Calculate the line integral $\int_L \mathbf{A} \cdot d\mathbf{r}$ using Greens formula

Question

Where $L$ is the curve $\frac{x^2}{4} + \frac{y^2}{9} = 1 $ for every $y \geq 0 $ and $x: 2 \rightarrow -2$. $\mathbf{A} = (y, -x)$.

My attempt at solution: I added the curve $L_1: y=0, x: -2 \rightarrow 2$ and noticed that $L + L_1$ is the boundary of the closed surface $S: \frac{x^2}{4} + \frac{y^2}{9} \leq 1 $ for every $y \geq 0$. $\mathbf{A}$ is continuously differentiable on $S$ so we can use Greens formula.

Am I right thus far? After doing the calculations I get $-\pi$ but the correct answer is $-6\pi$.

Answer 1

Using Green's theorem: $$A=(P, Q)= (y, -x)\Longrightarrow\int_L \mathbf{A} \cdot d\mathbf{r}=\displaystyle\int\int_{S}(Q_x-P_y)dA=-2\displaystyle\int\int_{S}dA,$$ where $ S:\, \frac{x^2}{4} + \frac{y^2}{9} \leq 1$. You can change the variables as $x=2u$ and $y=3v$ in which the Jacobian becomes $6$. Then it reduces to $$-2\displaystyle\int\int_{D}dA=-12\displaystyle\int\int_{D}dA,$$ where $D:\,u^2+v^2\leq 1$ is the upper semi unit circle in the $uv$-plane whose area is $\frac{\pi}{2}$. Thus the result is $-6\pi$.

Answer 2

Your parametrization is correct. You missed the jacobian factor during the change of variable. $\int_{-2}^{2} \int_{0}^{3\sqrt{1-x^2/4}}-2dxdy=-2.\frac{1}{2}$area of allipse=$-6\pi$ ($\because \frac{x^2}{a^2}+\frac{y^2}{b^2}$, area enclosed by the closed ellipse=$\pi a b$)

Answer 3

You can use Green's formulas:

$\int_{S}1=\frac{1}{2}\oint_{\partial S}xdy-ydx$

Your vector field is exactly the one described in the RHS, so, defining $S: \frac{x^2}{4}+\frac{y^2}{9}=1; y\geq 0$:

$\oint_{L}A=-2\cdot Area (S)=-2\cdot\frac{1}{2}\pi\cdot2\cdot3=-6\pi$

(You get the minus since $L$ has inverted orientation respecting that of S)

In your computation you were wrong in the change of variables: $dxdy=6rdrd\theta$