10,11

A rhombus, with its smaller diagonal equal to its side, which is 1, rotates around a line passing through the end of the larger diagonal and perpendicular to this diagonal. Find the volume of the solid of revolution.
Let the line $l$ pass through vertex $D$ of the rhombus $ABCD$ perpendicular to the larger diagonal $BD$. Extend $BA$ and $BC$ of the rhombus to intersect line $l$ at points $P$ and $Q$ respectively. Let $M$ and $N$ be the feet of the perpendiculars dropped from points $A$ and $C$ to line $l$. Since the smaller diagonal of the rhombus is equal to its side, the rhombus is divided by the smaller diagonal into two equilateral triangles with side length 1. Thus, $AM$ and $CN$ are the heights of the equilateral triangles $APD$ and $CQD$. Let $V_1$ be the volume of the solid obtained by rotating the equilateral triangle $PBQ$ about the line $l$. Then $V_1$ is equal to twice the volume of a cone with height $DP$ and base radius $DB$, i.e.,

$$
V_1 = 2 \cdot \frac{1}{3} \pi \cdot BD^2 \cdot DP = 2 \cdot \frac{1}{3} \pi \cdot (\sqrt{3})^2 \cdot 1 = 2 \pi
$$

Let $V_2$ be the volume of each of the four equal cones obtained by rotating the right triangles $APM$, $ADM$, $CQN$, and $CDN$ about the line $l$. Then

$$
V_2 = \frac{1}{3} \pi \cdot AM^2 \cdot PM = \frac{1}{3} \pi \cdot \left(\frac{\sqrt{3}}{2}\right)^2 \cdot \frac{1}{2} = \frac{\pi}{8}.
$$

If $V$ is the volume of the desired solid of revolution, then

$$
V = V_1 - 4V_2 = 2\pi - \frac{\pi}{2} = \frac{3\pi}{2}
$$

## Answer

$\frac{3\pi}{2}$.