of on ) (volume of ) = flux of out of

and

average divergence of on

Taking the limit of both sides as
shrinks to
we have

In this sense **divergence is outward flux per unit volume.**

Think of as the velocity of a fluid. Negative divergence at signals an accumulation of fluid near

at Pflux out of net flow into .

Positive divergence at signals a flow of liquid away from

at Pflux out of net flow out of .

Points at which the divergence is negative are called **sinks;**
points at which the divergence is positive are called
**sources.** If the divergence of
is
throughout, then the flow has no sinks and no sources and
is called **solenoidal.**

The divergence theorem, stated for solids bounded by a single closed surface,
can be extended to solids bounded by several closed surfaces. Suppose, for
example, that we start with a solid bounded by a closed surface
and extract from the interior of that solid a solid bounded by a closed
surface
.
The remaining solid, call it
has a boundary
that consists of two pieces: an outer piece
and an inner piece
.
The key here is to note that the outer normal for
points out of
but into
.
The divergence theorem can be proven for
by slicing
into two pieces
and
and applying the divergence theorem to each piece:

The triple integrals over
and
add up to the triple integral over
.
When the surface integrals are added together, the integrals along the common
cut cancel (because the normals are in opposite directions) and therefore only
the integrals over
and
remain. Thus the surface integrals add up to the surface integral over
and the divergence theorem still holds:

Consider a point charge
somewhere in space. This charge creates around itself an electric field
,
which in turn exerts an electric force on every other nearby charge. If we
center our coordinate system at
,
then the electric field at
can be written

(This is found experimentally.) Note that this field has exactly the same form
as the gravitational field: a constant multiple of
.
It follows from the formula

that

We are interested in the flux of
out of a closed surface
.
We assume that
does not pass through
.

If the charge is outside of , then is continuously differentiable on the region bounded by , and, by the divergence theorem,

flux of out of =

If is inside of , then the divergence theorem does not apply to directly because is not differentiable on all of . We can circumvent this difficulty by surrounding by a small sphere of radius and applying the divergence theorem to the region bounded on the outside by and on the inside by .

Since
is continuously differentiable on
,

Since
on
,
the triple integral on the left is zero and therefore

flux of out of = flux of out of .

The quantity on the right is easy to calculate: on
,
and therefore

Thus

flux of out of area of

It follows that the flux of E out of .

In summary, is the electric field of a point charge and is a closed surface that does not pass through , then

flux of out of

Calculate the flux out of the unit ball by applying the divergence theorem.

1.

2.

Calculate the flux out of the unit cube for the following vector fields.

3.

4.

Use the divergence theorem to find the total flux out of the given solid.

5.

Calculate the total flux of out of the ball

Use the Divergence Theorem to evaluate where is the outer unit normal to

6. is the surface of the cube bounded by the coordinate planes and the planes and

7. where is the sphere

8. is the surface of the cylindrical solid bounded by and

9. is the surface of the cylindrical solid bounded by and .

10. is the surface of the tetrahedron in the first octant bounded by and the coordinate planes.

11. is the surface of the conical solid bounded by and

12. is the surface of the solid bounded by and

13. Suppose that is a differentiable function of one variable and Determine and

14. Show that, if and is twice differentiable, then

Evaluate .

15. is the portion of the

surface above the xy-plane, oriented by upward normals.

16. is the upper hemisphere given by oriented by upward unit normals.

17. is the upper hemisphere , oriented by upward unit normals.

18. is the portion of the cone between the planes and oriented by upward unit normals.

19. is the portion of the parabolold below the plane oriented by downward unit normals.

20. where is the sphere oriented by outward unit normals.

Use Stokes' Theorem to evaluate the integral

1. is the intersection of the paraboloid and the plane with a counterclockwise orientation looking down the positive z-axis.

2. is the circle in the xy-plane with counterclockwise orientation looking down the positive z-axis.

3. Use Stokes' Theorem to evaluate over the circle in the xy-plane traversed counterclockwise looking down the positive z-axis. \lbrack Hint: Find a surface whose boundary is .

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