$A=\int_0^r\sqrt{r^2-x^2}\,dx$ is the area of a quarter-circle. Show why, with a graph and thin rectangles. Calculate this integral by substituting $x=r\sin\theta$ and $\,dx = r\cos\theta\,d\theta$.
As shown below, the area can be seen as a collection of thin horizontal strips $\sqrt{r^2-x^2}$, $0 \le x \le r$.
By substituting $x=r\sin\theta$ and $dx = r\cos\theta\,d\theta$, we have $$\begin{align} \int_0^r\sqrt{r^2-x^2}\,dx&=\int_0^rr^2\cos^2\theta\,d\theta\\ &=r^2\left(\frac14\sin(2\theta)+\frac12\theta\right)|_0^r\\ &=\frac14r^2\sin(2r)+\frac12r^3\\ \end{align}$$
Why did we not get the area $\frac14\pi r^2$?
$\endgroup$1 Answer
$\begingroup$When you do the substitution $x = r \sin \theta$, the limits of integration of your integral must change as well. That is,
when $x = 0 $, then $\sin \theta = 0 \implies \theta = 0$
when $x = r$, then $\sin \theta =1 \implies \theta = \frac{ \pi }{2} $
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