12.419 A rectangle rotates around an axis passing through its vertex parallel to the diagonal. Find the surface area of the solid of revolution, if the area of the rectangle is $S$, and the angle between the diagonals is $\alpha$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-298.jpg?height=283&width=391&top_left_y=568&top_left_x=148)

Let \(ABCD\) be the given rectangle rotating around the line \(l\), where \(l \parallel BD\). The surface of the solid of revolution \(T\) consists of the surfaces of two cones with axis \(l\) and generators \(BA\) and \(AD\), as well as two frustums of cones with axis \(l\) and generators \(BC\) and \(CD\). Drop perpendiculars \(BN\), \(CK\), and \(DM\) to the line \(l\).

Then the area \(S_T\) of the surface of the solid \(T\) is:

\(\widehat{S_T = \pi \cdot AB} \cdot BN + \pi \cdot AD \cdot DM + \pi \cdot BC(BN + CK) + \pi \cdot CD(DM + CK)\).

But \(BN = DM\), \(AB = CD\), \(AD = BC\), and \(CK = 2 \cdot DM\), since \(BD\) is the midline of \(\triangle ACK\) and, consequently, divides \(CK\) in half \(\Rightarrow S_T = 4 \pi (AB + AD) \cdot DM\). Note that \((AB + AD)^2 = AB^2 + AD^2 + 2 \cdot AB \cdot AD = BD^2 + 2S\);

\(\frac{1}{2} DM \cdot BD = S_{ABD} = \frac{S}{2} \Rightarrow DM = \frac{S}{BD}; S = \frac{1}{2} BD^2 \sin \alpha \Rightarrow \frac{2S}{BD^2} = \sin \alpha\).

Thus, we have:

\(S_T = 4 \pi \frac{S}{BD} \sqrt{BD^2 + 2S} = 4 \pi S \sqrt{1 + \frac{2S}{BD^2}} = 4 \pi S \sqrt{1 + \sin \alpha} = 4 \sqrt{2} \pi S \cdot \sin \left(\frac{\alpha}{2} + \frac{\pi}{4}\right)\).

Answer: \(4 \sqrt{2} \pi S \cdot \sin \left(\frac{\alpha}{2} + \frac{\pi}{4}\right)\).