Area of intersection using integration 
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?
 A: 
Consider $x=f(y)$ instead of $y=f(x)$. 
Then the area of intersection is just
\begin{align} 
\int_0^1 \sqrt{R^2-y^2}\,dy
&=\left.\left(
\tfrac12\,y\sqrt{R^2-y^2}
+\tfrac12\,R^2\arctan\frac{y}{\sqrt{R^2-y^2}}
\right)\right|_0^1
\\
&=
\left(
\tfrac12\,\sqrt{R^2-1}
+\tfrac12\,R^2\arctan\frac{1}{\sqrt{R^2-1}}
\right)
\\
&=
\left(
\tfrac12\,\sqrt{\tfrac{9}{4}-1}
+\tfrac12\,\tfrac{9}{4}\arctan\frac{1}{\sqrt{\tfrac{9}{4}-1}}
\right)
\\
&=\tfrac{\sqrt5}4+\tfrac98\,\arctan(\tfrac{2\sqrt5}5)
\approx 1.38
.
\end{align}  
A: The area of the rectangle is defined by $A_r=w*h\implies A_r=2*1\implies A_r=2$.  However, that is immaterial, as all you want is the area of the circle bounded by $y=0$, $y=1$, $x=0$ and $y^2=(1.5)^2-x^2$.  Solve the circle equation for $x$ and take the definite integrate for $y$ between $0$ and $1$.
NOTE: the circle is not of radius $1$, so you will not get a quarter circle.  You could rescale everything (multiply by $\frac{2}{3}$ so $1.5$ becomes $1$ and the $y=0$ and $y=1$ become $y=0$ and $y=\frac{2}{3}$); then integrate and multiply the resultant area by $(\frac{3}{2})^2$ to put the area back to scale.
