This explanation is trying to get out the presence (or lack) of a
scaling factor (e.g., the *r* in the polar coordinates area
element *dA* = *r* *dr* *dtheta*)
when we parameterize a surface to integrate over it. We'll use the
same surface as in the divergence theorem example
page. Which doesn't matter for the purposes of the calculation
here, of course.

**The Problem**: write the surface integral of a function
*f*(*x*,*y*,*z*) over the surface shown to the
right. It's given by
*z* = 4 -
*x*^{2} -
*y*^{2}, for
*z* >= 3.

**A First Parameterization**:
We can parameterize this by thinking about cylindrical coordinates.
The vector giving the surface is
**r** = <*x*, *y*, *z*> =
<*x*, *y*, 4 -
*x*^{2} -
*y*^{2}>.
In cylindrical coordinates, *x* = *r* cos(*theta*)
and *y* = *r* sin(*theta*), so the surface is
parameterized by

What are the limits on *r* and *theta*? We want
*z* >= 3, so
4 - *r*^{2} >= 3,
and *r* <= 1. We want to get the complete circle, so
0 <= *theta* <= 2*pi*.

Then we can find the surface area element
*dS* =
| **r**_{r} x
**r**_{theta}| *dr* *dtheta*
by calculating the partials of **r** and taking the cross product.
Check this on your own; the result should be

So *dS* =
sqrt( 4 *r*^{4}
(cos^{2}(*theta*) +
sin^{2}(*theta*)) +
*r*^{2} ) *dr* *dtheta*,
and we can write the surface integral as

Why isn't there an extra factor of *r*? *Because we didn't
convert from rectangular to cylindrical coordinates* -- here
the scaling factor is the factor of
sqrt( 4 *r*^{4} +
*r*^{2} ),
which serves to scale from the *r* *theta* world to the
world on the surface *S*. (In other words, it gives the amount
the little element *dr* *dtheta* is stretched as it gets
moved up onto the surface *S*.)

*Also note*: because we're interested in the value of the
function *f* *on the surface*, we evaluate it on **r**:
*f*(*r* cos(*theta*), *r* sin(*theta*),
4 - *r*^{2}).

**A Second Parameterization**: we can also formulate this
problem using a "parameterization" in *x* and *y*. Then we have

with -1 <= *x* <= 1 and -sqrt(1 -
*x*^{2}) <=
*y* <=
sqrt(1 - *x*^{2}).

Then we can calculate *dS* the same way we did above:
*dS* =
| **r**_{x} x
**r**_{y} | *dx* *dy*,
which comes out to (check that you get the same thing)
*dS* = sqrt(4 *x*^{2} +
4 *y*^{2} +
1) *dx* *dy*.

Then we can formulate our flux integral as before:

Here the scaling factor is
sqrt( 4 *x*^{2} +
4 *y*^{2} +
1)
which gives us the scaling to get from the *x* *y*
world (which is the projection of the surface onto the
*x* *y*-plane) to the world on the surface *S*.

Now, suppose that we converted this integral to *polar*
coordinates. Then *x* = *r* cos(*theta*) and
*y* = *r* sin(*theta*), and
*dy* *dx* = *r* *dtheta* *dr*,
so that the integral becomes

And the two formulations are in fact the same!

*Note what happened here*: the factor of *r* that shows up
when we convert from *dx* *dy* to
*dr* *dtheta* is the *scaling factor* for how
the area element stretches as we move between the
*x* *y* and *r* *theta* worlds. This is
analagous to the factors that represent the stretching as we move from
*x* *y* (or, more generally, *u* *v*) to
the world on the surface.

It is worth noting that in section 16.6 of the book, the surface area
element is derived in somewhat less generality. There, the surface
area element for a surface *z* = *f*(*x*,*y*)
is found to be

Let's compare this with what we did. If *z* =
*f*(*x*,*y*), then
**r**(*x*,*y*) =
<*x*, *y*, *f*(*x*,*y*)>, and

= | <-

= sqrt(

(Check that you can duplicate all of those steps!) Then in our case
*f*(*x*,*y*) =
4 - *x*^{2} -
*y*^{2}, so

which is the same as we got before.

In general we won't memorize this special case, and will instead
use the formulation in terms of the cross product of the partial
derivatives of **r**.

surface integrals and scaling factors

Last modified: Tue Dec 7 09:40:12 EST 2004

Comments to:glarose@umich.edu

©2004 Gavin LaRose, UM Math Dept.