Calculus of a Single Variable, Chapter 7, 7.4, Section 7.4, Problem 42

The function y = sqrt(9-x^2) is rotated about the x-axis and the surface area that is created in this way is a surface of revolution.
The area to be calculated is definite, since we consider only the region of the x-axis x in [0,3] , ie x between 0 and 3.
The formula for a surface of revolution A is given by
A = int_a^b (2pi y) sqrt(1 + (frac(dy)(dx))^2) dx
The circumference of the surface at each point along the x-axis is 2pi y and this is added up (integrated) along the x-axis by cutting the function into infinitessimal lengths of sqrt(1 + (frac(dy)(dx))^2) dx
ie, the arc length of the function in a segment of the x-axis dx in length, which is the hypotenuse of a tiny triangle with width dx , height dy . These lengths are then multiplied by the circumference of the surface at that point, 2pi y to give the surface area of rings around the x-axis that have tiny width dx yet have edges that slope towards or away from the x-axis. The tiny sloped rings are added up to give the full sloped surface area of revolution. In this case,
frac(dy)(dx) = -frac(x)(sqrt(9 -x^2))
and since the range over which to take the arc length is [-2,2] we have a = -2 and b=2 . Therefore, the area required, A, is given by
A= int_(-2)^2 2pi sqrt(9-x^2) sqrt(1 + frac(x^2)(9-x^2)) dx
which can be simplified to
A = int_(-2)^2 2pi sqrt(9-x^2 + x^2) dx = int_(-2)^2 6 pi dx
so that
A = 6pi x|_(-2)^2 = 6 pi (2 + 2) = 24pi