Solving an integral without calculus

I've always been a fan of solving integrals using goemetry or probability even if they are pretty trivial and / or straightforward with calculus. The content of this post is such one such integral which I encountered while studying the Kepler problem.

The subject of this post will be the following integrals.

$\displaystyle \int_b^a\sqrt{(a-r)(r-b)}\,dr$, $\displaystyle \int_b^a\frac{\,dr}{\sqrt{(a-r)(r-b)}}$ and $\displaystyle \int_b^a\frac{r\,dr}{\sqrt{(a-r)(r-b)}}$.

Consider the two concentric circles as shown above with center $D$ such that $\vert BC \vert=b$ and $\vert AC \vert=a$. The radius of the larger (red) circle is then $(a+b)/2$ and that of the smaller (orange) circle is $(a-b)/2$.

Two concentric circles

Now let $r=\vert BH \vert$. Then with simple geometry it can be seen that $\vert GH \vert=\sqrt{(a-r)(r-b)}$.

Then the $\vert GH \vert\,dr=\sqrt{(a-r)(r-b)}\,dr$ is the area of the thin infinitesimal rectangular strip with of dimensions $\vert GH \vert$ and $dr$. It is now easy to see that the first integral is just half the area of orange circle. That is,

$\displaystyle \int_b^a\sqrt{(a-r)(r-b)}\,dr=\frac{1}{2}\pi\left(\frac{a-b}{2}\right)^2$

Again from simple geometry, we have

$\displaystyle \frac{\vert GH \vert}{\vert DG \vert}=\frac{-d(\vert DH \vert)}{ds}$

where $ds$ is the infinitesimal arclength of the orange circle i.e. $ds=\vert DG \vert d\theta$ where $\theta=\angle BDF$.

Because $\vert DH \vert=\vert DB \vert - r$, we have $-d(\vert DH \vert)=dr$. Using all these and substituting, we have,

$\displaystyle \,d\theta=\frac{\,dr}{\sqrt{(a-r)(r-b)}}$

As $H$ moves from $C$ to $C'$, $r$ goes from $b$ to $a$ and $\theta$ from $0$ to $\pi$. Therfore,

$\displaystyle \int_b^a\frac{\,dr}{\sqrt{(a-r)(r-b)}}=\pi$

Similarly,

$\displaystyle \frac{\vert DH \vert}{\vert DG \vert}=\frac{d(\vert GH \vert)}{ds} \implies \vert DB \vert - r=\frac{d(\vert GH \vert)}{d\theta} \implies \vert DB \vert d\theta - r d\theta=d(\vert GH \vert)$

Summing up the infinitesimal changes of $\vert GH \vert$ as $\theta$ moves from $0$ to $\pi$ results in $0$ as $G$ rises in height from the diametric line $DB$ and comes back to the same line. Substituting for $d\theta$ from above, we have,

$\displaystyle \vert DB \vert \pi-\int_b^a \frac{r\,dr}{\sqrt{(a-r)(r-b)}}=0$ (or) $\displaystyle \int_b^a \frac{r\,dr}{\sqrt{(a-r)(r-b)}}=\pi\left(\frac{a+b}{2}\right)$

Hope you enjoyed the discussion.

Until then

Your Aye

Me