What is the variance of the volumes of particles? According to Zimmels (1983), the sizes of particles used in sedimentation experiments often have a uniform distribution. In sedimentation involving mixtures of particles of various sizes, the larger particles hinder the movements of the smaller ones. Thus, it is important to study both the mean and the variance of particle sizes. Suppose that spherical particles have diameters that are uniformly distributed between $0.01$ and $0.05$ centimeters. Find the mean and variance of the volumes of these particles (Volume of the sphere is $\frac 4 3 \pi R^3$).
Book's solution:
Mean $= 0.0000065\pi$
Variance $= 0.0003525\pi^2$
My solution:
Volume $= \frac 4 3 \pi R^3 = \pi D^3$, where $D$ is diameter.
Mean $\displaystyle = E\left[\frac \pi 6 D^3\right] = \int_{0.01}^{0.05} \frac \pi 6  D^3 \cdot25$.
$= 0.0000065\pi$ (same answer with the book)
\begin{align}
\text{Variance} & = E\left[\frac{\pi^2}{36} D^6\right] - E\left[ \frac \pi 6  D^3\right]^2 \\[8pt]
& = \int_{0.01}^{0.05} \left[ \frac{\pi^2}{36} D^6\right] \cdot 25 - 4.225\times(10^{-11})\times \pi^2 \\[8pt]
& = 3.5254 \times (10^{-11})\times \pi^2 \text{ (about $10^7$ times smaller than the book solution)}
\end{align}
Is my solution wrong?
 A: I'm going to use units of hundredths of a centimeter, then convert back to centimeters.
The diameter of a given particle is modeled as a random variable $$D \sim \operatorname{Uniform}(1,5);$$ namely, $$f_D(x) = \begin{cases} \frac{1}{4}, & 1 \le x \le 5 \\ 0, & \text{otherwise}. \end{cases}$$  If $$V = \frac{4}{3}\pi R^3$$ where $R = D/2$, this suggests we wish to find $\operatorname{E}[D^3]$ and $\operatorname{E}[D^6]$.  To this end, we write $$\operatorname{E}[D^k] = \int_{x=1}^5 x^k f_D(x) \, dx = \frac{1}{4} \left[ \frac{x^{k+1}}{k+1} \right]_{x=1}^5 = \frac{5^{k+1} - 1}{4(k+1)},$$ for any positive integer $k$.  For $k = 3, 6$, we get $$\operatorname{E}[D^3] = 39, \quad \operatorname{E}[D^6] = \frac{19531}{7}.$$  Therefore, $$\operatorname{E}[V] = \operatorname{E}\left[\frac{4}{3} \pi \frac{D^3}{8} \right] = \frac{13}{2}\pi,$$ and $$\operatorname{Var}[V] = \operatorname{E}\left[\frac{\pi^2}{36} D^6 \right] - \operatorname{E}[V]^2 = \left( \frac{19531}{252} - \frac{169}{4}\right) \pi^2 = \frac{2221}{63}\pi^2.$$  Now converting these to centimeters gives $$\operatorname{E}[V] = \frac{13\pi}{2000000} \approx 2.04204 \times 10^{-5}$$ and $$\operatorname{Var}[V] = \frac{2221\pi^2}{63000000000000} \approx 3.47943 \times 10^{-10}.$$
A: Your variance seems correct, based on quick computations
and a simulation of a million such particles. (Also, your
method using $E(V^2) - [E(V)]^2$ looks sound to me, especially now that it has
been edited by @MichaelHardy and is easier to read.)
 .0003525*pi^2
 ## 0.003479036          # book's answer
 3.5254*(10^-11)*pi^2
 ## 3.47943e-10          # your answer

 d = runif(10^6, .01, .05)
 v = (4/3)*pi*(d/2)^3
 var(v)
 ## 3.482039e-10         # simulated (correct to about 3 signif digits)

