Chapter 4.4, Problem 57E

To determine

Find the mean volume of the particles.

Find the variance of the volume of the particles.

Answer to Problem 57E

The mean volume of the particles is $E\left(V\right)=0.00002041$.

The variance of the volume of the particles is $V\left(V\right)=3.49\times {10}^{-10}$.

Explanation of Solution

The mean volume of the particles is obtained below:

The probability density function for the uniform distribution on interval $\left({\theta }_1,{\theta }_2\right)$ is as follows:

$f\left(y\right)=\frac{1}{{\theta }_2-{\theta }_1}{\theta }_1\le y\le {\theta }_2$

It is given that the spherical particles have diameters that are uniformly distributed between 0.01 and 0.05 centimeters.

Let Y be the random variable associated with the diameter.

The random circle has a radius that is uniformly distributed on the interval (0, 1). Thus, the probability density function is given below:

$\begin{array}{l}f\left(y\right)=\{\begin{array}{l}\frac{1}{25}0.01<y<0.05\\ 0\text{elsewhere}\end{array}\\ \end{array}$

Let V be the random variable associated with the volume and the random variable associated with the radius be R.

The volume of a sphere is given as follows:

Let V be denoted as the volume and R be denoted as the radius.

$\begin{array}{c}V=\frac{4\pi R^3}{3}\\ Y=2R\\ =\frac{\pi Y^3}{6}\end{array}$

The required value is calculated below:

$\begin{array}{c}E\left(Y^3\right)=E\left(\frac{\pi Y^3}{6}\right)\\ =E\left(Y^3\right)\frac{\pi }{6}\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}^{3}f\left(y\right)dy}\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}^{3}25dy}\end{array}$

             $\begin{array}{c}=25{\left[\frac{y^4}{4}\right]}_{0.01}^{0.05}\\ =\frac{39}{1,000,000}\\ =0.000039\end{array}$

Then,

$\begin{array}{c}E\left(V\right)=\frac{\pi }{6}E\left(Y^3\right)\\ =\frac{3.14}{6}\left(0.000039\right)\\ =0.00002041\end{array}$

The mean of the area of the particle is $E\left(V\right)=0.00002041$.

The variance of the area of the particle is calculated below:

The formula for variance is $V\left(Y\right)=E\left(Y^2\right)-{\left[E\left(Y\right)\right]}^2$.

$\begin{array}{c}V\left(V\right)=E{\left(V\right)}^2-{\left(E\left(V\right)\right)}^2\\ =E\left[{\left(\frac{\pi Y^3}{6}\right)}^2\right]-{\left(E\left(V\right)\right)}^2\end{array}$

The required value is obtained as follows:

$\begin{array}{c}E\left[{\left(\frac{\pi Y^3}{6}\right)}^2\right]=E\left[Y^6\left(\frac{{\pi }^2}{36}\right)\right]\\ =E{\left(Y\right)}^6\left(\frac{{\pi }^2}{36}\right)\end{array}$

The required value is obtained as follows:

$\begin{array}{c}E\left[{\left(\frac{\pi Y^3}{6}\right)}^2\right]=E{\left(Y\right)}^6\left(\frac{{\pi }^2}{36}\right)\\ ={\displaystyle \underset{0.01}{\overset{0.05}{\int }}{y}^6\left(\frac{{\pi }^2}{36}\right)25dy}\\ =\left(\frac{{\pi }^2}{36}\right){\displaystyle \underset{0.01}{\overset{0.05}{\int }}{y}^{6}25dy}\\ =25\left(\frac{{\pi }^2}{36}\right){\left[\frac{y^7}{7}\right]}_{0.01}^{0.05}\end{array}$

                        $\begin{array}{c}=\left[\frac{6,241y^7}{6,300}\right]\\ =\frac{121,892,971}{157500000000000000}\\ =7.7392\times {10}^{-10}\end{array}$

Then,

$\begin{array}{c}V\left(V\right)=E\left[{\left(\frac{\pi Y^3}{6}\right)}^2\right]-{\left(E\left(V\right)\right)}^2\\ =7.65\times {10}^{-10}-{\left(0.00002041\right)}^2\\ =3.49\times {10}^{-10}\end{array}$

Thus, the variance of the area of the particle is $V\left(V\right)=3.49\times {10}^{-10}$.