Answer
\[ I = 3 \]
Work Step by Step
We have the path of integration given by: \[ y^2 = 3x, \quad (3,3) \to (0,0) \] Introducing the parametrization: \[ x = \frac{(3-t)^2,}{3} \quad y = 3-t, \quad 0 \leq t \leq 3 \] We have for the integral: \[ I = \int_{0}^{3}\frac{2}{3} (3-t)^{3/2} (3-t)dt + \frac{1}{3}(3-t)^2(-dt) \] \[ I = \int_{0}^{3} \left[\frac{2}{3}(3-t)^2 - \frac{1}{3}(3-t)^2\right]dt \] \[ I = \frac{1}{3} \int_{0}^{3} (3-t)^2 dt \] \[ I = -\frac{1}{9}(3-t)^3 \Bigg|_{0}^{3} \] \[ I = -\frac{1}{9}(0-27) = 3 \] Result: \[ I = 3 \]
