Green's Theorem — Question 7

Question 7 – Solution

Strategy. Only the slanted edge contributes directly; then use $0\le y\le 1-x/2$ for the interior.

Step 1: Boundary calculation The horizontal and vertical coordinate-axis edges contribute zero. Parametrize the slanted edge from $(2,0)$ to $(0,1)$ by $$\mathbf r(t)=\langle 2-2t,t\rangle,\qquad 0\le t\le 1.$$ Then $dx=-2dt$, $dy=dt$, and $$\int_{\mathrm{slant}}P\,dx+Q\,dy =\int_0^1\left[2t^2+4(1-t)^2\right]dt=\boxed{2}.$$

Step 2: Interior calculation Since $Q_x-P_y=2x+2y$, $$\begin{align*} \iint_D(2x+2y)\,dA &=\int_0^2\int_0^{1-x/2}(2x+2y)\,dy\,dx\\ &=\int_0^2\left(1+x-\frac{3x^2}{4}\right)dx=\boxed{2}. \end{align*}$$

Verification Both methods give $2$. The slanted parametrization runs from $(2,0)$ to $(0,1)$, which is the required counterclockwise direction after traversing the bottom edge.