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)
Calculus – Finding the Area Under a Quarter Circle by Integration
areacalculusdefinite integralsintegrationpi
Related Solutions
A figure helps a lot. The three segments labeled $r$ are all radii of the small circle. The diagonal of the square is $r\sqrt 2$, so $r(1+\sqrt 2)=x$ No, the tangency points do not bisect the radii of the big circle, they cut them in ratio $1: \sqrt 2$
Imagine a point $P$ moving along the unit circle from $0$ to $\pi/2$. It has a speed in the $x$-direction, and a speed in the $y$-direction (I'll call these $x$-speed and $y$-speed). Note that the $x$-speed of $P$ is constantly changing, and this is reflected in the graph that it draws (the unit circle): we are drawing the $y$-position, but with a changing $x$-speed.
Whereas, the sine graph is drawing the $y$-position with respect to the angle: this is a constant speed. Near an angle of $0$, the point $P$ has a very small $x$-speed, but the sine graph already has a significant $x$-speed. This results in the sine graph covering more area near the start of the movement of $P$ than $P$ does. Thus, the area under the sine wave is bigger than the area under the unit circle.
You could imagine taking the top half of the unit circle and stretching it at the ends where it hits the $x$-axis, to compensate for the slow $x$-speed, and turning it into the sine graph.
Best Answer
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 $$ \begin{align} A&=\int_0^{\pi/2}\sqrt{r^2-r^2\sin^2 \theta}\:r\cos \theta \:d\theta\\ &=r^2\int_0^{\pi/2}\sqrt{1-\sin^2 \theta}\:\cos\theta \:d\theta\\ &=r^2\int_0^{\pi/2}\sqrt{\cos^2 \theta}\:\cos\theta \:d\theta\\ &=r^2\int_0^{\pi/2}\cos^2 \theta \:d\theta\\ &=r^2\int_0^{\pi/2}\frac{1+\cos(2\theta)}2 \:d\theta\\ &=r^2\int_0^{\pi/2}\frac12 \:d\theta+\frac{r^2}2\underbrace{\left[ \frac12\sin(2\theta)\right]_0^{\pi/2}}_{\color{#C00000}{=\:0}}\\ &=\frac{\pi}4r^2. \end{align} $$