calculating an integral with green's theorem

Question

Let $ M:\{ (x,y) \in \mathbb{R}_+^{2} : x^2+ 4y^2 \leq 4, 1 \leq x^2-y^2 \} $

I want to use Green's Theorem:

$ I(M)= \frac{1}{2} \int_{\partial M} x_1 dx_2 -x_2 dx_1 $

M is the region where the ellipse and hyperbole are overlapping above the x axis, so $ \int_{\partial M }$ must be $ \int_1^2 $ right?

now, I am not sure what to use as function I need to integrate.I could not find any similar examples. Your help is very apreciated :-)

Answer 1

I think you mean $\displaystyle I( M) =\frac{1}{2}\int _{\partial M} xdy-ydx$

Taking $\displaystyle I( M)$ as written above, by green's theorem this is the area of region M

Blue shaded region is the Region M. \begin{gather*} \therefore Area\ of\ M=I( M) =2\int ^{\frac{3}{5}}_{-\frac{3}{5}}\int ^{2\sqrt{1-y^{2}}}_{^{\sqrt{y^{2} +1}}} dxdy\\ or\ I( M) =\left[ y\sqrt{1-y^{2}} -\frac{1}{2}\sqrt{y^{2} +1} +\arcsin( y) -\frac{1}{2}\ arcsinh( y)\right]^{y=\frac{3}{5}}_{y=-\frac{3}{5}}\\ or\ I( M) =\frac{24}{25} -\frac{3\sqrt{34}}{25} +2\arcsin\left(\frac{3}{5}\right) -\ arcsinh\left(\frac{3}{5}\right)\\ or\ I( M) \approx 0.9785 \end{gather*}

Answer 2

Green's theorem says that $$\int_{\partial S}{Pdx+Qdy}=\int\int_S(Q_y-P_x)dxdy$$ with suitable hypotheses on $P,Q,S$. So if we set $P(x,y)=0, Q(x,y)=y,$ we see that the area of $S$ is $$\int_{\partial S}{y dy}$$ You simply have to parametrize the curves bounding $S$ and integrate the expression for $ydy$.

$S$ breaks into two congruent regions, so it's enough to compute the area of one of them.

The region on the right is bounded below by a segment the x-axis. If we parameterize $y$ on this axis, we will have $y=0$ of course, so the line integral will be $0$ and we can ignore it. The blue boundary curve runs from $(2,0)$ to the red dot at $(\sqrt{1.6},\sqrt{.4})$ and we can parameterize along this curve as $y$ as $$ y=\sqrt{{4-x^2\over4}}\implies dy= {-xdx\over 2\sqrt{4-x^2}},$$ so you have to integrate ${-x\over4}dx$ from $x=2$ to x=$\sqrt{1.6}.$ The direction of the integral is determined because in Green's theorem, the contour of the line integral is oriented counter-clockwise.

We have $ydy={-x\over8}dx,$ so we compute $$\int_2^{\sqrt{.6}}{-x\over8}dx={-x^2\over16}\biggr\rvert_2^{\sqrt{.6}}={3.4\over16}$$

Do the same thing for the arc of the hyperbola that bounds the region on the right.