Chapter 12.4, Problem 32WE

a.

To determine

To find: the volume of cone inscribed in the sphere with radius $10$ .

Answer to Problem 32WE

The volume of inscribed cone is $\frac{1}{3}\pi \left(100-x^2\right)\left(10+x\right)$ .

Explanation of Solution

Given: A cone inscribed in the sphere with radius $10$ and perpendicular distance from center of sphere to the base of cone is $x$ .

  

Calculation:

Calculate the radius of base of inscribed cone say $r$ and height $h$ .

  $\begin{array}{l}r=\sqrt{{10}^2-x^2}\\ r=\sqrt{100-x^2}\end{array}$

And

  $h=10+x$

Calculate the volume of inscribed cone.

  $\begin{array}{l}V=\frac{1}{3}\pi r^{2}h\\ =\frac{1}{3}\pi {\left(\sqrt{100-x^2}\right)}^2\left(10+x\right)\\ =\frac{1}{3}\pi \left(100-x^2\right)\left(10+x\right)\end{array}$

Therefore, the volume of inscribed cone is $\frac{1}{3}\pi \left(100-x^2\right)\left(10+x\right)$ .

b.

To determine

To find: the maximum volume of cone by substituting the value of $x$ .

Answer to Problem 32WE

The maximum volume of inscribed cone is $\frac{32000\pi }{81}{\text{ unit}}^3$ .

Explanation of Solution

Given:

Value of $x$ to substitute:

  $x=\frac{10}{3}$

Calculation:

Calculate the volume of cone by substituting $x$ .

  $\begin{array}{l}V=\frac{1}{3}\pi \left(100-x^2\right)\left(10+x\right)\\ =\frac{1}{3}\pi \left(10+\frac{10}{3}\right)\left(100-{\left(\frac{10}{3}\right)}^2\right)\\ =\frac{1}{3}\pi \left(\frac{40}{3}\right)\left(100-\frac{100}{9}\right)\\ =\frac{1}{3}\pi \left(\frac{40}{3}\right)\left(\frac{900-100}{9}\right)\end{array}$

  $\begin{array}{l}=\frac{1}{3}\pi \left(\frac{40}{3}\right)\left(\frac{800}{9}\right)\\ =\frac{1}{3}\pi \left(\frac{32000}{27}\right)\\ =\frac{32000\pi }{81}{\text{ unit}}^3\end{array}$

Therefore, the maximum volume of inscribed cylinder is $\frac{32000\pi }{81}{\text{ unit}}^3$ .