Chapter 12.6, Problem 17PSD

To determine

To show: That the volumes of two cylindrical shells A and B are equal using Cavalieri’s Principle.

Explanation of Solution

Given Information:

Two cylindrical shells have equal heights

Formula used:

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

  $"r"$ is the radius of the circle

Proof:

In the cylindrical shell A, 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 $\left(A_1\right)=\pi {\left(r_1\right)}^2$

  $\begin{array}{l}\therefore A_1=\pi {\left(2r\right)}^2\\ \therefore A_1=4\pi r^2\mathrm{...}\left(\text{ii}\right)\end{array}$

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

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

Using (i), we get

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

In the cylindrical shell B, we need to find the area of a circle with radius $r\sqrt{3}$ .

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 $\left(A"\right)=\pi {\left(r_3\right)}^2$

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

From (iv) and (v), we get

The corresponding sections of the cylindrical shells A and B have the same area at the level mentioned.

Thus, by using Cavalieri’s principle we can conclude that the volumes of the cylindrical shells A and B are equal.