Let $a,\ b$ be positive constant numbers.Find the volume of the revolution of the region surrounded by the parabola $y=ax^{2}$ and the line $y=bx$ about  the line $y=bx$ as the axis on $xy$ plane.
To find the volume of the solid of revolution formed by rotating the region bounded by the parabola \( y = ax^2 \) and the line \( y = bx \) about the line \( y = bx \), we can use the method of cylindrical shells. 

1. **Find the intersection point:**
   The intersection point of the parabola \( y = ax^2 \) and the line \( y = bx \) is found by setting \( ax^2 = bx \):
   \[
   ax^2 = bx \implies x(ax - b) = 0 \implies x = 0 \text{ or } x = \frac{b}{a}
   \]
   Thus, the intersection points are \( (0, 0) \) and \( \left( \frac{b}{a}, \frac{b^2}{a} \right) \).

2. **Distance from a point to a line:**
   Consider a point \( P(x, ax^2) \) on the parabola and a point \( Q(x, bx) \) on the line. The distance from \( P \) to the line \( y = bx \) is given by the formula for the distance from a point to a line:
   \[
   \overline{PH} = \frac{|bx - ax^2|}{\sqrt{b^2 + 1}}
   \]

3. **Volume of revolution:**
   The volume \( V \) of the solid of revolution is given by integrating the area of the circular cross-sections perpendicular to the line \( y = bx \). The radius of each cross-section is \( \overline{PH} \), and the differential element of volume is \( \pi (\overline{PH})^2 \frac{d(\overline{OH})}{dx} \, dx \).

4. **Calculate \(\overline{OH}\):**
   Using the Pythagorean theorem, we have:
   \[
   \overline{OH}^2 = OP^2 - PH^2 = \frac{(abx^2 + x)^2}{b^2 + 1}
   \]
   Therefore,
   \[
   \overline{OH} = \frac{abx^2 + x}{\sqrt{b^2 + 1}}
   \]

5. **Set up the integral:**
   The volume \( V \) is:
   \[
   V = \pi \int_{0}^{\frac{b}{a}} \left( \frac{bx - ax^2}{\sqrt{b^2 + 1}} \right)^2 \frac{d(\overline{OH})}{dx} \, dx
   \]
   Simplifying, we get:
   \[
   V = \frac{\pi}{(b^2 + 1)^{3/2}} \int_{0}^{\frac{b}{a}} (ax^2 - bx)^2 (2abx + 1) \, dx
   \]

6. **Evaluate the integral:**
   Split the integral into two parts:
   \[
   V = \frac{\pi b}{(b^2 + 1)^{3/2}} \int_{0}^{\frac{b}{a}} (ax^2 - bx)^2 (ax^2 - bx)' \, dx + \frac{\pi}{\sqrt{b^2 + 1}} \int_{0}^{\frac{b}{a}} (ax^2 - bx)^2 \, dx
   \]
   The first integral vanishes because it is the integral of a derivative over its entire range. For the second integral, we rewrite it as:
   \[
   \int_{0}^{\frac{b}{a}} x^2 \left( x - \frac{b}{a} \right)^2 \, dx = \frac{1}{30} \left( \frac{b}{a} \right)^5
   \]

7. **Final volume:**
   Thus, the volume \( V \) is:
   \[
   V = \frac{\pi b^5}{30 a^3 \sqrt{b^2 + 1}}
   \]

The final answer is \(\boxed{\frac{\pi b^5}{30 a^3 \sqrt{b^2 + 1}}}\).