Chapter 12.6, Problem 18PSD

a.

To determine

To compare: The cross-sectional areas of the cylinder and hemisphere.

Answer to Problem 18PSD

The corresponding sections of the cylinder and hemisphere have the same area.

Explanation of Solution

Given Information:

Cylinder and hemisphere have equal heights.

Formula used:

Area of a circle $\left(A\right)=\pi r^2$

  $"r"$ is the radius of the circle

Calculation:

In the cylinder, we need to find the area of a circular ring.

Area of the circular ring $\left(A\text{'}\right)=$ Area of the bigger circle − Area of the smaller circle … (i)

We know that, Area of a circle $\left(A\right)=\pi r^2$

Let $r_1\text{and}{r}_2$ be the radii of the bigger circle and smaller circle respectively.

  $\therefore$ Area of the bigger circle $\left(A_1\right)=\pi {\left(r_1\right)}^2$

  $\begin{array}{l}\therefore A_1=\pi {\left(R\right)}^2\\ \therefore A_1=\pi R^2\mathrm{...}\left(\text{ii}\right)\end{array}$

Now, Area of the smaller circle $\left(A_2\right)=\pi {\left(r_2\right)}^2$

  $\begin{array}{l}\therefore A_2=\pi {\left(\frac{R}{2}\right)}^2\\ \therefore A_2=\frac{\pi R^2}{4}\mathrm{...}\left(\text{iii}\right)\end{array}$

Using (i), we get

  $\begin{array}{l}A\text{'}=A_1-A_2\\ \therefore A\text{'}=\pi R^2-\frac{\pi R^2}{4}\\ \therefore A\text{'}=\frac{3\pi R^2}{4}\mathrm{...}\left(\text{iv}\right)\end{array}$

In the hemisphere, we need to find the area of a circle with radius $R$

We know that, Area of a circle $\left(A\right)=\pi r^2$

Let $r_3$ be the radius of the circle.

  $\therefore$ Area of the circle $\left(A"\right)=\pi {\left(r_3\right)}^2$

  $\begin{array}{l}\therefore A"=\pi {\left(\frac{R\sqrt{3}}{2}\right)}^2\\ \therefore A"=\frac{3}{4}\pi R^2\mathrm{...}\left(\text{v}\right)\end{array}$

From (iv) and (v), we get

The corresponding sections of the cylinder and hemisphere have the same area at the level mentioned.

b.

To determine

To find: The volume of the hemisphere using Cavalieri’s Principal.

Answer to Problem 18PSD

The volume of the hemisphere is $\frac{2}{3}\pi R^3.$

Explanation of Solution

Given Information:

From part a, cross-sectional areas of cylinder and hemisphere are equal.

Radius of cylinder = radius of cone $\left(r\right)=R$

Height of cylinder = height of cone $=h$

Formula used:

Volume of a cylinder $\left(V\right)=\pi r^{2}h$

  $"r"\text{and}"h"$ are the radius and height of the cylinder

Volume of a cone $\left(V\text{'}\right)=\frac{1}{3}\pi {\left(r\text{'}\right)}^{2}h\text{'}$

  $r\text{'}\text{and}h\text{'}$ are the radius and height of the cone

Calculation:

The corresponding sections of the cylinder and hemisphere have the same area at the level mentioned.

Thus, by using Cavalieri’s principle we can conclude that the two solids will have equal volumes.

Volume of the $1^{\text{st}}$ solid $\left(V_1\right)=$ Volume of cylinder − volume of cone

We know that, Volume of a cylinder $\left(V\right)=\pi r^{2}h$

Volume of a cone $\left(V\text{'}\right)=\frac{1}{3}\pi {\left(r\text{'}\right)}^{2}h\text{'}$

  $\therefore V_1=V-V\text{'}$

  $\begin{array}{l}\therefore V_1=\pi \times R^2\times h-\frac{1}{3}\pi R^{2}h\\ \therefore V_1=\frac{2}{3}\pi R^{2}h\end{array}$

Volume of the hemisphere $\left(V_2\right)=$ Volume of $1^{\text{st}}$ solid

  $\therefore V_2=V_1$

  $\begin{array}{l}\therefore V_2=\frac{2}{3}\pi R^{2}h\\ \therefore V_2=\frac{2}{3}\pi R^{2}R\\ \therefore V_2=\frac{2}{3}\pi R^3\end{array}$

Hence, volume of the hemisphere is $\frac{2}{3}\pi R^3.$