Help in finding curve equation.

Question

What I have is length of the bottom line $L$ and area under parabolic curve $S$. How can I find this parabolic curve equation, depending on area under it? The following picture illustrates the problem.

Answer 1

Assume your equation is $y = ax^2$. Compute the area $S$ through integration to find the area under the curve from $x \in [0,L/2]$. You should be able to find $a$ in terms of $L,S$.

Answer 2

Let's assume that the parabola is concave and symmetric with respect to $y$ axis. In this case its equation would look like $y(x) = -ax^2+b$, and its graph would be similar to

Obviously, roots are $\{-\sqrt{\frac ba}, \sqrt{\frac ba}\}$. Also $L = 2\sqrt{\frac ba}$, from which $aL^2 = 4b$. Now, let's find the area under the parabola.

$\displaystyle S = 2\int_0^{\frac L2} \left(-ax^2+b\right)dx = \left.\left(-\frac {ax^3}3 + bx\right)\right|_0^{\frac L2} = \frac L2 \left(b-\frac {aL^2}{12}\right)$.

Now we can use the relation $aL^2 = 4b$, so $\displaystyle S = \frac L2 \left(b-\frac b3\right) = \frac {Lb}3$, and finally $\displaystyle b = \frac {3S}L$ and $\displaystyle a = \frac {12S}{L^3}$.

$\displaystyle y(x) = -\frac {12S}{L^3} x^2 + \frac {3S}L$