How much additional area does a ridge and furrow system create? Various parts of Europe used a Ridge and furrow system for their fields. One side effect is it provides additional land area.
Approximating the ridges and furrows using parts of a circle with radius, r, and, given a depth / height of h from the mean height, how much additional area does it create in comparison to flat land with surface area, A?
E.g. flat land, where r is infinite and h is zero, would give a surface area of A.
At the other extreme, land with half circles for the ridges and furrows, where r = h, would give a surface area of A x (circumference of semi circle / distance across semi circle) = A x (πr / 2r) = Aπ/2.
What is the general solution?

 A: 
We denote the area before ploughing and the area after ploughing as $A_{\mathrm{flat}}$ and $A_{\mathrm{tilled}}$ respectively. We shall write,
$$A_{\mathrm{flat}}= b\times\big[\mathrm{segment}\space AC\big]\quad \mathrm{and}\quad A_{\mathrm{tilled}}= b\times\big[\mathrm{arc}\space ABC\big],$$
where $b$ is the the length of the ridgelet.
Let the height and the radius of the ridgelet be $h$ and $r$ respectively. It follows from the diagram, that
$$BD=h, \quad OA=OB=r, \quad \mathrm{and}\quad OD=r-h.$$
$$\therefore\quad AD=\sqrt{OA^2-OD^2}=\sqrt{r^2-\left(r-h\right)^2}=\sqrt{h\left(2r-h\right)}\quad\rightarrow\quad AC=2AD=2\sqrt{h\left(2r-h\right)}$$
Using this result, we can express $A_{\mathrm{flat}}$ as
$$A_{\mathrm{flat}}= 2b\sqrt{h\left(2r-h\right)}.$$
$$\therefore\quad b=\frac{A_{\mathrm{flat}}}{2\sqrt{h\left(2r-h\right)}},\quad\mathrm{where}\space h\gt 0. \tag{1}$$
If the angle $\measuredangle AOB=\phi$ (Please note that $\phi$ is measured in radians), then we have,
$$\cos\left(\phi\right)=\frac{r-h}{r}.$$
$$\therefore\quad ABC=2r\phi=2r\times\cos^{-1}\left(\frac{r-h}{r}\right) \tag{2}$$
Using (1) and (2), we can express $A_{\mathrm{tilled}}$ in term of $A_{\mathrm{flat}}$ as shown below.
$$A_{\mathrm{tilled}}= 2br\phi=\frac{rA_{\mathrm{flat}}}{\sqrt{h\left(2r-h\right)}}\cos^{-1}\left(\frac{r-h}{r}\right),\quad\mathrm{where}\space h\gt 0$$
Now, if we denote the additional area created by ploughing as $\Delta A$, we can express it as,
$$\Delta A=A_{\mathrm{tilled}}-A_{\mathrm{flat}}=\left(\frac{r}{\sqrt{h\left(2r-h\right)}}\cos^{-1}\left(\frac{r-h}{r}\right)-1\right)A_{\mathrm{flat}},\quad\mathrm{where}\space h\gt 0. \tag{3}$$
You can check the validity of this formula by substituting $r$ in place of $h$. If you do so, you will get the value of $\Delta A$ corresponding to the semicircular ridgelet, i.e. $\left(\frac{\pi}{2}-1\right)A_{\mathrm{flat}}.$
If you have trouble comprehending trigonometric identities, we give below another formula to determine $\Delta A$, which, however, only yields an approximate value. Besides, the values of $\Delta A$ you obtain from this equation are good if and only if $h$ is almost equal to $r$.
$$\Delta A \approx \left(\frac{3\pi+9k-3k^2+k^3-7}{6\sqrt{k\left(2-k\right)}}-1\right)A_{\mathrm{flat}},\quad\mathrm{where}\quad k=\frac{h}{r}\gt 0 \tag{4}$$
A: Series expansion of YNK's answer in Mathematica
$$\frac{0.285763}{\sqrt{k}}+1.1321 \sqrt{k}-0.0615981 k^{3/2}+0.140062 k^{5/2}+0.0426328 k^{7/2}+0.017562 k^{9/2}+0.00772557 k^{11/2}+0.00350133 k^{13/2}+0.00161512 k^{15/2}+O\left(k^{17/2}\right).$$
