Exercises-Surface Integral

Evaluate the surface integrals.

1. MATH, where $S$ is the portion of the cone MATHbetween the planes $z=1$ and MATH

2. MATH where $S$ is the portion of the cylinder $x^{2}+y^{2}$ between the planes $y=0,y=1$, and above the xy-plane. MATH

3. MATH where $S$ is the portion of the plane $x+y=1$ in the first octant between $z=0$ and MATH

4. Evaluate MATH over the surface of the cube defined by the inequalities MATH [Hint: Integrate over each face separately.] $[9]$

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

Find the mass of the given lamina assuming the density to be a constant $\delta _{0}$.

6. The lamina that is the portion of the paraboloid $z=1-x^{2}-y^{2}$ above the $xy$-plane. MATH

7. The lamina that is the portion of the circular cylinder $x^{2}+z^{2}=4$ that lies directly above the rectangle MATH in the xy-plane. MATH

8. Find the mass of the lamina that is the portion of the surface $y^{2}=4-z$ between the planes $x=0,x=3,y=0,$ and $y=3$ if the density is MATH

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

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

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

THE VECTOR DIFFERENTIAL OPERATOR $\nabla $

Divergence $\nabla \cdot v,$ Curl $\nabla \times v$

The vector differential operator $\nabla $ is defined by setting
MATH
This is no ordinary vector. Its ``components'' are differentiation symbols. As the term ``operator'' suggests, $\nabla $ is to be thought of as something that ``operates'' on things. What sort of things? Scalar fields and vector fields.

Suppose that $f$ is a differentiable scalar field. Then $\nabla $ operates on $f$ as follows:
MATH
This is just the gradient of $f$, with which we are already familiar.

How does $\nabla $ operate on vector fields? In two ways. If MATH is a differentiable vector field, then by definition
MATH
and
MATH
The first ``product,'' $\nabla \cdot v$, defined in imitation of the ordinary dot product, is called the divergence of $v$:
MATH
The second ``product,'' $\nabla \times v$, defined in imitation of the ordinary cross product, is called the curl of $v$:
MATH

Interpretation of Divergence and Curl

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 $v$ as the velocity field of some fluid. The divergence of $v $ at a point $P$ gives us an indication of whether the fluid tends to accumulate near $P$ (negative divergence) or tends to move away from $P$ (positive divergence). The curl at $P$ measures the rotational tendency of the fluid.

Example 1. Set
MATH
Divergence:
MATH
Curl:
MATH
because the partial derivatives that appear in the expanded determinant are all zero.

The field of Example 1
MATH
can be viewed as the velocity field of a fluid in radial motion--toward the origin if $\alpha <0$, away from the origin if $\alpha >0$. Consider a point $(x,y,z),$ 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 $3\alpha $ is negative, then $\alpha $ 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 $3\alpha $ is positive, then $\alpha $ 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
MATH
Divergence:
MATH
Curl:
MATH
The field of Example 2
MATH
is the velocity field of uniform counterclockwise rotation about the $z$-axis with angular speed $\omega $. We can see this by noting that $v$ is perpendicular to $r:$
MATH
and the speed at each point is $\omega R$ where $R$ is the radius of rotation:
MATH
How is the curl, $2\omega k$, related to the rotation? The angular velocity vector is the vector $\omega k$. In this case then the curl of $v$ 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.

Basic Identities

For vectors we have $a\times a=0.$ Is it true that MATH? We make it true by defining
MATH

THEOREM (THE CURL OF A GRADIENT IS ZERO)

If $f$ is a scalar field with continuous second partials, then
MATH

Proof
MATH
by the equality of the mixed partials.

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

THEOREM (THE DIVERGENCE OF A CURL IS ZERO)

If the components of the vector field $v=v_1i+v_2j+v_3k$ have continuous second partials, then
MATH

Proof. Again the key is the equality of the mixed partials:
MATH
since for each component the mixed partials cancel.

The next two identities are product rules. Here $f$ is a scalar field and $v$ is a vector field.
MATH

MATH

We know from Example 1 that $\nabla \cdot r=3$ and $\nabla \times r=0$ at all points of space. Now we can show that, if $n$ is an integer, then, for all $\QTR{bf}{r\ne 0},$
MATH

Proof. Recall that MATH We have
MATH
Since MATHit follows that
MATH

The Laplacean

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

Example 3. If MATH then
MATH

MATH

Example 4. If $f(x,y,z)=e^{xyz},$ then
MATH

Example 5. To calculate MATH we could write
MATH
and proceed from there. The calculations are straightforward but lengthy. We will do it in a different way.

