Green's Theorem — Question 10

Question 10 – Solution

Strategy. Reverse Green’s Theorem with $P=y^2$ and $Q=x^3$, then use the supplied ellipse parametrization.

Step 1: Choose a boundary field Set $$P=y^2, \qquad Q=x^3.$$ Then $Q_x-P_y=3x^2-2y$, so $$\iint_D(3x^2-2y)\,dA=\oint_Cy^2\,dx+x^3\,dy.$$

Step 2: Substitute the boundary Since $dx=-2\sin t\,dt$ and $dy=3\cos t\,dt$, $$\begin{align*} \oint_Cy^2\,dx+x^3\,dy &=\int_0^{2\pi}\left[-18\sin^3t+24\cos^4t\right]dt. \end{align*}$$ The sine-cubed term integrates to zero over a full period, while $$\int_0^{2\pi}\cos^4t\,dt=\frac{3\pi}{4}.$$ Therefore $$\boxed{\iint_D(3x^2-2y)\,dA=24\left(\frac{3\pi}{4}\right)=18\pi}.$$

Verification The $-2y$ term is odd across the $x$-axis and contributes zero. Also $3x^2\ge 0$, so the positive result has the required sign.