An oblique cylinder has a characteristic cross-section that is a rhombus with side 2 cm and angle 60°. Find its volume.

Step-by-step solution

1. 1

   Recall that obliqueness does not change the volume formula. By Cavalieri's principle, for any cylinder — right or oblique —

   $V = (\text{base area}) \times (\text{height})$

   where the height is measured perpendicular to the bases, not along the slanted axis.

2. 2

   Understand what the characteristic cross-section is. It is the section cut by the plane through the axis and perpendicular to the bases. Its area is

   $A_{\text{char}} = (\text{base diameter}) \times (\text{height}) = 2r \cdot h$

3. 3

   Compute the rhombus area. For a rhombus with side $a$ and included angle $\varphi$,

   $A_{\text{rhombus}} = a^2 \sin\varphi = 2^2 \sin 60^\circ = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}\ \text{cm}^2$

4. 4

   Link the two. Setting $2rh = 2\sqrt{3}$ gives

   $rh = \sqrt{3}$

5. 5

   Use the rhombus condition to pin down $r$ and $h$ separately. In an axial section the two side lengths are the base diameter $2r$ and the generator length $L$. A rhombus has all four sides equal, so

   $2r = L = a = 2 \implies r = 1,\qquad L = 2$

   The perpendicular height is then $h = L\sin\varphi = 2\sin 60^\circ = \sqrt{3}$.

6. 6

   Substitute into the volume formula.

   $V = \pi r^2 h = \pi (1)^2 \sqrt{3} = \sqrt{3}\,\pi \approx 5.44\ \text{cm}^3$

   A quick consistency check: $2rh = 2(1)(\sqrt{3}) = 2\sqrt{3}$, which matches the rhombus area from step 3.