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 $$ \begin{align} A&=\int_0^{\pi/2}\sqrt{r^2-r^2\sin^2 \theta}\:r\cos \theta \:d\theta\\\ &=r^2\int_0^{\pi/2}\sqrt{1-\sin^2 \theta}\:\cos\theta \:d\theta\\\ &=r^2\int_0^{\pi/2}\sqrt{\cos^2 \theta}\:\cos\theta \:d\theta\\\ &=r^2\int_0^{\pi/2}\cos^2 \theta \:d\theta\\\ &=r^2\int_0^{\pi/2}\frac{1+\cos(2\theta)}2 \:d\theta\\\ &=r^2\int_0^{\pi/2}\frac12 \:d\theta+\frac{r^2}2\underbrace{\left[ \frac12\sin(2\theta)\right]_0^{\pi/2}}_{\color{#C00000}{=\:0}}\\\ &=\frac{\pi}4r^2. \end{align} $$