# Area under quarter circle by integration

How would one go about finding out the area under a quarter circle by integrating. The quarter circle's radius is r and the whole circle's center is positioned at the origin of the coordinates. (The quarter circle is in the first quarter of the coordinate system)

From the equation $x^2+y^2=r^2$, you may express your area as the following integral $$A=\int_0^r\sqrt{r^2-x^2}\:dx.$$ Then substitute $x=r\sin \theta$, $\theta=\arcsin (x/r)$, to get \begin{align} A&=\int_0^{\pi/2}\sqrt{r^2-r^2\sin^2 \theta}\:r\cos \theta \:d\theta\\ &=r^2\int_0^{\pi/2}\sqrt{1-\sin^2 \theta}\:\cos\theta \:d\theta\\ &=r^2\int_0^{\pi/2}\sqrt{\cos^2 \theta}\:\cos\theta \:d\theta\\ &=r^2\int_0^{\pi/2}\cos^2 \theta \:d\theta\\ &=r^2\int_0^{\pi/2}\frac{1+\cos(2\theta)}2 \:d\theta\\ &=r^2\int_0^{\pi/2}\frac12 \:d\theta+\frac{r^2}2\underbrace{\left[ \frac12\sin(2\theta)\right]_0^{\pi/2}}_{\color{#C00000}{=\:0}}\\ &=\frac{\pi}4r^2. \end{align}
• Yes, we have, for $0<x<r$, $\frac{d\theta}{dx}=\frac{1}{\sqrt{r^2-x^2}}>0$, $0=\arcsin (0/r) \leq \theta (r)\leq \arcsin (r/r)=\pi/2$. Thanks! Oct 8 '15 at 18:45
Here is a quicker solution. The area can be seen as a collection of very thin triangles, one of which is shown below. As $d\theta\to0$, the base of the triangle becomes $rd\theta$ and the height becomes $r$, so the area is $\frac12r^2d\theta$. The limits of $\theta$ are $0$ and $\frac\pi2$. $$\int_0^\frac\pi2\frac12r^2d\theta=\frac12r^2\theta|_0^\frac\pi2=\frac14\pi r^2$$
let circle: $x^2+y^2=r^2$ then consider a slab of area $dA=ydx$ then the area of quarter circle $$A_{1/4}=\int_0^r ydx=\int_0^r \sqrt{r^2-x^2}dx$$ $$=\frac12\left[x\sqrt{r^2-x^2}+r^2\sin^{-1}\left(x/r\right)\right]_0^r$$
$$=\frac12\left[0+r^2(\pi/2)\right]=\frac{\pi}{4}r^2$$ or use double integration: $$=\iint rdr d\theta= \int_0^{\pi/2}\ d\theta\int_0^R rdr=\int_0^{\pi/2}\ d\theta(R^2/2)=(R^2/2)(\pi/2)=\frac{\pi}{4}R^2$$