# calculating an integral with green's theorem

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 :-)

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

• thank you for your answere! Since $(x,y) \in \mathbb{R^+}^2$ are the bounds not quite right, right? Where does the Green's Theorem flows in ? Commented Nov 27, 2018 at 20:54
• @wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem Commented Nov 28, 2018 at 4:53
• I am sorry, but thats the way i would normally integrate it..I thought about it and I still can not explain it with the formular $I(M)$ above. would you mind explaining it? Commented Nov 28, 2018 at 19:27
• You haven't used Green's theorem to evaluate the area. If you use Green's theorem, you will evaluate a line integral, not a double integral. Commented Nov 28, 2018 at 19:38

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.

• thanks for the detailled explanation! It is still quite hard for me to get it..and I need to integrate only for positive $x,y$ Commented Nov 28, 2018 at 21:27
• I'll add a few more details. Wait a bit. Commented Nov 28, 2018 at 21:48
• I added some details. Note that the contour satisfies $x\ge0, y\ge0$ so we are considering only positive $x$ and $y$. What makes you think we aren't? Commented Nov 28, 2018 at 22:01