Recall that
MATH

MATH
Using these relations, we have
MATH

MATH

THE DIVERGENCE THEOREM

Green's theorem enables us to express a double integral over a Jordan region $\Omega $ with a piecewise-smooth boundary $C$ as a line integral over $C:$
MATH
In vector terms this relation can be written
MATH
Here $n$ is the outer unit normal and the integral on the right is taken with respect to arc length.

Proof. Set $v=Qi-Pj.$ Then
MATH
All we have to show then is that
MATH
For $C$ traversed counterclockwise, $n=T\times k,$ where $T$ is the unit tangent vector. Thus
MATH
Since MATH we have MATH. Therefore
MATH

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

THEOREM THE DIVERGENCE THEOREM

Let $T$ be a solid bounded by a closed surface $S$ which, if not smooth, is piecewise smooth. If the vector field $v=v(x,y,z)$ is continuously differentiable throughout $T$, then
MATH
where $n$ is the outer unit normal.

Proof. We will carry out the proof under the assumption that $S$ is smooth and that any line parallel to a coordinate axis intersects $S$ at most twice. Our first step is to express the outer unit normal $n$ in terms of its direction cosines:
MATH
Then
MATH
The idea of the proof is to show that
MATH
All three equations can be verified in much the same manner. We will carry out the details only for the third equation.

Let $\Omega _{xy}$ be the projection of $T$ onto the $xy$-plane. If MATH then, by assumption, the vertical line through $(x,y)$ intersects $S$ in at most two points, an upper point $P^{+}$ and a lower point $P^{-}$. (If the vertical line intersects $S$ at only one point $P$, we set $P=P^{+}=P^{-}$.) As $(x,y)$ ranges over $\Omega _{xy}$, the upper point $P^{+}$ describes a surface
MATH
and the lower point describes a surface
MATH
By our assumptions,$f^{+}$ and$f^{-}$ are continuously differentiable, $S=S^{+}\cup S^{-}$, and the solid $T$ is the set of all points $(x,y,z)$ with
MATH
Now let $\gamma $ be the angle between the positive $z$-axis and the upper unit normal. On $S^{+}$ the outer unit normal $n$ is the upper unit normal. Thus on $S^{+}$
MATH
On $S^{-}$ the outer unit normal $n$ is the lower unit normal. In this case
MATH
Thus,
MATH
and
MATH
It follows that
MATH

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

Divergence as Outward Flux per Unit Volume

Choose a point $P$ and surround it by a closed ball $N_\epsilon $, of radius MATH According to the divergence theorem
MATH
Thus

(average divergence of $v$ on $N_\epsilon $) $\times $ (volume of $N_\epsilon $) = flux of $v$ out of $N_\epsilon $

and

average divergence of $v$ on $N_\epsilon =$ MATH

Taking the limit of both sides as $\epsilon $ shrinks to $0,$ we have
MATH

In this sense divergence is outward flux per unit volume.

Think of $v$ as the velocity of a fluid. Negative divergence at $P$ signals an accumulation of fluid near $P:$

$\nabla \cdot v<0$ at P$\Rightarrow $flux out of $N_\epsilon $ $<0\Rightarrow $net flow into $N_\epsilon $.

Positive divergence at $P$ signals a flow of liquid away from $P:$

$\nabla \cdot v>0$ at P$\Rightarrow $flux out of $N_\epsilon $ $>0\Rightarrow $net flow out of $N_\epsilon $.

Points at which the divergence is negative are called sinks; points at which the divergence is positive are called sources. If the divergence of $v$ is $0$ throughout, then the flow has no sinks and no sources and $v$ is called solenoidal.

Solids Bounded by Two or More Closed Surfaces

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 $S_{\QTR{group}{1}}$ and extract from the interior of that solid a solid bounded by a closed surface $S_{2}$. The remaining solid, call it $T$ has a boundary $S$ that consists of two pieces: an outer piece $S_{1}$ and an inner piece $S_{2} $. The key here is to note that the outer normal for $T$ points out of $S_{\QTR{group}{1}}$ but into $S_{2}$. The divergence theorem can be proven for $T$ by slicing $T$ into two pieces $T_{\QTR{group}{1}}$ and $T_{2}$ and applying the divergence theorem to each piece:
MATH
The triple integrals over $T_{\QTR{group}{1}}$ and $T_{2}$ add up to the triple integral over $T$. 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 $S_{\QTR{group}{1}}$ and $S_{2}$ remain. Thus the surface integrals add up to the surface integral over MATH and the divergence theorem still holds:
MATH

An Application to Static Charges

Consider a point charge $q$ somewhere in space. This charge creates around itself an electric field $\QTR{bf}{E}$, which in turn exerts an electric force on every other nearby charge. If we center our coordinate system at $q$, then the electric field at $\QTR{bf}{r}$ can be written
MATH
(This is found experimentally.) Note that this field has exactly the same form as the gravitational field: a constant multiple of $r^{-3}\QTR{bf}{r}$. It follows from the formula
MATH
that
MATH
We are interested in the flux of $\QTR{bf}{E}$ out of a closed surface $S$. We assume that $S$ does not pass through $q$.

