Question

A right circular cylinder is inscribed in a sphere of radius r. Find the largest possible volume of such a cylinder.

Short answer

The maximum volume of the cylinder in a sphere is \(\frac{{4\pi {r^3}}}{{3\sqrt 3 }}\).

Step-by-step solution

Step 1-Find the function for the volume of the cylinder

The figure below represents the sketch of the cylinder inside the sphere of radius r as:

The volume of the cylinder is:

\(V = \pi {y^2}\left( {2x} \right)\)

The equation of the circle is \({x^2} + {y^2} = {r^2}\), so the equation of the volume is:

\(\begin{aligned}{c}V = \pi \left( {{r^2} - {x^2}} \right)\left( {2x} \right)\\ = 2\pi \left( {x{r^2} - {x^3}} \right)\end{aligned}\)

Step 2-Differentiate the function of volume

Differentiate the function \(V\left( x \right) = 2\pi \left( {x{r^2} - {x^3}} \right)\).

\(\begin{aligned}{c}V'\left( x \right) = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {2\pi \left( {x{r^2} - {x^3}} \right)} \right)\\ = 2\pi \left( {{r^2} - 3{x^2}} \right)\\ = 2\pi {r^2} - 6\pi {x^2}\end{aligned}\)

Step 3-Find the values of x at which volume is maximum

Find the roots of \(V'\left( x \right) = 0\).

\(\begin{aligned}{c}2\pi {r^2} - 6\pi {x^2} = 0\\{x^2} = \frac{{{r^2}}}{3}\\x = \frac{r}{{\sqrt 3 }}\end{aligned}\)

So, according to the first derivative test, the value of area will be maximum when \(x = \frac{r}{{\sqrt 3 }}\).

Step 4-Find the maximum volume of the cylinder

Substitute \(\frac{r}{{\sqrt 3 }}\) for x in the equation \(V = 2\pi {r^2} - 6\pi {x^2}\).

\(\begin{aligned}{c}V = 2\pi \left( {\left( {\frac{r}{{\sqrt 3 }}} \right){r^2} - {{\left( {\frac{r}{{\sqrt 3 }}} \right)}^3}} \right)\\ = 2\pi \left( {\frac{{{r^3}}}{{\sqrt 3 }} - \frac{{{r^3}}}{{3\sqrt 3 }}} \right)\\ = \frac{{2\pi {r^3}}}{{\sqrt 3 }}\left( {\frac{2}{3}} \right)\\ = \frac{{4\pi {r^3}}}{{3\sqrt 3 }}\end{aligned}\)

Thus, the maximum volume of the cylinder in the sphere is \(\frac{{4\pi {r^3}}}{{3\sqrt 3 }}\).