# area of spherical zone

Let us consider the circle

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

with radius $r$ and centre  $(r,\,0)$.  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

 $\displaystyle A\;=\;2\pi\!\int_{a}^{b}\!y\,\sqrt{1+\left(\frac{dy}{dx}\right)^% {2}}\,dx$ (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

 $A\;=\;2\pi\!\int_{a}^{b}\sqrt{rx\!-\!x^{2}}\,\sqrt{1+\left(\frac{r\!-\!x}{% \sqrt{rx\!-\!x^{2}}}\right)^{2}}\,dx\;=\;2\pi\!\int_{a}^{b}r\,dx\;=\;2\pi r(b% \!-\!a).$

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

 $\displaystyle A\;=\;2\pi rh.$ (2)

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

 $\displaystyle A\;=\;4\pi r^{2}.$ (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 ($h=\frac{r}{2}$).

Title area of spherical zone AreaOfSphericalZone 2013-03-22 18:19:05 2013-03-22 18:19:05 pahio (2872) pahio (2872) 9 pahio (2872) Derivation msc 26B15 msc 53A05 msc 51M04 area of spherical calotte AreaOfTheNSphere CentreOfMassOfHalfDisc