11.194. 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.
## Solution.

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. 11.81), and two cones obtained by rotating segments $A B$ and $A D$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0890.jpg?height=666&width=507&top_left_y=63&top_left_x=154)

Fig. 11.80

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0890.jpg?height=481&width=571&top_left_y=260&top_left_x=671)

Fig. 11.81

Thus, $S=\pi(K B+A C) B C+\pi(M D+A C) C D+\pi K B \cdot A B+\pi M D \cdot A D$. Considering that $A D=B C, C D=A D$ and $K B+M D=A C$, we have $\pi(K B \cdot B C+A C \cdot B C+M D \cdot A B+A C \cdot A B+K B \cdot A B+M D \cdot B C)=$ $=\pi((K B+M D) B C+(K B+M D) A B+A C \cdot B C+A C \cdot A B)=$ $=\pi(2 B C+2 A B) A C$.

Given $A C=d, 2 B C+2 A B=2 p$, we have $S=2 \pi d p$.

Answer: $2 \pi d p$.