`y = 3x , 0

Monday, December 17, 2012

$\displaystyle{y}={3}{x},{0}$

The quantity to be calculated is the area of what is called a surface of revolution. The function $\displaystyle{y}={3}{x}$ is rotated about the x-axis and the surface that is created in this way is a surface of revolution. The area to be calculated is definite, since we consider only the region of the x-axis $\displaystyle{x}\in{\left[{0},{3}\right]}$ , that is, $\displaystyle{x}$ between 0 and 3.

The formula for a surface of revolution (which is an area, A) is given by

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

The circumference of the surface at each point along the x-axis is $\displaystyle{2}\pi{y}$ and this is added up (integrated) along the x-axis by cutting the function into tiny lengths of $sqrt(1 + (frac(dy)(dx))^2) dx$

ie, the arc length of the function in a segment of the x-axis $\displaystyle{\left.{d}{x}\right.}$ in length, which is the hypotenuse of a tiny triangle with width $\displaystyle{\left.{d}{x}\right.}$ and height $\displaystyle{\left.{d}{y}\right.}$ . These lengths are then multiplied by the circumference of the surface at that point $\displaystyle{2}\pi{y}$ to give the surface area of rings around the x-axis that have tiny width $\displaystyle{\left.{d}{x}\right.}$ yet have edges that slope towards or away from the x-axis. The tiny sloped rings are added up to give the full sloped surface area of revolution.

In this case, $frac(dy)(dx) = 3$ and since the range over which to take the arc length is $\displaystyle{\left[{0},{3}\right]}$ we have $\displaystyle{a}={0}$ and $\displaystyle{b}={3}$ . Therefore, the area required, A, is given by

$\displaystyle{A}={\int_{{0}}^{{3}}}{6}\pi{x}\sqrt{{{10}}}{\left.{d}{x}\right.}={3}\sqrt{{{10}}}\pi{{x}}^{{2}}{{\mid}_{{0}}^{{3}}}={27}\sqrt{{{10}}}\pi$