\section*{Problem 1 - 151031}

Given a triangle \(A B C\) with side length \(B C=a\) and height \(A D=h_{a}\). The line \(g\) is the parallel to \(B C\) through \(A\).

Calculate the volume \(V\) of the body that is formed by rotating the triangular area \(A B C\) around \(g\), in terms of \(a\) and \(h_{a}\)!
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1363.jpg?height=397&width=345&top_left_y=607&top_left_x=273)

If the triangle is rotated around the line \(g\), a cylindrical body is formed from which a cone is cut out on the sides of the base and top faces. The cylinder \(Z\) and both cones \(K_{1}, K_{2}\) have the height \(h_{a}\) as their radius. The height of the cylinder is \(s\), and the heights of the two cones are \(x\) and \(a-x\). For the volume of the desired body, we then get:

\(V=V_{Z}-V_{K_{1}}-V_{K_{2}}=\pi \cdot h_{a}^{2} \cdot a-\frac{1}{3} \pi \cdot h_{a}^{2} \cdot x-\frac{1}{3} \pi \cdot h_{a}^{2} \cdot(a-x)=\frac{2}{3} \pi \cdot h_{a}^{2} \cdot a\)