Page:Calculus Made Easy.pdf/236

to the work done in suddenly compressing the gas) from volume $v_{2}$ to volume $v_{1}$.

Here we have

${\begin{aligned}{\text{area}}&=\int _{v=v_{1}}^{v=v_{2}}cv^{-n}\cdot dv\\&=c\left[{\frac {1}{1-n}}v^{1-n}\right]_{v_{1}}^{v_{2}}\\&=c{\frac {1}{1-n}}(v_{2}^{1-n}-v_{1}^{1-n})\\&={\frac {-c}{0.42}}\left({\frac {1}{v_{2}^{0.42}}}-{\frac {1}{v_{1}^{0.42}}}\right).\end{aligned}}$

An Exercise.

Prove the ordinary mensuration formula, that the area $A$ of a circle whose radius is $R$, is equal to $\pi R^{2}$.

Consider an elementary zone or annulus of the surface (Fig 59), of breadth $dr$, situated at a distance

Fig. 59.

$r$ from the centre. We may consider the entire surface as consisting of such narrow zones, and the whole area $A$ will simply be the integral of all