Find $\int^r_{-r} \sqrt{r^2 - x^2} \:dx$ where r is a positive constant.

Question

Find $\int^r_{-r} \sqrt{r^2 - x^2} \:dx$ where $r$ is a positive constant.

Using $x = r\sin(\theta)$ it follows that $dx = r\cos(\theta) \:d\theta$. In addition we need to take into account that $x = \pm r$ results in $\theta = \pm \pi/2$. Therefore we get:

$\int^r_{-r} \sqrt{r^2 - x^2} \:dx = \int^{\pi/2}_{-\pi/2} r\cos(\theta)r\cos(\theta)\:d\theta$.

I'm confused as to how $x = \pm r$ results in $\theta = \pm \pi/2$. $r$ is only said to be a positive constant, so there doesn't seem to be any reason to assume the values $\pi/2$ and $-\pi/2$ for the limits of integration. Why not just use two other values that result in positive and negative values for $\sin$?

I would greatly appreciate it if people could please take the time to explain what's going on here.

Answer 1

If $x=r\sin\theta$ then $\theta=\arcsin(\frac{x}{r})$. Therefore if $x=r$ then $\theta=\arcsin(1)=\frac{\pi}{2}$, while if $x=-r$ then $\theta=\arcsin(-1)=-\frac{\pi}{2}$.

It's worth noting that there's a faster way to evaluate the integral: if $y=\sqrt{r^2-x^2}$ then $x^2+y^2=r^2$. Therefore the integral is equal to the area of the upper half of a disk of radius $r$ (since $y\geq 0$), hence is equal to $\frac{\pi r^2}{2}$.

Answer 2

$ \sqrt{r^2 - x^2} $ this curve represents a semicircle above x-axis with centre at origin and radius $r$. So $\int^r_{-r} \sqrt{r^2 - x^2} \:dx$ is the area of semicircle with radius $r$. Hence $\int^r_{-r} \sqrt{r^2 - x^2} \:dx=\frac{\pi r^2}{2}$