Area of intersection using integration

Question

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?

Answer 1

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}

Answer 2

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.