Let $a$ be a constant number such that $0<a<1$ and $V(a)$ be the volume formed by the revolution of the figure 

which is enclosed by the curve $y=\ln (x-a)$, the $x$-axis and two lines $x=1,x=3$ about the $x$-axis.

If $a$ varies in the range of  $0<a<1$,  find the minimum value of $V(a)$.
1. The volume \( V(a) \) formed by the revolution of the figure enclosed by the curve \( y = \ln(x-a) \), the \( x \)-axis, and the lines \( x = 1 \) and \( x = 3 \) about the \( x \)-axis can be given by:
   \[
   V(a) = \pi \int_1^3 \ln^2(x-a) \, dx
   \]

2. To evaluate the integral \( \int \ln^2(x-a) \, dx \), we use integration by parts. Let:
   \[
   u = \ln^2(x-a) \quad \text{and} \quad dv = dx
   \]
   Then:
   \[
   du = \frac{2\ln(x-a)}{x-a} \, dx \quad \text{and} \quad v = x
   \]

3. Applying integration by parts:
   \[
   \int \ln^2(x-a) \, dx = x \ln^2(x-a) - \int x \cdot \frac{2\ln(x-a)}{x-a} \, dx
   \]
   Simplifying the integral:
   \[
   \int x \cdot \frac{2\ln(x-a)}{x-a} \, dx = 2 \int \ln(x-a) \, dx
   \]

4. We need to evaluate \( \int \ln(x-a) \, dx \). Using integration by parts again, let:
   \[
   u = \ln(x-a) \quad \text{and} \quad dv = dx
   \]
   Then:
   \[
   du = \frac{1}{x-a} \, dx \quad \text{and} \quad v = x
   \]

5. Applying integration by parts:
   \[
   \int \ln(x-a) \, dx = x \ln(x-a) - \int x \cdot \frac{1}{x-a} \, dx
   \]
   Simplifying the integral:
   \[
   \int x \cdot \frac{1}{x-a} \, dx = \int \left(1 + \frac{a}{x-a}\right) \, dx = x + a \ln|x-a|
   \]

6. Therefore:
   \[
   \int \ln(x-a) \, dx = x \ln(x-a) - (x + a \ln|x-a|) = -a \ln|x-a|
   \]

7. Substituting back, we get:
   \[
   \int \ln^2(x-a) \, dx = x \ln^2(x-a) - 2(-a \ln|x-a|) = x \ln^2(x-a) + 2a \ln|x-a|
   \]

8. Evaluating the definite integral from 1 to 3:
   \[
   V(a) = \pi \left[ (3-a)\left(\ln^2(3-a) + 2a \ln|3-a|\right) - (1-a)\left(\ln^2(1-a) + 2a \ln|1-a|\right) \right]
   \]

9. To find the minimum value of \( V(a) \), we take the derivative \( V'(a) \) and set it to zero:
   \[
   V'(a) = \pi \left[ \ln^2(1-a) - \ln^2(3-a) \right] = 0
   \]

10. Let \( x = \ln(1-a) \) and \( y = \ln(3-a) \). The equation \( \ln^2(1-a) = \ln^2(3-a) \) implies:
    \[
    x^2 = y^2 \quad \Rightarrow \quad x = \pm y
    \]

11. **Case 1:** \( x = y \) has no solutions because \( \ln(1-a) = \ln(3-a) \) is not possible for \( 0 < a < 1 \).

12. **Case 2:** \( x = -y \) implies:
    \[
    \ln(1-a) = -\ln(3-a) \quad \Rightarrow \quad \ln(1-a) + \ln(3-a) = 0 \quad \Rightarrow \quad \ln((1-a)(3-a)) = 0
    \]
    \[
    (1-a)(3-a) = 1 \quad \Rightarrow \quad 3 - 4a + a^2 = 1 \quad \Rightarrow \quad a^2 - 4a + 2 = 0
    \]

13. Solving the quadratic equation:
    \[
    a = \frac{4 \pm \sqrt{16 - 8}}{2} = 2 \pm \sqrt{2}
    \]
    Since \( 2 + \sqrt{2} \) is extraneous (greater than 1), the valid solution is:
    \[
    a = 2 - \sqrt{2}
    \]

14. Finally, we evaluate \( V(a) \) at \( a = 2 - \sqrt{2} \):
    \[
    V\left(2 - \sqrt{2}\right) = \pi \left[ (\sqrt{2} + 1)\left(\ln^2(\sqrt{2} + 1) + 2(2 - \sqrt{2}) \ln(\sqrt{2} + 1)\right) - (\sqrt{2} - 1)\left(\ln^2(\sqrt{2} - 1) + 2(2 - \sqrt{2}) \ln(\sqrt{2} - 1)\right) \right]
    \]

The final answer is \( \boxed{ V\left(2 - \sqrt{2}\right) } \)