Page:Calculus Made Easy.pdf/237

such elementary zones from centre to margin, that is, integrated from $r=0$ to $r=R$.

We have therefore to find an expression for the elementary area $dA$ of the narrow zone. Think of it as a strip of breadth $dr$, and of a length that is the periphery of the circle of radius $r$, that is, a length of $2\pi r$. Then we have, as the area of the narrow zone,

$dA=2\pi r\,dr$.

Hence the area of the whole circle will be:

$dA=2\pi r\,dr$.

Now, the general integral of $r\cdot dr$ is ${\tfrac {1}{2}}r^{2}$. Therefore,

${\begin{aligned}A&=2\pi {\bigl [}{\tfrac {1}{2}}r^{2}{\bigr ]}_{r=0}^{r=R};\\or\quad \quad \quad \quad A&=2\pi {\bigl [}{\tfrac {1}{2}}R^{2}-{\tfrac {1}{2}}(0)^{2}{\bigr ]};\\whence\quad \quad \quad \quad A&=\pi R^{2}.\end{aligned}}$.

Another Exercise.

Let us find the mean ordinate of the positive part of the curve $y=x-x^{2}$, which is shown in Fig. 60.

Fig. 60.

To find the mean ordinate, we shall have to find the area of the piece $OMN$, and then divide it by the