\section*{Problem 16 - V01116}

A trench with a parabolic cross-section is to be excavated. Its width is 3 meters, and its depth is \(b\) meters.

Calculate the cross-sectional area of the trench!
}

To calculate the cross-sectional area of the ditch, we first need to determine a parabolic function equation that passes through the points \(P_{1}(-1.5 ; 0), P_{2}(0, b)\) and \(P_{3}(1.5 ; 0)\). We start with the quadratic function equation \(f(x)=\widetilde{a} x^{2}+\widetilde{b}+\widetilde{c}\). This leads to the linear system of equations

\[
\begin{aligned}
2.25 \cdot \widetilde{a}-1.5 \widetilde{b}+\widetilde{c} & =0 \\
\widetilde{c} & =b \\
2.25 \cdot \widetilde{a}+1.5 \widetilde{b}+\widetilde{c} & =0
\end{aligned}
\]

It holds that \(\widetilde{c}=b\) and by setting the remaining two equations equal, we get \(\widetilde{b}=0\). It remains that \(\widetilde{a}=-\frac{4}{9} b\). Thus, the function equation is

\[
f(x)=-\frac{4 b}{9} x^{2}+b
\]

To determine the cross-sectional area, the integral \(\int_{-1.5}^{1.5} f(x) \mathrm{d} x\) must be solved. It holds that

\[
\begin{aligned}
\int_{-1.5}^{1.5} f(x) \mathrm{d} x & =\int_{-1.5}^{1.5}\left(-\frac{4 b}{9} x^{2}+b\right) \mathrm{d} x=\left[-\frac{4 b}{27} x^{3}+b x\right]_{-1.5}^{1.5} \\
& =-\frac{4 b}{27} \cdot(1.5)^{3}+1.5 b+\frac{4 b}{27} \cdot(-1.5)^{3}+1.5 b=3 b-\frac{8 b}{27} \cdot\left(\frac{3}{2}\right)^{3}=3 b-b=2 b
\end{aligned}
\]

Thus, the cross-sectional area corresponds to twice the depth.