>Consider the region bounded above by y = 1/x. Compute the area under
>this curve, above the x-axis, between x = 1 and x = 00 (infinity).
>The "area" of this region is given by fnInt(1/x,x,1,00) and is
>infinite (the improper integral diverges).
>
>However the volume obtained when we rotate this region about the x-axis
>is given by pi*fnInt(1/x*x,x,1,00) = pi.
>
>How do you expain to students that the area is infinite but the volume
>is finite, since the area fits inside the volume?
There have been a few messages about paint and I don't think that is the
right analogy to make. In fact, I don't think a physical analogy is the
best. Mathematical objects don't have the same qualities as physical
objects. For instance the area described in the integral
/int_1fty{1/xis finite. We can say that if that entire area were
reshaped it would fit in a rectangle of some predetermined size. However
the boundary is infinite in length. It would not be appropriate to say
that if the boundary were string that the string could not be placed in a
box of any finite size. String and a mathematical curve do not have the
same properties: string has a radius and a curve does not. We cannot talk
about a mathematical object with one local dimension filling up another
mathematical object with two local dimensions.
The paint analogy suffers from the same problems, physical versus
mathematical properties.
I don't know a great way of explaining finiteness and infiniteness to
students. I try to stick to the explanations that use the definition of
infinity. For instance, the area above fits into a finite rectangle but
the area described in /int_1fty{1/x} does not. In fact, for any
rectangle of any size that I choose the students can find an N large enough
so that /int_1/x} is larger than that area. Usually I do this with
sums instead of integrals because the students can get into the adding
blocks of width one together better than a shape such as the integrals
above describe. We write a program in class on the TI-85 to find out how
large N has to be so that /sum_1/k} exceeds a certain size. The
program is
Code Comments
Prompt AREA This is the target area
0->SM This is the accumulated area of the partial sums and it
starts at 0
1->N This is the starting value of the index of the sum
While SM<AREA While the partial sum is less than the targeted area
1/N+SM->SM Calculate the next partial sum
N+1->N Increment the index
End End the loop that will find the N where the partial
sums exceed the target area
Disp N This is how many terms of the sum are needed for the
partial sums to exceed the target area
Beware: for a target area of 6 square units the index is 228 (this takes
just a few seconds of time to run) but for a target area of 20 square units
the index exceeds 7000 and this takes quite a few minutes to run (in fact,
after 4 minutes the SM was only 9.4 and the index was 800 so I'm not sure
how long it would have taken). I tell my students that given enough time,
the calculator can do any target area, unfortunately the batteries might
run out before then.
If a student can handle the infinite boundary, finite area problem, then I
think they should grasp the infinite surface area, finite volume problem
also.
Thank you for allowing me to ramble a bit.
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Murphy Waggoner
Department of Mathematics
Simpson College
701 North C Street
Indianola, IA 50125
waggoner@storm.simpson.edu
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