I am trying to evaluate the integral $\int_{0,2}^{2,0}x^2ydy$ along C. With C being the portion of the curve $x^2+y^2=4$ lying in the first quadrant.
So far I have $x=\sqrt{4-y^2}$ suggesting the parametric representation C:$z=\sqrt{4-y^2}+iy(-2\le y\le2)$ however I feel as though I have already gone down the wrong track
This question has come from the Brown & Churchill 8th Edition of complex variables and applications. Unfortunately only the odd answers are provided.
$$\int_{0.2}^{2.0}x^2ydy=\int_{0.2}^{2.0}(4-y^2)ydy=\left[2y^2-\frac14 y^4\right]_{0.2}^{2.0}=2(2.0)^2-\frac14 (2.0)^4-2(0.2)^2+\frac14 (0.2)^4=3.9204$$ Correction after Bob Jones's comment : $$\int_{(x=0,y=2)}^{(x=2,y=0)}x^2ydy=\int_{2}^{0}(4-y^2)ydy=\left[2y^2-\frac14 y^4\right]_{2}^{0}=2(0)^2-\frac14 (0)^4-2(2)^2+\frac14 (2)^4=-4$$