Chapter 1.3, Problem 29E

To determine

To find: The ratio of the volume of the larger tree to that of the smaller tree.

Answer to Problem 29E

The ratio of the volume of the larger tree to that of the smaller tree is 1.132.

Explanation of Solution

Given information:

The height of the first tree is

  $h1=23.1m$ and the height of the second tree is $h2=24.1$ m.

Formula used:

The volume of a cylinder is $v=\pi h{(r)}^2$

The volume of a sphere is $v^1=\frac{4\pi }{3}{r}^3$

Calculation:

The height of the first tree is $h1=23.1\text{m}$

The height of the second tree is $h2=24.1\text{m}$

The radius of both trees is $r=0.5m$

The volume of a cylinder is $v=\pi h{(r)}^2$

Where, $h$ is a height and $r$ is radius

The volume of a sphere is $v^1=\frac{4\pi }{3}{r}^3$

Where $r$ is radius

The half of the height of the first tree is a cylindrical trunk. Therefore,

  $h1=\frac{23.1}{\begin{array}{l}2\\ \end{array}}$

Hence, the volume $V_1$ of the cylindrical part of the first tree is

  $\begin{array}{l}{v}_1=\pi h_1{r}^2\\ =\pi \times \frac{23.1}{2}\text{m}\times {\left(0.5\text{m}\right)}^2\\ =3.14\times 11.55\text{m}\times 0.5\times 0.5{\text{m}}^2\mathrm{..............}\left(use\pi \approx 3.14\right)\\ =9.07{\text{m}}^3\end{array}$

Since half of the height of the tree is a spherical blob, we get

  $\begin{array}{l}{h}_1=\frac{23.1}{2}\\ =11.55\end{array}$

Therefore, the radius of the sphere is as follows

  $\begin{array}{l}r=\frac{11.55}{2}\\ =5.78\end{array}$

So, the volume of the spherical blob part of the first tree is shown below

  $\begin{array}{l}{v}_1^{\text{'}}=\frac{4\pi }{3}{\left(5.78\text{m}\right)}^3\\ =\frac{4\times 3.14}{3}\times 5.78\times 5.78\times 5.78{\text{m}}^3\mathrm{........}(use\pi \approx 3.14)\\ =808.45{\text{m}}^3\end{array}$

Therefore, the volume of the first tree is

  $\begin{array}{l}{v}_1+v_1^{\text{'}}=9.07{\text{m}}^3+808.45{\text{m}}^3\\ =817.52{\text{m}}^3\end{array}$

Since half of the height of the tree is a cylindrical trunk, we find that

  $h_2=\frac{24.1}{2}$

The volume of cylindrical part of the second tree is shown below

  $\begin{array}{l}{v}_2=\pi h_2{r}^2\\ =\pi \times \frac{24.1}{2}\text{m}\times {\left(0.5\text{m}\right)}^2\\ =3.14\times 12.05\text{m}\times 0.5\times 0.5{\text{m}}^2\mathrm{................}(Use\pi \approx 3.14)\\ =9.46{\text{m}}^3\end{array}$

Since half of the height of the tree is spherical blob, So

  $\begin{array}{l}{h}_2=\frac{24.1}{2}=12.05\\ \end{array}$

Therefore the radius of sphere can be calculate as

  $r=\frac{12.05}{2}=6.025$

The volume of the second tree with half of the height in the spherical blob is

  $\begin{array}{l}{v}_2^{\text{'}}=\frac{4\pi }{3}{\left(6.025\text{m}\right)}^3\\ =\frac{4\times 3.14}{3}\times 6.025\times 6.025\times 6.025{\text{m}}^3\mathrm{.............}(use\pi \approx 3.14)\\ =915.67{\text{m}}^3\end{array}$

The volume of the second tree is $\begin{array}{l}{v}_2+v_2^{\text{'}}=9.46{\text{m}}^3+915.67{\text{m}}^3\\ =925.13{\text{m}}^3\end{array}$

Since $v_2+v_2^{\text{'}}>v_1+v_1^{\text{'}}$ , then the second tree is larger and the first tree is smaller

The ratio is $\begin{array}{l}\frac{\text{volume of the larger tree}}{\text{volume of the smaller tree}}=\frac{925.13}{817.52}\\ =1.132\end{array}$

Hence, the ratio $1.132$