Computing Area of Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1$ with Green's Theorem

Question

Since it is obvious that $\int\int_D1\space\mathrm{d}A$ is the area of $A$, my book suggests we can use green's theorem $\int_C{P\mathrm{d}x+Q\mathrm{d}y}=\int\int_d{(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathrm{d}A}$, to compute area where $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$ to compute the area.

The book gives a list of conditions for which this is true, most of them fairly trivial (ie, $P(x,y)=0 \space \mathrm{and} \space Q(x,y) = x$).

Assuming the veracity of green's theorem itself, this is not hard to understand. However, I can't reconcile this with the example that follows immediately:

Find the area enclosed by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

It makes use of the formula $A=\frac{1}{2}\int_c{x\mathrm{d}y-y\mathrm{d}x}$.

Sadly, I have become hopelessly stuck trying to show $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$ in this case; therefore, while I could apply this method mechanically, I have no idea when it is appropriate to do so.

Would someone be so kind as to explain this?

Answer 1

With $A=\frac 1 2 \int_C xdy - ydx$ we have $P=-\frac 12 y$ and $Q=\frac 12 x$.

What do you get if you calculate $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ of that?

Answer 2

Look, there's nothing deep here: in the integral $$\int_C x\,dy-y\,dx$$ we have $P=-y$ and $Q=x$, so $$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1-(-1)=2$$ and so by Green's theorem we get $$\int_C x\,dy-y\,dx=\int\int_D2\,dA=2A$$ So if you want the area $A$, you must take half the integral.

Where are you "hopelessly stuck trying to show $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$"? You haven't explained what you've done or tried or where you got stuck. What did you calculate $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}$ to be?

Answer 3

It suffices to take $Q =0$ and $P =-y$ then $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$ and by Green formula we get, $$ \iint_{\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1} 1dxdy =\iint_{\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1}\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}dxdy= -\oint_{\frac{x^2}{a^2}+\frac{y^2}{b^2}= 1} ydx =\\\color{blue}{- \int_0^{2\pi} b\sin td(a\cos t)}= ab\int_0^{2\pi} \sin^2 t dt= ab\int_0^{2\pi} \frac{1-\cos (2t)}{2}dt= \color{red}{ ab\pi}$$ Where we used, $$x= a\cos t~~~and ~~y =b\sin t$$