Surface Area of a Sphere

In a previous article, we saw how the area of a circle, and the volume of a sphere are computed using Kerala School methods. We had not looked at the surface area of a sphere at that point, since we had not looked at jyā yet. Now that that is familiar to us, it is time to see how we use saṅkalita to compute the surface area of a sphere.

Segmenting a Sphere

As we did when we computed the volume of a sphere, we divide it into segments. However, instead of dividing it into segments of equal thickness, which is convenient for volume calculation, we divide it into segments of equal arc lengths \(\alpha\), as seen from a vertical slice through the sphere. This makes it more convenient to compute surface area, as each roughly trapezoidal segment on the surface ends up with constant width \(\alpha\). In the limit of large number of segments, each trapezoid becomes a rectangle of area \(C_j.\alpha\) where \(C_j\) is the circumference of the jth segment. If r is the radius of the sphere, and C the circumference of any of its great circles (eg: the equator of the sphere), we can use त्रैराशिकम् to compute the circumference of the jth segment as \(C_j = \frac{C}{r}.B_j\) where \(B_j\) is the jth jyā. Thus, the surface area of the jth segment is \( S_j = C_j.\alpha = \frac{C}{r}.B_j.\alpha\).

Summing up the area of the upper hemisphere to get half the surface area of the sphere, we have

\(\frac{S}{2} = \frac{C}{r}.(B_1 + B_2 + … B_n).\alpha \)

Recollecting the second differential of the jyā from an earlier article, we note that \((B_j – B_{j-1}) – (B_{j+1}-B_j) = (\frac{\alpha}{r})^2.B_j\)

Thus:

\(
\begin{align*}
\frac{S}{2} &= \frac{C}{r}.(B_1 + B_2 + … B_n).\alpha \\
&= \frac{C}{r}.(\frac{r}{\alpha})^2((B_1-0) – (B_2 -B_1) + (B_2-B_1) – (B_3 – B_2) + (B_3 – B_2) – … – (B_{n-1}-B_{n-2}) + (B_{n-1}-B_{n-2}) – (B_n – B_{n-1})).\alpha \\
& = \frac{C}{r}.(\frac{r}{\alpha})^2 (B_1 – (B_n – B_{n-1})) .\alpha \\
&= C.r \\
S &= 2.C.r = C.d && (=4.\pi.r^2)
\end{align*}
\)

Using \(B_1 \approx \alpha\) and \((B_n – B_{n-1}) \approx 0 \) for large n.