This problem will be a challenge to describe without diagrams, but I'll
try.
Consider a sphere. Imagine its surface described as the earth's with
latitude and longitude, equator and poles. This (I hope) will aid the
description.
As is often seen in physics and math books, one can find the surface area
and volume of geometric solids by strategically subdividing them into
infinitesimal pieces and adding up the pieces with an integration.
The area of a triangle may be found by a method if narrow strips. It turns
out to be one half the base times the height for any shape of triangle.
The area of a circle, likewise, but it's more trouble. The area of the
circle can be more easily be found by splitting it into sectors of vertex
angle d-theta. In the limit, these are are skinny triangles of base dL and
height R, where dL is the arc length subtended and R is the circle radius.
Add them up, get 2x(pi)xRxR/2 = (pi)R^2. So far, so good.
The volume of a sphere can be found by splitting it into slaps parallel to
latitudes, of thickness dh and area (pi)r^2 where r is the radius of the
slab at that latitude. Add them up; result correct.
The area of that sphere can be found by defining ribbons around a latitude
of infinitesimal thickness. The thickness is Rx(d-phi) where phi is the
polar angle of that latitude. A common error here is to use dy as the
thickness where y is measured in the direction of the polar axis. It
matters!
Now suppose we get innovative. Drunk with our success in the case of the
circle, using skinny triangles, we try triangles here. Imagine a triangle
in one hemisphere, with one vertex at the pole and the opposite side along
the equator. The length of the small base of the triangle can be called
dL. Its sides are along two longitude lines, its base is a small arc along
the equator. Its height is c/4 where c is the circumference of a great
circle. Granted it's a segment of a sector, not a triangle, but in the
limit as dL goes to zero, it approaches a triangle with two right angles
and the angle at the pole is zero, and can be peeled away from the sphere
so it's flat. So, in the limit, its area can be taken as one-half base
times height. So add them up around the hemisphere and then double to get
the whole sphere area, which comes out to be pi^2 times r^2. That isn't
correct, of course. It's too large.
The volume of a cone can be found to be 1/3 base times height, by the
method of slabs. No problem.
The volume of a sphere can be found by splitting it into small cones with
vertices at the center, and base area dA, and volume (1/3)RdA. Add them up
and you get (1/33)R(4)(pi)(R^2) = (4/3)(pi)(R^3). Correct, and you can do
it in your head.
Now I'm sure you all have spotted the flaw in the one which gave the wrong
answer for the area of the sphere. My students didn't however. Now here's
the question I pose to you. From a student's point of view this 'failure'
to get the right answer makes all such subdivision techniques suspect. How
does the student know *why* the ribbon method is ok, and the 'triangle'
method is ok in the case of area of a circle, but *not* ok in the case of
the sphere area? What is the fundamental difference? Be specific. What
criterion can one use to guarantee that the subdivision is valid? Or
should *all* such methods be abandoned as flawed, and the double or triple
integral methods be used instead? This strikes at the heart of the
student's understanding of fundamental calculus.
Does the difficulty lie in some change in the finite small geometric
entity as one shrinks it to an infinitesimal entity? That spherical
triangle on the sphere seems intuitively to shrink to a perfect triangle,
which can be peeled away from the sphere to form a flat triangle. But does
it really approach triangularity in the limit? How can we tell? How can we
explain it to students?
I have one interesting insight into why this particular case didn't work,
but I'll hold back with that till you folk have had a crack at it.
How do we know, when we choose a subdivision into finite entities, what
shape *is* the limit?
-- Donald
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Dr. Donald E. Simanek Office: 717-893-2079
Prof. of Physics Internet: dsimanek@eagle.lhup.edu
Lock Haven University, Lock Haven, PA. 17745 CIS: 73147,2166
Home page: http://www.lhup.edu/~dsimanek FAX: 717-893-2047
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