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,

Hence the area of the whole circle will be:

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

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.



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