Notes: `y = sqrt(9-x^2) , -2

Friday, 30 January 2015

$\displaystyle{y}=\sqrt{{{9}-{{x}}^{{2}}}},-{2}$

The function $\displaystyle{y}=\sqrt{{{9}-{{x}}^{{2}}}}$ is rotated about the x-axis and the surface area 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 $\displaystyle{x}\in{\left[{0},{3}\right]}$ , ie $\displaystyle{x}$ between 0 and 3.

The formula for a surface of revolution A is given by

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

The circumference...

The function $\displaystyle{y}=\sqrt{{{9}-{{x}}^{{2}}}}$ is rotated about the x-axis and the surface area 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 $\displaystyle{x}\in{\left[{0},{3}\right]}$ , ie $\displaystyle{x}$ between 0 and 3.

The formula for a surface of revolution 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 infinitessimal 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 $\displaystyle{\left.{d}{x}\right.}$ , 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,

$frac(dy)(dx) = -frac(x)(sqrt(9 -x^2))$

and since the range over which to take the arc length is $\displaystyle{\left[-{2},{2}\right]}$ we have $\displaystyle{a}=-{2}$ and $\displaystyle{b}={2}$ . Therefore, the area required, A, is given by

$A= int_(-2)^2 2pi sqrt(9-x^2) sqrt(1 + frac(x^2)(9-x^2)) dx$

which can be simplified to

$\displaystyle{A}={\int_{{-{2}}}^{{2}}}{2}\pi\sqrt{{{9}-{{x}}^{{2}}+{{x}}^{{2}}}}{\left.{d}{x}\right.}={\int_{{-{2}}}^{{2}}}{6}\pi{\left.{d}{x}\right.}$