If the charge $q$ is outside of $S$, then $\QTR{bf}{E}$ is continuously differentiable on the region $T$ bounded by $S$, and, by the divergence theorem,

flux of $\QTR{bf}{E}$ out of $S$ =MATH

If $q$ is inside of $S$, then the divergence theorem does not apply to $T$ directly because $\QTR{bf}{E}$ is not differentiable on all of $T$. We can circumvent this difficulty by surrounding $q$ by a small sphere $S_a$ of radius $a$ and applying the divergence theorem to the region $T^{\prime }$ bounded on the outside by $S$ and on the inside by $S_a$.

Since $\QTR{bf}{E}$ is continuously differentiable on $T^{\prime }$,
MATH

MATH
Since MATH on $T^{\prime }$, the triple integral on the left is zero and therefore

flux of $\QTR{bf}{E}$ out of $S$ = flux of $\QTR{bf}{E}$ out of $S_a$.

The quantity on the right is easy to calculate: on $S_a$, MATH and therefore
MATH
Thus

flux of $\QTR{bf}{E}$ out of MATHarea of $S_a)=4\pi q.$

It follows that the flux of E out of $S=4\pi q$.

In summary, $\QTR{bf}{E}$ is the electric field of a point charge $q$ and $S$ is a closed surface that does not pass through $q$, then

flux of $\QTR{bf}{E}$ out of $S$ MATH

Exercise-Divergence Theorem

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

1.MATH

2.MATH

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

3.MATH

4.MATH

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

5.MATH

Calculate the total flux of MATH out of the ball MATH

Use the Divergence Theorem to evaluate MATH where $n$ is the outer unit normal to $S.$

6. MATH $S$ is the surface of the cube bounded by the coordinate planes and the planes $x=1,y=1,$ and MATH

7. MATH where $S$ is the sphere MATH

8. MATH $S$ is the surface of the cylindrical solid bounded by MATH and MATH

9. MATH is the surface of the cylindrical solid bounded by $x^{2}+y^{2}=4,z=0,$ and $z=3$. $[108\pi ]$

10.MATH is the surface of the tetrahedron in the first octant bounded by $x+y+z=1$ and the coordinate planes. $[\frac{1}{24}]$

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

12.MATH is the surface of the solid bounded by MATH and MATH

13. Suppose that $f$ is a differentiable function of one variable and $v(x,y,z)=f(x)i.$ Determine $\nabla \cdot v$ and MATH

14. Show that, if MATH and $g(r)$ is twice differentiable, then MATH

Evaluate MATH.

15. MATH is the portion of the

surface $z=1-x^{2}-y^{2}$ above the xy-plane, oriented by upward normals. $[2\pi ]$

16. $F(x,y,z)=z^{2}k;S$ is the upper hemisphere given by MATH oriented by upward unit normals. $[\pi /2]$

17. MATH is the upper hemisphere MATH, oriented by upward unit normals. $[54\pi ]$

18. MATH is the portion of the cone $z^{2}=x^{2}+y^{2}$ between the planes $z=1$ and $z=2,$ oriented by upward unit normals. $[\frac{14}{3}\pi ]$

19. $F(x,y,z)=xk;S$ is the portion of the parabolold $z=x^{2}+y^{2}$ below the plane $z=y,$ oriented by downward unit normals. $[0]$

20. $F(x,y,z)=xi+yj+zk,$ where $S$ is the sphere MATH oriented by outward unit normals. $[4\pi a^{3}]$

Exercise-Stokes's Theorem

Use Stokes' Theorem to evaluate the integral $\int_{C}F\cdot dr.$

1. MATH is the intersection of the paraboloid $z=x^{2}+y^{2}$ and the plane $z=y$ with a counterclockwise orientation looking down the positive z-axis. $[0]$

2. MATH is the circle $x^{2}+y^{2}=a^{2}$ in the xy-plane with counterclockwise orientation looking down the positive z-axis. $[\pi a^{2}]$

3. Use Stokes' Theorem to evaluate MATH over the circle $x^{2}+y^{2}=1$ in the xy-plane traversed counterclockwise looking down the positive z-axis. \lbrack Hint: Find a surface whose boundary is $C$. MATH

This document created by Scientific WorkPlace 4.0.