area of spherical zone

Let us consider the circle

$${(x-r)}^2+y^2=r^2$$

with radius $r$ and centre  $(r,\mathrm{\hspace{0.17em}0})$.  A spherical zone may be thought to be formed when an arc of the circle rotates around the $x$-axis.  For finding the are of the zone, we can use the formula

$A=\mathrm{\hspace{0.33em}2}\pi {\displaystyle {\int }_a^b}y\sqrt{1+{\left({\displaystyle \frac{dy}{dx}}\right)}^2}𝑑x$(1)

of the entry area of surface of revolution.  Let the ends of the arc correspond the values $a$ and $b$ of the abscissa such that  $b-a=h$  is the of the spherical zone.  In the formula, we must use the solved form

$$y=(\pm )\sqrt{rx-x^2}$$

of the equation of the circle.  The formula then yields

$$A=\mathrm{\hspace{0.33em}2}\pi {\int }_a^b\sqrt{rx-x^2}\sqrt{1+{\left(\frac{r-x}{\sqrt{rx-x^2}}\right)}^2}𝑑x=\mathrm{\hspace{0.33em}2}\pi {\int }_a^{b}r𝑑x=\mathrm{\hspace{0.33em}2}\pi r(b-a).$$

Hence the area of a spherical zone (and also of a spherical calotte) is

$A=\mathrm{\hspace{0.33em}2}\pi rh.$(2)

From here one obtains as a special case  $h=2r$  the area of the whole sphere:

$A=\mathrm{\hspace{0.33em}4}\pi r^2.$(3)

Remark.  The formula (2) implies that the centre of mass of a half-sphere is at the halfway point of the axis of symmetry ($h=\frac{r}{2}$).