Chapter 6, Problem 7P

To determine

To find : The height of $P$ and the volume of the intersection of sphere ( $S$ )and pyramid (

  $P$ )

Answer to Problem 7P

The required volume is $V=\left(\frac{28}{27}\sqrt{6}-2\right)2b^3$

Explanation of Solution

Given information :

The pyramid has a base of side $2b$ .

The sphere (S) its center on the base of pyramid (P) and S is tangent to all eight edges of P.

Formula used :

  $V_{cap}=\frac{\pi }{3}{H}^2\left(3R-H\right)$

R= base radius

H=height of the spherical segment.

Calculation :

  

The figure-I shows a cross section of pyramid passing through the top and through two opposite corners of the square base.

  

  

Now, $\left|BD\right|=b$ , since it is a radius of the sphere which has diameter $2b$ since it is tangent to the opposite sides of the square base. Also $\left|AD\right|=b$ since $△ABD$ is isosceles. So, the height is $\left|AB\right|=\sqrt{a^2+b^2}=\sqrt{2}b$

We first observe that the shaded volume is equal to (half the volume of the sphere) − (sum of the four equal volumes cut off by the triangular faces of the pyramid).

To find the volume of four spherical caps we need the distance $h$ in figure II.

We have $h=b-d$

Now from similar triangles $△ABC$ and $△BCD$

  $\frac{d}{b}=\frac{\left|AB\right|}{\left|AC\right|}=\frac{\sqrt{2}b}{\sqrt{b^2+2b^2}}=\frac{\sqrt{6}}{3}\text{ [as AB is height of isosceles }△ABC]$

Therefore, $d=\frac{\sqrt{6}}{3}$

So, $h=b-d=\frac{3-\sqrt{6}}{3}b$

So, volume of each cap $=\pi h^2\left(r-\frac{h}{3}\right)$

Here, $r=b$

  $\begin{array}{c}{V}_{cap}=\pi {\left(\frac{3-\sqrt{6}}{3}\right)}^2{b}^2\left(b-\frac{3-\sqrt{6}}{9}b\right)\\ =\left(\frac{2}{3}-\frac{7}{27}\sqrt{6}\right)\pi b^3\end{array}$

So, required volume is

  $\begin{array}{c}V=\frac{1}{2}\frac{\pi }{3}\pi b^3-4\left(\frac{2}{3}-\frac{7}{27}\sqrt{6}\right)\pi b^3\\ =\left(\frac{28}{27}\sqrt{6}-2\right)2b^3\end{array}$