Question

(a) Determine the more wood content of the ring from two rings.

(b) Check part (a) answer by computing the volume of a napkin ring using cylindrical shells.

Short answer

(a) The both napkin rings have the same quantity of wood.

(b) The volume of a napkin ring is \(\frac{{\pi {h^3}}}{6}\).

Step-by-step solution

Given data

The height of the napkin ring is \(h\). The radius of the napkin ring is \(r\). The radius of the sphere is \(R\).

Concept of Volume

Volume is the measure of the capacity that an object holds.

Determine the volume

(a)

The volume only depends up on the height of the napkin ring. Hence, the both napkin rings have the same quantity of wood.

Determine the volume and substitute the value

(b)

Consider the equation of napkin ring as follows: \(y = \sqrt {{R^2} - {x^2}} \) …….. (1)

The region lies between \(x = r\) and \(x = R\).

Sketch the solid region as shown below in Figure 1.

Calculate the volume and use the method of cylindrical shell as follows

\(V = 2 \times \int_a^b 2 \pi x(f(x))dx\) …….. (2)

Substitute \(r\) for a, R for \(b\), and \(\sqrt {{R^2} - {x^2}} \) for \((f(x))\) in equation (2).

\(\begin{aligned}{}V = 2\int_r^R 2 \pi x\sqrt {{R^2} - {x^2}} dx\\V = 2\pi \int_r^R 2 x\sqrt {{R^2} - {x^2}} dx\end{aligned}\) …….. (3)

Consider \(u = {R^2} - {x^2}\) ……. (4)

Differentiate the equation

Differentiate both sides of the equation.

\(\begin{aligned}{}du = - 2xdx\\ - du = 2xdx\end{aligned}\)

Calculate the lower limit value of \(u\) and use equation (4).

Substitute \(r\) for \(x\) in equation (4).

\(u = {R^2} - {r^2}\)

Calculate the upper limit value of \(u\) and use equation (4).

Substitute \(R\) for \(x\) in equation (4).

\(\begin{aligned}{}u = {R^2} - {R^2}\\u = 0\end{aligned}\)

Apply lower and upper limits for \(u\) in equation (3).

Substitute and integrate the equation

Substitute \(u\) for \(\left( {{R^2} - {x^2}} \right)\) and \(( - du)\) for 2xdxdx in equation (3).

\(\begin{aligned}{l}V = 2\pi \int_{{R^2} + {r^2}}^0 {\sqrt u } ( - du)\\V = - 2\pi \int_{{R^2} + {r^2}}^0 {{u^{\frac{1}{2}}}} du\end{aligned}\)

Integrate equation (5).

\(\begin{aligned}{}V = - 2\pi \left( {\frac{{{u^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}}} \right)_{{R^2} + {r^2}}^0\\V = - 2\pi \left( {\frac{{{u^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right)_{{R^2} + {r^2}}^0\\V = - 2\pi \left( {\frac{{2{{(0)}^{\frac{3}{2}}}}}{3} - \frac{{2{{\left( {{R^2} + {r^2}} \right)}^{\frac{3}{2}}}}}{3}} \right)\\V = \frac{4}{3}\pi {\left( {{R^2} + {r^2}} \right)^{\frac{3}{2}}}\end{aligned}\)

Consider \({R^2} + {r^2} = {\left( {\frac{h}{2}} \right)^2}\)

Substitute \({\left( {\frac{h}{2}} \right)^2}\) for \({R^2} + {r^2}\) in equation (6).

\(\begin{aligned}{}V = \frac{4}{3}\pi {\left( {\frac{h}{2}} \right)^{2 \times \frac{3}{2}}}\\V = \frac{4}{{24}}\pi {h^3}\\V = \frac{1}{6}\pi {h^3}\end{aligned}\)

Hence, the volume of a napkin ring is \(\frac{{\pi {h^3}}}{6}\).