Notes: `y = x/2 + 3 , 1

Sunday, 10 September 2017

$\displaystyle{y}=\frac{{x}}{{2}}+{3},{1}$

To find the area of this surface, we rotate the function $\displaystyle{y}=\frac{{x}}{{2}}+{3}$ about the y-axis (not the x-axis) in the range $\displaystyle{1}\le{x}\le{5}$ and this way create a finite surface of revolution.

A way to approach this problem is to swap the roles of $\displaystyle{x}$ and $\displaystyle{y}$ , essentially looking at the page side-on, 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 $\displaystyle{x}$ and $\displaystyle{y}$ )

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

Since we are swapping the roles of $\displaystyle{x}$ and $\displaystyle{y}$ , we need the function $\displaystyle{y}=\frac{{x}}{{2}}+{3}$ written as $\displaystyle{x}$ in terms of $\displaystyle{y}$ as opposed to $\displaystyle{y}$ in terms of $\displaystyle{x}$ . So we have

$\displaystyle{x}={2}{y}-{6}$

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 length $\displaystyle{\left.{d}{y}\right.}$ of the y-axis, which can be thought of as 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, $\displaystyle{x}={2}{y}-{6}$ , that $\displaystyle\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}={2}$

and since the range (in $\displaystyle{y}$ ) over which to take the integral is $\displaystyle{1}\le{x}\le{5}$ , or equivalently $\displaystyle\frac{{7}}{{2}}\le{y}\le\frac{{11}}{{2}}$ we have $\displaystyle{a}=\frac{{7}}{{2}}$ and $\displaystyle{b}=\frac{{11}}{{2}}$ .

Therefore, the area required, A, is given by

$\displaystyle{A}={\int_{{\frac{{{7}}}{{{2}}}}}^{{\frac{{{11}}}{{{2}}}}}}{2}\pi{\left({2}{y}-{6}\right)}\sqrt{{{5}}}{\left.{d}{y}\right.}$ $= 2sqrt(5)pi int_((7)/(2))^((11)/(2)) 2y - 6 \quad dy$

$\displaystyle={2}\sqrt{{{5}}}\pi{y}{\left({y}-{6}\right)}{{\mid}_{{\frac{{{7}}}{{{2}}}}}^{{\frac{{{11}}}{{{2}}}}}}=\frac{{\sqrt{{{5}}}}}{{{2}}}\pi{\left[{11}{\left({11}-{6}\right)}-{7}{\left({7}-{6}\right)}\right]}$