# y = 1-x^2/4 , 0<=x<=2 Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the y-axis.

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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))