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))
http://mathworld.wolfram.com/SurfaceofRevolution.html