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

Wednesday, March 7, 2012

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

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

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

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).

For the range of interest $\displaystyle{0}\le{x}\le{a}$ , the zone of interest is specifically a spherical cap on the sphere of radius $\displaystyle{r}$ . The range of interest for $\displaystyle{y}$ corresponding for that for $\displaystyle{x}$ is $\displaystyle\sqrt{{{{r}}^{{2}}-{{a}}^{{2}}}}\le{y}\le{r}$ .

The equivalent on planet Earth of the surface area of such a spherical cap could be, for example, the surface area of a polar region. This of course makes the simplifying assumption that the Earth is perfectly spherical, which is not the case.

To calculate the surface area of this cap of a sphere with radius $\displaystyle{r}$ , 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 $\displaystyle\alpha\le{y}\le\beta$ is given by

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

Here, we have that $\displaystyle\alpha=\sqrt{{{{r}}^{{2}}-{{a}}^{{2}}}}$ and $\displaystyle\beta={r}$ . Also, we have that

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

so that the cap of interest has area $A = int_sqrt(r^2-a^2)^r 2pi sqrt(r^2-y^2) sqrt(1+(y^2)/(r^2-y^2)) \quad dy$

which can be simplified to

$A =2pi int_sqrt(r^2-a^2)^r sqrt((r^2-y^2) + y^2) \quad dy$

$\displaystyle={2}\pi{\int_{\sqrt{{{{r}}^{{2}}-{{a}}^{{2}}}}}^{{r}}}{r}{\left.{d}{y}\right.}$ $\displaystyle={2}\pi{r}{y}{{\mid}_{\sqrt{{{{r}}^{{2}}-{{a}}^{{2}}}}}^{{r}}}={2}\pi{r}{\left({r}-\sqrt{{{{r}}^{{2}}-{{a}}^{{2}}}}\right)}$

So that the zone (specifically cap of a sphere) area of interest A =

$\displaystyle=\pi{\left({2}{{r}}^{{2}}-{2}{r}\sqrt{{{{r}}^{{2}}-{{a}}^{{2}}}}\right)}$

This marries up with the formula for the surface area of a spherical cap

$\displaystyle{A}=\pi{\left({{h}}^{{2}}+{{a}}^{{2}}\right)}$

where $\displaystyle{a}$ is the radius at the base of the spherical cap and $\displaystyle{h}$ is the height of the cap. The value of $\displaystyle{h}$ is the range covered on the y-axis, so that

$\displaystyle{h}={r}-\sqrt{{{{r}}^{{2}}-{{a}}^{{2}}}}$ and

$\displaystyle{{h}}^{{2}}={2}{{r}}^{{2}}-{2}{r}\sqrt{{{{r}}^{{2}}-{{a}}^{{2}}}}-{{a}}^{{2}}$ and

$\displaystyle{{h}}^{{2}}+{{a}}^{{2}}={2}{{r}}^{{2}}-{2}{r}\sqrt{{{{r}}^{{2}}-{{a}}^{{2}}}}$

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Wednesday, March 7, 2012

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

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Wednesday, March 7, 2012