2.54. A parallelogram with a perimeter of $2 p$ rotates around an axis perpendicular to a diagonal of length $d$ and passing through its end. Find the surface area of the solid of revolution.
2.54. The surface $S$ of the body of revolution consists of the lateral surfaces of two truncated cones obtained by rotating segments $B C$ and $C D$ (Fig. $2.44$), and two cones obtained by rotating segments $A B$ and $A D$. Thus,

$$
\begin{gathered}
S=\pi(K B+A C) B C+ \\
+\pi(M D+A C) C D+ \\
+\pi K B \cdot A B+\pi M D \cdot A D
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-484.jpg?height=454&width=585&top_left_y=417&top_left_x=567)

Fig. 2.44

Let's transform this expression, considering that $A D=B C, C D=A B$ and $K B+M D=A C$. We have

$$
\begin{gathered}
\pi(K B \cdot B C+A C \cdot B C+M D \cdot A B+A C \cdot A B+K B \cdot A B+M D \times \\
\times B C)=\pi((K B+M D) B C+(K B+M D) A B+A C \cdot B C+A C \cdot A B)= \\
=\pi(2 B C+2 A B) A C
\end{gathered}
$$

Since by condition $A C=d, 2 B C+2 A B=2 p$, then $S=2 \pi d p$.

Answer: $2 \pi d p$.