## Stokes' Theorem

Stokes' Theorem relates line integrals of vector
fields to surface integrals of vector
fields.

Consider the surface S described by the parabaloid z=16-x^2-y^2
for z>=0, as shown in the figure below.

Let **n** denote the
unit normal vector to S with positive z component. The intersection
of S with the z plane is the circle x^2+y^2=16. Let C denote
this circle traversed in the counter clockwise direction.
If **F** is a vector field with continuous partial derivatives
in some region containing S, then Stokes' Theorem states

In general C is the boundary of S and is assumed to be piecewise
smooth.

For the above equality to hold the direction of the normal
vector **n** and the direction in which C is traversed must be
consistent. Suppose that **n** points in some direction and consider
a person walking on the curve C with their head pointing in the same
direction as **n**. For consistency C must be traversed in such a way
so that the surface
is always on the left.

**Example**

Verify Stokes' Theorem for the surface S described above and the
vector field **F**=<3y,4z,-6x>.

Let us first compute the line integral. The curve C can be
parameterized by the vector function
**r**(t)=<4cos(t),4sin(t),0>
and **r**'(t)=<-4sin(t),4cos(t),0> for 0<=t<=2*pi.
Recall
that

In this case, **F**=<12sin(t),0,-24cos(t)>. The line integral is
given
by

Now, let us compute the surface integral.
Recall that for
**F**=<P,Q,R>,

For our vector field, the curl is <-4,6,-3>. For z=g(x,y)=16-x^2-y^2
the normal vector to S pointing in
the positive z direction is given
by <-g_x,-g_y,1>=<2x,2y,1>. Using the formula for the surface
integral
of a vector field, we have

where R is the disk 0<=x^2+y^2<=16, the projection of S onto the xy
plane. The expression on the right is

a double integral.

It is convenient to convert to polar
coordinates to compute the
double integral. The region of integration is 0<=r<=4 and
0<=theta<=2*pi.
Recall, that x=rcos(theta) and y=rsin(theta). The
double integral
becomes

The inner integral is

The final result for the inner integral is

The outer integral is

This result is same as that for the line integral!

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