Creative way to find this area Let's say We have a circle with center at $(0,0)$ with radius $r$
and we have the line $y=a$ where $0 \leq a \leq r$.
the question is what is the area that between the circle and the line $y=a$(the area that above the line).
illustration for $r=5$ and $a=4$: http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%3D25+and+y%3D4
We will end with the area as a function of $a$. So far I have tried to do integration to get the area ... it is not pretty at all.

Using, geometry we have the equation of the circle centered at the origin as:
$$x^2+y^2=r^2$$
Now, substitute $y=a$ in the above equation
$$x^2+a^2=r^2 \implies x=\pm \sqrt{r^2-a^2}$$
Now, join the points of intersection $(\sqrt{r^2-a^2}, a),\,(-\sqrt{r^2-a^2}, a)$ to the origin, to get a sector with radius $r$ and an isosceles triangle with sides $r,\,r,\,2\sqrt{r^2-a^2}$
The aperture angle $\alpha$ of the sector is calculated as
$$\alpha=2\cos^{-1}\left(\frac{a}{r}\right)$$ 
Now, the area between the circle and the line is 
$$
\begin{align}
&\frac{1}{2}\cdot\alpha r^2-\frac{1}{2}(a)(2\sqrt{r^2-a^2})\\
=\:&\frac{1}{2}\left(2\cos^{-1}\left(\frac{a}{r}\right)\right)r^2-\frac{1}{2}a\left(2\sqrt{r^2-a^2}\right)\\
=\:&r^2\cos^{-1}\left(\frac{a}{r}\right)-a\sqrt{r^2-a^2}
\end{align}
$$
Where, $0\leq a\leq r$ 
Hence, as you mentioned for $r=5,\,a=4$. substituting these values in the expression, we get the area:
$$
\begin{align}
&5^2\cos^{-1}\left(\frac{4}{5}\right)-4\sqrt{5^2-4^2}\\
=\:&25\cos^{-1}\left(\frac{4}{5}\right)-12\\
\approx\:&4.08752772 \space \text{unit}^2
\end{align}
$$
