Calculus of a Single Variable, Chapter 7, 7.4, Section 7.4, Problem 57

The function y = sqrt(9 -x^2) describes a circle centred on the origin with radius 3.
If we rotate this function in the range 0 <=x <=2 about the y-axis we obtain a surface of revolution that is specifically a zone of a sphere with radius 3.
A zone of a sphere is the surface area between two heights on the sphere (surface area of ground between two latitudes when thinking in terms of planet Earth).
When 0 <=x<=2 for this given sphere (which can be written equivalently as x^2 + y^2 = 9 ) the corresponding range for y is sqrt(5) <=y <= 3
Since this range includes the top of the sphere, the zone we are considering is more specifically a cap. The equivalent on planet Earth would be a polar region.
To calculate the surface area of this cap of a sphere with radius 3, we require the formula for the surface area of revolution of a function x = f(y) (note, I have swapped the roles of x and y for convenience, as the formula is typically written for rotating about the x-axis rather than about the y-axis as we are doing here).
The formula for the surface area of revolution of a function x = f(y) rotated about the y-axis in the range a <=y <=b is given by
A = int_a^b 2pi x sqrt(1+ ((dx)/(dy))^2) \quad dy
Here, we have that a = sqrt(5) and b = 3 . Also, we have that
(dx)/(dy) = -y/sqrt(9-y^2)
so that the cap of interest has area
A = int_sqrt(5)^3 2pi sqrt(9-y^2) sqrt(1+(y^2)/(9-y^2)) \quad dy
which can be simplified to
A = 2pi int_sqrt(5)^3 sqrt((9-y^2) + y^2) \quad dy = 2pi int_sqrt(5)^3 3 dy = 6pi y |_sqrt(5)^3
So that the zone (specifically cap of a sphere) area of interest A = pi(18 - 6sqrt(5))
This can also be calculated using the formula for calculating the surface area of the cap of a sphere as A = pi (a^2 + h^2) where a is the radius at the lower limit of the cap and h is the perpendiculat height of the cap. Here this would give A = pi (4 + (3-sqrt(5))^2) = pi (4 + 9 -6sqrt(5) + 5) = pi(18 -6sqrt(5)) (ie the same result).