Posted: 03.04.2015

Let us examine the above calculation made by the *Dhavalākāra *in the light of modern mathematical method. In order to find out the value of the *adholoka *(in *mṛdañgākāra loka*) we know that it is in the shape of a chopped off right conical pyramid. (See figure no. 2).

When we extend the hypotaneouses of the triangles to meet at an apex, we get the figure of the complete pyramid. (See figure no. 3)

We know that distances:

ZY = 7 *rajjus *= h

CD = 7 *rajjus *= d_{2}

CY = 7/2 *rajjus *= r_{2}

AB = 1 *rajju *= d_{1}

AZ = 1/2 *rajju *= r_{1}

We have to find out the volume of the whole pyramid XCD

and also the small pyramid XAB. Then we get the volume of *adholoka *as -

= Volume of Pyramid **XCD **- Volume of Pyramid **XAB**

First we have to calculate the height XZ, say h_{1}.

Let Q denote the ∠XAZ, which is equal to ∠XCY. Then

∴ h1 = 14/12 = 7/6

Thus, XZ = 7/6 *rajjus*

and XY = 7 + 7/6 *rajjus*

= 49/6 *rajjus*

Volume of a right conical pyramid is given by

V = 1/3 πr^{2}h

Thus, volume of pyramid XAB

V1 = 1/3 π(½)^{2} x (7/6)

Taking the value of π as 355/113, we get

By comprising this value with the value of the volume of *adholoka *given by Vīrasenācārya, which is 106 261/1356 cubic *rajjus*, we find that the latter is not exact. This is for two reasons:

(1) The value of π taken is 371/113 (= 3.28...... approximately) instead of actual value of π taken by us which is 355/113 (= 3.14......approximately).

If we also take

π = 371/113

We will get

V = 109 335/4068

In the same way, we can also find out the volume of *ūrdhvaloka*.

The *ūrdhvaloka *can be divided into two equal halves.

In the upper half (see fig. 4), the distance XZ (h_{1}) can be found as before thus:

∴ The total volume of *ūrdhvaloka *2 x V = 56 1267/1356 cubic *rajjus*. Here, also the difference in the volume found by Vīrasenācārya and that found through modern mathematical method is worth noting.