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.

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- Jan 6th 2011, 12:53 AMalexmahoneVolume of revolution
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. - Jan 6th 2011, 03:22 AMearboth
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: $\displaystyle V_f=\dfrac13 \cdot h_f \left(r_c^2+r_c \cdot r_{f2}+r_{f2}^2 \right)$