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.


  1. Here’s a video of this from my MOOCulus course.

    Comment by Jim Fowler — August 2, 2013 @ 7:13 pm

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

Blog at

%d bloggers like this: