In $ xy$ plane, find the minimum volume of the solid by rotating the region boubded by the parabola $ y \equal{} x^2 \plus{} ax \plus{} b$ passing through the point $ (1,\ \minus{} 1)$ and the $ x$ axis about the $ x$ axis
1. **Identify the quadratic function:**
   The quadratic function is given by \( y = x^2 + ax + b \). Since it passes through the point \((1, -1)\), we can substitute these coordinates into the equation to find \(b\):
   \[
   -1 = 1^2 + a \cdot 1 + b \implies -1 = 1 + a + b \implies b = -a - 2
   \]
   Thus, the quadratic function becomes:
   \[
   y = x^2 + ax - (a + 2)
   \]

2. **Find the roots of the quadratic function:**
   To find the points where the parabola intersects the \(x\)-axis, we solve for \(x\) when \(y = 0\):
   \[
   x^2 + ax - (a + 2) = 0
   \]
   Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we get:
   \[
   x = \frac{-a \pm \sqrt{a^2 + 4(a + 2)}}{2} = \frac{-a \pm \sqrt{a^2 + 4a + 8}}{2}
   \]
   Let \(\alpha = \frac{-a - \sqrt{a^2 + 4a + 8}}{2}\) and \(\beta = \frac{-a + \sqrt{a^2 + 4a + 8}}{2}\) with \(\alpha < \beta\).

3. **Express the volume of the solid of revolution:**
   The volume \(V\) of the solid obtained by rotating the region bounded by the parabola and the \(x\)-axis about the \(x\)-axis is given by:
   \[
   V = \pi \int_{\alpha}^{\beta} y^2 \, dx
   \]
   Substituting \(y = (x - \alpha)(x - \beta)\), we get:
   \[
   V = \pi \int_{\alpha}^{\beta} (x - \alpha)^2 (x - \beta)^2 \, dx
   \]

4. **Simplify the integral:**
   The integral can be simplified using the binomial expansion and integration:
   \[
   V = \pi \int_{\alpha}^{\beta} (x - \alpha)^2 (x - \beta)^2 \, dx = \frac{\pi}{30} (\beta - \alpha)^5
   \]
   Since \(\beta - \alpha = \sqrt{a^2 + 4a + 8}\), we have:
   \[
   V = \frac{\pi}{30} (a^2 + 4a + 8)^{\frac{5}{2}}
   \]

5. **Minimize the volume:**
   To find the minimum volume, we need to minimize the expression \((a^2 + 4a + 8)^{\frac{5}{2}}\). The expression inside the parentheses is a quadratic function:
   \[
   f(a) = a^2 + 4a + 8
   \]
   The minimum value of a quadratic function \(ax^2 + bx + c\) occurs at \(x = -\frac{b}{2a}\). Here, \(a = 1\) and \(b = 4\), so:
   \[
   a = -\frac{4}{2 \cdot 1} = -2
   \]
   Substituting \(a = -2\) into \(f(a)\):
   \[
   f(-2) = (-2)^2 + 4(-2) + 8 = 4 - 8 + 8 = 4
   \]
   Therefore, the minimum volume is:
   \[
   V = \frac{\pi}{30} (4)^{\frac{5}{2}} = \frac{\pi}{30} \cdot 32 = \frac{32\pi}{30} = \frac{16\pi}{15}
   \]

The final answer is \(\boxed{\frac{16\pi}{15}}\).