Calculus problem?

Question

Consider the surface and solid of revolution of function y = √(2x-x²). From x = 0 to x = a, a > 0, the total volume is 0. What is the surface area?

When you consider quantum wierdness in physics, you have to wonder about saying things like, "the physicality of the problem". Maybe unreal things in mathematics is more real than we thought?

Answer 1

I see that you were inspired by that circle and sphere problem. I'm also sure you have some short cut method. But here's a purely calculus approach, which ignores the physicality of the problem.

First to find a.
V = ∫{0,a} πy^2 dx = a^2 - a^3/3 = 0
a = 3

S = ∫{0,a} 2π√[1 + (y')^2] dx
We can show that 1 + (y')^2 = 1/y^2
So S = ∫{0,a} 2π dx
= 2aπ = 6π

Of course for x>2, the argument of hte radical becomes -ve, but let's pretend that calculus doesn't know that.

Answer 2

What is the function for values of x greater than 2? This question, as stated, has no answer, unless you allow y to take on complex values. In that case, one could understand negative volume to mean what results when complex values of a function are revolved about the x-axis. But then, do we assume the boundary of negative-volume regions has positive, or negative surface area?

Answer 3

The volume = 0? It is impossible.
If you say the integral, from 0 to a, of pi (x-x^2) dx = 0, then you can go from there.

Answer 4

Not sure, I don't know how to manually do that. This question has made me want to take my caculator home over these holidays. Haha.