Question

Suppose f(x) is continuous on [a; b] and is revolved about the x-axis. The surface area of the
resulting gure is given by

integral from a to be of 2(pi)y * sqrt(1+(dy/dx)^2)

a) Explain why this formula makes intuitive sense. Hint: Think of the surface as being sliced up
into rings, and remember the formula for arc-length.

Answer 1

\[\int\limits_{a}^{b}2 \pi y \sqrt{1+(dy/dx)^2 }\]

Answer 2

The 2pi would come from the fact that it spans from 0 to 2pi (a circle) and y would be multiplying the height by the arc length, then integrating

Answer 3

and what is the dy/dx? maybe you can help me with the next one
-consider the function f(x) = 1=x on the interval [1; 1). Show that the surface area of f
revolved about the x-axis on the same interval is innite.

Answer 4

dy/dx is the derivative, it's part of the arc length formula

Answer 5

I don't really remember the surface area for single variable problems. I'm doing multivariable flux and surface integrals now so I could solve it with a double surface integral, but in order to do that I would be using the cross product and double integrals which probably is a bit beyond what you've learned