# Either of two unit spheres passes through the center of the other. Without using integration, how do you prove that the the common volume is nearly 1.633 cubic units?

Sep 20, 2016

See explanation for proof.

#### Explanation:

The two spherical surfaces meet along a small circle of radius

$\frac{\sqrt{3}}{2}$ units, with center as the midpoint of the line joining the

centers of the spheres.

Use the formula:

The volume of a conical part of the unit sphere, with conical (semi-

vertical) angle $\alpha$ radian and vertex at the center of the

sphere

$= \frac{4}{3} \alpha \sin \alpha$ cubic units.

If the radius is a, this will be $\frac{4}{3} {a}^{3} \alpha \sin \alpha$

Here, the common volume

= 2( volume of the cone-ice like part of the unit sphere with angle

$\frac{\pi}{3}$ less the volume of its part in the form of a right circular cone

of height 1/2 unit and radius of the base $\frac{\sqrt{3}}{2}$ units)

$= 2 \left(\left(\frac{4}{3}\right) \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{3}\right) - \frac{1}{3} \pi {\left(\frac{\sqrt{3}}{2}\right)}^{2} \left(\frac{1}{2}\right)\right)$ cubic units

$= \frac{2}{3} \pi \left(\frac{2}{3} \sqrt{3} - \frac{3}{8}\right)$ cubic units

$= 1.633001$ cubic units, nearly.

Graph for two conjoined unit spheres, each passing through the center

of the other.

graph{((x-0.5)^2+y^2-1)((x+0.5)^2+y^2-1)=0[-2 2 -1.2 1.2]}

For this matter, I give graphs for 4 and 8 conjoined unit spheres

resting on a Table such that each passes through the center of the

opposite sphere. These reveal the middle planar section of the

common-to-all space.

graph{((x-0.5)^2+y^2-1)((x+0.5)^2+y^2-1)((y-0.5)^2+x^2-1)((y+0.5)^2+x^2-1)=0[-4 4 -2.2 2.2]}
graph{((x-0.5)^2+y^2-1)((x+0.5)^2+y^2-1)((y-0.5)^2+x^2-1)((y+0.5)^2+x^2-1)((x-0.3536)^2+(y-0.3536)^2-1)((x+0.3536)^2+(y-0.3536)^2-1)((y+0.3536)^2+(x+0.3536)^2-1)((y+0.3536)^2+(x-0.3536)^2-1)=0[-1 1 -.6 .6]}