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Suppose, for the sake of illustration, that $AF=2$, where $F\mathbb{'s}$ coordinates are $(2,0)$ and $A$ is the point of origin; $E=(0,1)$ and circle $A$ has $R=1.5$

How do I solve the area of the intersection of the circle and rectangle **by integration**?

If we let $f(x)=1$, then the area of the rectangle is defined by:
$$\int\_0^Ff(x)dx$$ 

If we let $g(x)=\sqrt{R^2-x^2}$, then the area of the circle is defined by:
$$4\int\_0^Rg(x)dx$$
Now my best attempt to find the area is through:
$$\int\_0^Rg(x)dx-\int\_0^{\sqrt{R^2-1}}(g(x)-1)dx$$

The first integral gives me a quarter of the area circle $R$, the second gives me the area of the circle that is outside the rectangle, because the circle and the top line intersects at $x=\sqrt{R^2-1}$

Is my solution correct?