In the previous article in this series, we saw how sankalitas are the Kerala calculus analogue to integrals in modern calculus, and how they were computed in the limit of large n. In this article, we can take a look at how these were used to compute the volume of a sphere.

### Area of a circle

To derive the volume of a sphere, we must first derive the area of a circle of radius *r* and circumference *C*.

To find the area of such a circle, we first divide it into radial segments as in the figure, and then reassemble them as below. As the number of segments grows more and more, the assembled figure moves closer to a rectangle. At the limit, we get a true rectangle, and the result \( A = \frac{C}{2}.r \). We can see that this is equivalent to the modern result \( A = \pi . r^2 \). Sankalita doesn't appear directly in this derivation, but is implied in the expression for circumference, which we calculate using (for example), व्यासे वारिधिनिहते …

### त्रैराशिकम् – The principle of proportionality

त्रैराशिकम् (rule of three) is fundamental to the way Indian mathematicians thought. Aryabhata says:

त्रैराशिकफलराशिं तमथेच्छाराशिना हतं कृत्वा ।

लब्धं प्रमाणभजितं तस्मादिच्छाफलमिदं स्यात् ।।

If you have a quantity प्रमाणम्, that produces a proportional फलम्, and we wish to find what the corresponding इच्छाफलम् would be for a different quantity इच्छा, we use the relation इच्छाफलम् = फलम् * इच्छा / प्रमाणम्

For example, all circles have the same proportionality between radius and circumference. If we know a circle with radius *r* and circumference *C*, another circle with radius r_{1} will have circumference \(C_1 = \frac{C . r_1}{r}\).

Proportionality relationships with more than two elements can be calculated using rules of five, seven etc. Bhaskarācārya discusses the methodology for this in detail in *Līlāvatī*.

### Volume of a sphere

To compute the volume of a sphere of radius *r*, diameter \(d = 2.r\) and great-circle circumference *C*, we divide it into thin horizontal circular segments, slicing parallel to the horizontal axis. Each slice has a finite thickness. We add up the slice volumes to approximate the volume of the sphere, which becomes accurate at the limit of atomically thin slices.

We divide the upper hemisphere of the sphere (of radius *r* and great-circle circumference *C) *into *n* slices of equal thickness \(r/n\). We denote the vertical distance from the centre of the jth slice (numbering from the pole) by \(K_j\), and the radius of the jth slice as \(J_j\) (for कोटिः and ज्या, which we have seen before). The figure represents a vertical great-circle slice through the sphere, showing the radius, \(J_j\) and \(K_j\). The volume of the jth slice \(V_j\) is related to its circumference, using the area formula derived in the previous section.

\( \begin{align*}

C_j &= \frac{C}{r}.J_j && \text{(}C_j\text{ is the circumference of the jth slice, and trairāśikam)} \\

V_j &= \frac{C_j}{2} . J_j . \frac{r}{n} && \text{(using}\quad A = \frac{C}{2}.r \quad\text{)} \\

&= \frac{C}{2.r}. J_j^2. \frac{r}{n} \\

& = \frac{C}{2.n}. J_j^2.

\end{align*}\)

Summing up the upper hemisphere and considering *n* to be very large, we get

\( \begin{align*}

\frac{V}{2} &= \frac{C}{2.n}. (J_1^2+J_2^2+…J_n^2) \\

V &= \frac{C}{n} (J_1^2+J_2^2+…J_n^2) \\

&= \frac{C}{n} . (r^2+r^2+…r^2) \\

&\quad-\frac{C}{n} . (K_1^2+K_2^2+…K_n^2) && \text{(using jya-koti-karna nyaya)} \\

&= C.r^2 – \frac{C}{n} .\frac{r^2}{n^2} (1^2+2^2+…n^2) \\

&= C.r^2 – \frac{C}{n} .\frac{r^2}{n^2}.\frac{n^3}{3} && \text{(varga-sankalita)} \\

&= C.r^2 – C .\frac{r^2}{3} \\

&= \frac{2}{3}.C.r^2 &&(=\frac{4}{3}.\pi.r^3)\\

& = \frac{C}{6}.d^2

\end{align*}\)

And we see that the same result that is derived in modern textbooks using integration can be obtained by the large-n sankalita method.

*This is a slightly modified version of the yukti in Yuktibhāṣā. Instead of the ज्याकोटिकर्णन्यायः (aka “Pythagoras Theorem”) which is known to it, it uses the intersecting-chords theorem (वृत्ते समवर्गो …) to similar effect. We use the former, since it is more familiar to modern readers*.

### Summary of Results

Area of a circle: \( A = \frac{C}{2}.r \)

Volume of a sphere: \( V = \frac{2}{3}.C.r^2 = \frac{C}{6}.d^2 \)

In both cases, we calculate C from r or d using व्यासे वारिधिनिहते or any of the alternatives we will see in later articles.