College Math Teaching

April 18, 2012

Construction of the antiderivative of sqrt(r^2 – x^2) without trig substitution

Filed under: calculus, integrals, integration by substitution, pedagogy, popular mathematics — collegemathteaching @ 1:15 am

The above figure shows the graph of $y = \sqrt{r^2 - x^2}$ for $r = 1$.
Let $A(t) = \int^t_0 \sqrt{r^2 - x^2} dx$ This gives the area bounded by the graphs of $y=0, x = t, x =0, y = \sqrt{r^2 - x^2}$.

We break this down into the wedge shaped area and the triangle.
Wedge shaped area: has area equal to $\frac{1}{2} \theta r^2$ where $\theta$ is the angle made by the $y$ axis and the radius.
Triangle shaped area: has area equal to $\frac{1}{2}xy = \frac{1}{2}t\sqrt{r^2 - t^2}$
But we can resolve $\theta = arcsin(\frac{t}{r})$ so the wedge shaped area is $\frac{1}{2}r^2arcsin(\frac{t}{r})$ hence:
$A(t) = \int^t_0 \sqrt{r^2 - x^2} dx = \frac{1}{2}r^2arcsin(\frac{t}{r}) + \frac{1}{2}t\sqrt{r^2 - t^2}$

We can verify this by computing $A'(t) = \frac{1}{2} \frac{1}{r}\frac{r^2}{\sqrt{ 1- (\frac{t}{r})^2}} + \frac{1}{2}\sqrt{r^2 - t^2}-t^2\frac{1}{2}\frac{1}{\sqrt{ r^2- t^2}}$
$= \frac{1}{2} \frac{r^2}{\sqrt{ r^2- t^2}}-t^2\frac{1}{2}\frac{1}{\sqrt{ r^2- t^2}} + \frac{1}{2}\sqrt{r^2 - t^2}$
$= \frac{1}{2} \frac{r^2 - t^2}{\sqrt{ r^2- t^2}} + \frac{1}{2}\sqrt{r^2 - t^2} = \sqrt{r^2 - t^2}$ as required.