Green's Theorem — Question 5

Question 5 – Solution

Strategy. Replace the two curved boundary integrals with one Type I double integral over the enclosed region.

Step 1: Derivative difference Here $$P=-y, \qquad Q=x^2, \qquad Q_x-P_y=2x+1.$$

Positive orientation travels from $(0,0)$ to $(1,1)$ along the lower parabola and returns along the upper line.

Step 2: Set up and evaluate $$\begin{align*} \oint_C P\,dx+Q\,dy &=\int_0^1\int_{x^2}^{x}(2x+1)\,dy\,dx\\ &=\int_0^1(2x+1)(x-x^2)\,dx\\ &=\int_0^1(x+x^2-2x^3)\,dx\\ &=\frac 12+\frac 13-\frac 12=\boxed{\frac 13}. \end{align*}$$

Verification On $0\le x\le 1$, both $2x+1$ and the vertical thickness $x-x^2$ are nonnegative, so the positive result is required.