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).**

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