Notes: Find the area of the zone of a sphere formed by revolving the graph of `y=sqrt(9-x^2) , 0

Monday, 23 September 2013

Find the area of the zone of a sphere formed by revolving the graph of $\displaystyle{y}=\sqrt{{{9}-{{x}}^{{2}}}},{0}$

The function $\displaystyle{y}=\sqrt{{{9}-{{x}}^{{2}}}}$ describes a circle centred on the origin with radius 3.

If we rotate this function in the range $\displaystyle{0}\le{x}\le{2}$ about the y-axis we obtain a surface of revolution that is specifically a zone of a sphere with radius 3.

A zone of a sphere is the surface area between two heights on the sphere (surface area of ground between two latitudes when thinking in terms of planet Earth).

When $\displaystyle{0}\le{x}\le{2}$ for this given sphere (which can be written equivalently as $\displaystyle{{x}}^{{2}}+{{y}}^{{2}}={9}$ ) the corresponding range for $\displaystyle{y}$ is $\displaystyle\sqrt{{{5}}}\le{y}\le{3}$

Since this range includes the top of the sphere, the zone we are considering is more specifically a cap. The equivalent on planet Earth would be a polar region.

To calculate the surface area of this cap of a sphere with radius 3, we require the formula for the surface area of revolution of a function $\displaystyle{x}={f{{\left({y}\right)}}}$ (note, I have swapped the roles of $\displaystyle{x}$ and $\displaystyle{y}$ for convenience, as the formula is typically written for rotating about the x-axis rather than about the y-axis as we are doing here).

The formula for the surface area of revolution of a function $\displaystyle{x}={f{{\left({y}\right)}}}$ rotated about the y-axis in the range $\displaystyle{a}\le{y}\le{b}$ is given by

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

Here, we have that $\displaystyle{a}=\sqrt{{{5}}}$ and $\displaystyle{b}={3}$ . Also, we have that

$\displaystyle\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}=-\frac{{y}}{\sqrt{{{9}-{{y}}^{{2}}}}}$

so that the cap of interest has area

$A = int_sqrt(5)^3 2pi sqrt(9-y^2) sqrt(1+(y^2)/(9-y^2)) \quad dy$

which can be simplified to

$A = 2pi int_sqrt(5)^3 sqrt((9-y^2) + y^2) \quad dy$ $\displaystyle={2}\pi{\int_{\sqrt{{{5}}}}^{{3}}}{3}{\left.{d}{y}\right.}={6}\pi{y}{{\mid}_{\sqrt{{{5}}}}^{{3}}}$

So that the zone (specifically cap of a sphere) area of interest A = $\displaystyle\pi{\left({18}-{6}\sqrt{{{5}}}\right)}$

This can also be calculated using the formula for calculating the surface area of the cap of a sphere as A = $\displaystyle\pi{\left({{a}}^{{2}}+{{h}}^{{2}}\right)}$ where a is the radius at the lower limit of the cap and h is the perpendiculat height of the cap. Here this would give A = $\displaystyle\pi{\left({4}+{{\left({3}-\sqrt{{{5}}}\right)}}^{{2}}\right)}=\pi{\left({4}+{9}-{6}\sqrt{{{5}}}+{5}\right)}=\pi{\left({18}-{6}\sqrt{{{5}}}\right)}$ (ie the same result).