`y = 1-x^2/4 , 0

To find the area of this surface, we rotate the function `y = 1 - x^2/4 ` about the y-axis (not the x-axis!) and this way create a surface of revolution. It is a finite area, since we are looking only at a section of the x-axis and hence y-axis.

The range of the x-axis we are interested in is  ` ``0 <= x <=2 `  and hence the range of the y-axis we are interested in is `0 <=y <=1 `

It is easiest to swap the roles of `x ` and `y `, essentially turning the page so that we can use the standard formulae that are usually written in terms of `x ` (ie, that usually refer to the x-axis).

The formula for a surface of revolution A is given by (interchanging the roles of x and y)

  `A = int_a^b (2pi x) sqrt(1 + (frac(dx)(dy))^2) dy`

Evidently, we need the function `y = 1 - x^2/4 ` written as `x ` in terms of `y ` rather than `y ` in terms of `x ` . So we have

`x = pm2sqrt(1 - y) `

This describes a parabola, which is two mirror image sqrt curves when considered in terms of the y-axis. But we need only one half, the positive or the negative, to rotate the graph about the y-axis because the other half will be part of the resulting roatated object anyway. Without loss of generality (wlog for short) we can take the function to rotate about the y-axis as

`x = 2sqrt(1-y) `

To obtain the area required by integration, we are effectively adding together tiny rings (of circumference `2pi x ` at a point `y ` on the y-axis) where each ring takes up length `dy ` on the y-axis. The distance from the circular edge to circular edge of each ring is `sqrt(1 + (frac(dx)(dy))^2) dy`

This is the arc length of the function `x = f(y) ` in a segment of the y-axis `dy ` in length, which is the hypotenuse of a tiny triangle with width `dy ` and height `dx `. These distances from edge to edge of the tiny rings are then multiplied by the circumference of the surface at that point, `2pi x `, to give the surface area of each ring. The tiny sloped rings are added up to give the full sloped surface area of revolution.

We have for this function, `x = 2sqrt(1-y) `, that

`frac(dx)(dy) = -1/sqrt(1-y)`

and since the range (in `y `) over which to take the integral is `[0,1] ` we have `a=0 ` and `b=1 `.

Therefore, the area required, A, is given by

`A = int_0^1 4pi sqrt(1-y)sqrt(1 + 1/((1-y))) dy `

` `This can be simplified to give

`A = 4pi int_0^1 sqrt((1-y) + 1) \quad dy`

`= 4pi int_0^1 sqrt(2-y) \quad dy = - frac(8)(3) pi (2-y)^(3/2)|_0^1` 

So that the surface area of rotation A is given by

`A = -8/3 pi (1 - 2sqrt(2)) `