Hi, would just like to know if what I have done is correct.
Find the area integral of the surface in polar form lying over the annulus
The equation in polar form is
Integrating with respect to
and with respect to , I get the area as
Does this look ok?
That person probably thinks that any triple integral expresses volume and only volume. It's hard for some people to understand that a single, double or triple integral can express both area and volume. After all, a double integral is a simplified triple integral and a single integral is a simplified double integral.
Oops, I think there is indeed a problem.
What you did was to find the volume between the annulus and the surface. But it looks as though the question was actually asking for the surface area of that part of the suface lying above the annulus.
The formula for surface area says that this area is given by , where A denotes the annulus.
If then , , and . So the answer to the question is
. . . . .
I thought that the question was not very clearly worded, but the fact that those 3/2 powers both turn out to be integers makes me fairly sure that "area integral" is supposed to mean "surface area" in this question.
Since you were never one for numbers, what brings you to a math website?
Since this is the second time (besides this thread) that I haven't gotten an answer to my question leads me to believe that no such formula exists to transform a double integral into a single integral. Can you prove me wrong?
Green's Theorem relates a surface integral to a line integral. Since surface integrals often decompose into a double iterated integral, and line integrals often decompose into a single integral, you could say that Green's Theorem can reduce some double integrals into a single integral. I'm not sure there is a way, short of actually computing the inner integral, to reduce all double integrals to single integrals.