Find the minimum area of the part bounded by the parabola $ y\equal{}a^3x^2\minus{}a^4x\ (a>0)$ and the line $ y\equal{}x$.
1. **Find the points of intersection:**
   The parabola \( y = a^3 x^2 - a^4 x \) and the line \( y = x \) intersect where:
   \[
   a^3 x^2 - a^4 x = x
   \]
   Rearrange and factorize:
   \[
   a^3 x^2 - (a^4 + 1)x = 0 \implies x(a^3 x - (a^4 + 1)) = 0
   \]
   This gives:
   \[
   x = 0 \quad \text{or} \quad x = \frac{a^4 + 1}{a^3}
   \]

2. **Determine the area between the curves:**
   The area \( A \) between the curves from \( x = 0 \) to \( x = \frac{a^4 + 1}{a^3} \) is given by:
   \[
   A = \int_0^{\frac{a^4 + 1}{a^3}} (x - (a^3 x^2 - a^4 x)) \, dx
   \]
   Simplify the integrand:
   \[
   A = \int_0^{\frac{a^4 + 1}{a^3}} (x - a^3 x^2 + a^4 x) \, dx = \int_0^{\frac{a^4 + 1}{a^3}} x(1 + a^4) - a^3 x^2 \, dx
   \]

3. **Evaluate the integral:**
   \[
   A = \int_0^{\frac{a^4 + 1}{a^3}} (x(1 + a^4) - a^3 x^2) \, dx
   \]
   Split the integral:
   \[
   A = \int_0^{\frac{a^4 + 1}{a^3}} x(1 + a^4) \, dx - \int_0^{\frac{a^4 + 1}{a^3}} a^3 x^2 \, dx
   \]
   Evaluate each part separately:
   \[
   \int_0^{\frac{a^4 + 1}{a^3}} x(1 + a^4) \, dx = (1 + a^4) \int_0^{\frac{a^4 + 1}{a^3}} x \, dx = (1 + a^4) \left[ \frac{x^2}{2} \right]_0^{\frac{a^4 + 1}{a^3}} = (1 + a^4) \frac{\left( \frac{a^4 + 1}{a^3} \right)^2}{2}
   \]
   \[
   = (1 + a^4) \frac{(a^4 + 1)^2}{2a^6}
   \]
   \[
   \int_0^{\frac{a^4 + 1}{a^3}} a^3 x^2 \, dx = a^3 \int_0^{\frac{a^4 + 1}{a^3}} x^2 \, dx = a^3 \left[ \frac{x^3}{3} \right]_0^{\frac{a^4 + 1}{a^3}} = a^3 \frac{\left( \frac{a^4 + 1}{a^3} \right)^3}{3}
   \]
   \[
   = a^3 \frac{(a^4 + 1)^3}{3a^9} = \frac{(a^4 + 1)^3}{3a^6}
   \]
   Combine the results:
   \[
   A = (1 + a^4) \frac{(a^4 + 1)^2}{2a^6} - \frac{(a^4 + 1)^3}{3a^6}
   \]
   \[
   = \frac{(a^4 + 1)^3}{2a^6} - \frac{(a^4 + 1)^3}{3a^6}
   \]
   \[
   = \left( \frac{1}{2} - \frac{1}{3} \right) \frac{(a^4 + 1)^3}{a^6} = \frac{1}{6} \frac{(a^4 + 1)^3}{a^6}
   \]
   \[
   = \frac{(a^4 + 1)^3}{6a^6}
   \]

4. **Find the minimum area:**
   To find the minimum area, differentiate \( A \) with respect to \( a \) and set the derivative to zero:
   \[
   A = \frac{(a^4 + 1)^3}{6a^6}
   \]
   \[
   \frac{dA}{da} = \frac{d}{da} \left( \frac{(a^4 + 1)^3}{6a^6} \right)
   \]
   Using the quotient rule:
   \[
   \frac{dA}{da} = \frac{6a^3(a^4 + 1)^2 \cdot a^6 - (a^4 + 1)^3 \cdot 6a^5}{36a^{12}}
   \]
   Simplify:
   \[
   \frac{dA}{da} = \frac{6a^9(a^4 + 1)^2 - 6a^5(a^4 + 1)^3}{36a^{12}} = \frac{a^4(a^4 + 1)^2 - (a^4 + 1)^3}{6a^7}
   \]
   \[
   = \frac{(a^4 + 1)^2 (a^4 - (a^4 + 1))}{6a^7} = \frac{(a^4 + 1)^2 (a^4 - a^4 - 1)}{6a^7} = \frac{-(a^4 + 1)^2}{6a^7}
   \]
   Set the derivative to zero:
   \[
   \frac{-(a^4 + 1)^2}{6a^7} = 0
   \]
   This implies:
   \[
   a^4 + 1 = 0
   \]
   Since \( a > 0 \), this is not possible. Therefore, we need to check the boundary conditions. For \( a = 1 \):
   \[
   A = \frac{(1^4 + 1)^3}{6 \cdot 1^6} = \frac{8}{6} = \frac{4}{3}
   \]

The final answer is \(\boxed{\frac{4}{3}}\).