Step 1:

Area of circle $x^2+y^2=a^2$

$\quad=4 \times$ area in first quadrant bounded by $x^2+y^2=a^2,x=0,x=a$

Step 2:

$A= 4 \int \limits _0^a ydx=4 \int \limits_0^a \sqrt {a^2-x^2}dx$

$\quad= 4 \bigg[\large\frac{a^2}{2} $$\sin ^{-1} \large\frac{x}{a} -\frac{x}{2} $$\sqrt {a^2-x^2}\bigg]_0^a$

$\quad= 4 \bigg[\large\frac{a^2}{2} $$\sin ^{-1} 1\bigg]$

$\quad=4. \large\frac{a^2}{2}.\large\frac{\pi}{2}$

$\quad= \pi a^2 sq.units$