A cube C of edge length 1 is rotated around a line passing through two opposite vertices, thereby sweeping out a solid S of revolution. Find the volume of S.
Any hints would be much appreciated.
1. Draw a sketch. Basically you are dealing with a rectangle whose length is $\displaystyle l=\sqrt{2}$ and whose width is $\displaystyle w = 1$.
Split this rectangle into two congruent right triangles.
2. By rotating the rectangle about it's diaogonal you'll get a solid composed of two cones and two frustrums of cones.
The height of the solid is $\displaystyle H=\sqrt{3}$.
The base radius of the two cones is: $\displaystyle r_c=\dfrac{1 \cdot \sqrt{2}}{\sqrt{3}}=\dfrac13 \cdot \sqrt{6}$
The height of the two cones is: $\displaystyle h_c=\dfrac13 \cdot \sqrt{3}$
The second radius of the two frustrums of cones is: $\displaystyle r_{f2}=\dfrac14 \cdot \sqrt{6}$
The height of the two frustrums of cones is: $\displaystyle h_f=\dfrac16 \cdot \sqrt{3}$.
3. To calculate the complete volume of the solid use the formulas of the volume of a cone: $\displaystyle V_c = \dfrac13 \cdot \pi \cdot r_c^2 \cdot h_c$
and the formula of the volume of a drustrum of a cone: $\displaystyle V_f=\dfrac13 \cdot h_f \left(r_c^2+r_c \cdot r_{f2}+r_{f2}^2 \right)$