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

Monday, December 28, 2009

$\displaystyle{y}={1}-\frac{{{x}}^{{2}}}{{4}},{0}$

To find the area of this surface, we rotate the function $\displaystyle{y}={1}-\frac{{{x}}^{{2}}}{{4}}$ 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 $\displaystyle{0}\le{x}\le{2}$ and hence the range of the y-axis we are interested in is $\displaystyle{0}\le{y}\le{1}$

It is easiest to swap the roles of $\displaystyle{x}$ and $\displaystyle{y}$ , essentially turning the page so that we can use the standard formulae that are usually written in terms of $\displaystyle{x}$ (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 $\displaystyle{y}={1}-\frac{{{x}}^{{2}}}{{4}}$ written as $\displaystyle{x}$ in terms of $\displaystyle{y}$ rather than $\displaystyle{y}$ in terms of $\displaystyle{x}$ . So we have

$\displaystyle{x}=\pm{2}\sqrt{{{1}-{y}}}$

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

$\displaystyle{x}={2}\sqrt{{{1}-{y}}}$

To obtain the area required by integration, we are effectively adding together tiny rings (of circumference $\displaystyle{2}\pi{x}$ at a point $\displaystyle{y}$ on the y-axis) where each ring takes up length $\displaystyle{\left.{d}{y}\right.}$ 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 $\displaystyle{x}={f{{\left({y}\right)}}}$ in a segment of the y-axis $\displaystyle{\left.{d}{y}\right.}$ in length, which is the hypotenuse of a tiny triangle with width $\displaystyle{\left.{d}{y}\right.}$ and height $\displaystyle{\left.{d}{x}\right.}$ . These distances from edge to edge of the tiny rings are then multiplied by the circumference of the surface at that point, $\displaystyle{2}\pi{x}$ , 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, $\displaystyle{x}={2}\sqrt{{{1}-{y}}}$ , that

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

and since the range (in $\displaystyle{y}$ ) over which to take the integral is $\displaystyle{\left[{0},{1}\right]}$ we have $\displaystyle{a}={0}$ and $\displaystyle{b}={1}$ .