1. ## cyclinder

two cylinders are inscribed in a cube of side length 2. what is the volume of the solid that the two cylinders enclose? Use horizontal slices.

I found the solution known to Archimedes but it does not do it by horizontal slices.

Any suggestions or tips are greatly appreciated!!!

2. Originally Posted by ca27brin
two cylinders are inscribed in a cube of side length 2. what is the volume of the solid that the two cylinders enclose? Use horizontal slices.

I found the solution known to Archimedes but it does not do it by horizontal slices.

Any suggestions or tips are greatly appreciated!!!
I have some difficulties to imagine how the two cylinders are placed in the cube. I've sketched a possible situation. (see attachment)

If this is correct you have to calculate the painted area which is the base of the volume surrounding the two cylinders.

I consider only the half of the base square of the cube. Then the base circle of one cylinder is the inscribed circle to the isosceles right triangle forming the half of the base square.
Let s denote the side of the square and d the diagonal of the square.
Then $\displaystyle d = s \cdot \sqrt{2}$

Then the radius of this circle can be calculated by:

$\displaystyle r = \frac12 \cdot s \cdot \sqrt{2} \cdot \tan(22.5^\circ)$

Since $\displaystyle \tan(22.5^\circ) = \frac{\sin(45^\circ)}{1+\cos(45^\circ)} = \frac{\frac12 \cdot \sqrt{2}}{1+\frac12 \cdot \sqrt{2}} = \sqrt{2} - 1$ the radius is:

$\displaystyle r = \frac12 \cdot s \cdot \sqrt{2} \cdot (\sqrt{2} - 1) = s (1-\frac12 \cdot \sqrt{2})$

With s = 2 the radius $\displaystyle r = 2 \cdot (1-\frac12 \cdot \sqrt{2}) \approx 0.585786...$

The area of the horizontal slice becomes:

$\displaystyle a = s^2 - 2 \cdot \pi \cdot (s (1-\frac12 \cdot \sqrt{2}))^2$

With s = 2 the area is $\displaystyle a \approx 1.84395..$

And therefore the volume surrounding the two cylinders is:

$\displaystyle V = a \cdot s~\implies~V \approx 3.6879..$