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From: JEMebius on 8 Aug 2010 16:35
Michael Robinson wrote:
> "Michael Robinson" <nospam(a)billburg.com> wrote in message
> news:zyy7o.50983$f_3.12262(a)newsfe17.iad...
>> I have this surface integral I can't get to come out right.
>> I posted it in a place where I can use math characters. Here's a link:
>> http://burlington.craigslist.org/forums/?forumID=21
>> if you have time to look at it. Otherwise I will ask my math teacher
>> tomorrow.
>> It is a problem from a book, but it's not homework. I'm practicing.
> That link only takes you to the forum. The post is titled
> "surface integral," posted early morning 08/08.
>
>

Let me assume you are practicing for obtaining the outcome with the least effort, rather
than practicing for coordinate transformations, Jacobians and all that.

Then there is a approach very much in the vein of Archimedes and of the 18th-century
infinitesimal calculus. Only one-dimensional integration is involved.

First of all, observe that Z = x^2 + y^2 is a surface of revolution, and that the part
inside the cylinder of radius 2 (x^2 + y^2 = R^2 = 4) is a surface of revolution too, i.e.
the upper rim is flat and is perpendicular to the Z-axis, the axis of revolution.
In short: revolve the parabolic arc {z = x^2 | 0 <= x <= 2} around the Z-axis and obtain
the surface to be calculated. Let we denote this surface by PS (parabolic Surface).
Anyhow, I guess this is the problem at hand.

Now think of PS as being subdivided into narrow circular strips by cylinders concentric to
the outer cylinder x^2 + y^2 = 4. Consider such a strip in-between cylinders with radii r
and r+dr. One can also view this strip as lying in-between horizontal planes at heights
z = r^2 and z+dz = (r+dr)^2 = r^2 + 2.r.dr + (dr)^2.
A first approximation to its area is dS = circumference x width. The circumference is
2pi.r, the width is sqrt(dr^2 + dz^2) = sqrt(1 + 4r^2).dr, so its area is

dS = 2pi.r.sqrt(1 + 4r^2).dr,

save for second- and higher-order quantities.
At the bottom one has a small almost flat parabolic cap rather than a strip. Actually it
is a transformed copy of the original surface, so we seem to have a circular problem at
hand. But fortunately, its area is O(dr^2), so it disappears in the usual process of
transition from sum to integral.

We have to calculate

Integral from 0 to 2 of 2pi.r.sqrt(1 + 4r^2).dr,

so we need the primitive function F(r) of f(r) = 2pi.r.sqrt(1 + 4r^2).

By means of the substitution u = 4r^2 and the usual algebraic calculations we find the
primitive function

F(r) = 2pi.sqrt((1 + 4r^2)^3)/12.

Finally, PS has the area F(2) - F(0) = 2pi.(sqrt(17^3) - 1)/12 = 36.176903...

Curious fact: area calculations on the paraboloid involve pi and algebraic functions; no
logarithms or whatever transcendental functions; just the single transcendental number pi.

This problem was for me a nice recall of my schoolboy's and freshman's years!
I checked and double-checked this stuff (BTW: without using Maple or similar) and I do
hope that not mistakes crept in.

Happy practicing and good luck: Johan E. Mebius
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