Yuktibhāṣā does not explicitly calculate the volumes of pyramids or cones, but it would be instructive to use the method of Sankalita to do so, to demonstrate the generalizability of this method beyond what is explicitly shown.

## Volume of a Pyramid

Consider a square pyramid of side *a*, and height *h*. We divide it into *n* slices of equal thickness such that \(\frac{h}{n}\) is the thickness of the slice. The base of each slice will have varying dimensions. Let \(a_k\) be the side of slice number *k*.

By *trairāśika, *we have \(a_k = \frac{a.k}{n}\). Thus, the volume of slice number *k* is

\begin{align*}

V_k &= \frac{h}{n} . a_k^2 \\

&= \frac{h}{n}. {(\frac{a.k}{n})}^2 \\

& = \frac{h.a^2.k^2}{n^3} \\

\end{align*}

\)

The total volume is the sum of all such segments. Taking *n* to *parardhā*, and using the value of vargasankalita seen earlier,

\begin{align*}

V &= \sum_{0}^{n} V_k \\

&= \sum_{0}^{n} \frac{h.a^2.k^2}{n^3} \\

&= \frac{h.a^2}{n^3}.S_n^{(2)} \\

&= \frac{h.a^2}{n^3}.\frac{n^3}{3} \\

&= \frac{h.a^2}{3}.

\end{align*}

\)

Which is identical to the modern result. A rectangular pyramid can be handled similarly.

## Volume of a Cone

As it turns out, the volume of a cone is also quite similar. Taking the base of the cone to have radius *r* and using a similar slicing methodology, the volume of slice number *k* becomes \(V_k = \pi. {(\frac{k.r}{n})}^2\), and the volume of the cone becomes

\begin{align*}

V &= \sum_{0}^{n} V_k \\

&= \frac{\pi . h . r^2}{n^3}.S_n^{(2)} \\

&= \frac{\pi . h . r^2}{n^3}.\frac{n^3}{3} \\

&= \frac{\pi . h . a^2}{3}.

\end{align*}

\)

again, identical to the modern result.

Clearly then, the Sankalita is equivalent to the modern integral, at least for powers (and by extension, polynomials).