Question

3. Use Green's theorem to find the area of an ellipse with semi-axes a and b.

Answer 1

Let us consider the ellipse to be with origin as its center, and its $a$ being along X-axis.

Let us consider the first quadrant part of the ellipse, say $E$, and let $A$ be the area of this part. Then the area of the complete ellipse is $4A$ (by symmetry of the ellipse). Note that $E=\{(a\cos t, b\sin t):t\in [0,\pi/4]\}$ . Let $B$ be the region bounded by $E$ and the coordinate axes.

Now, the area inside the ellipse is that of $B$, and is the integral

$\iint_BdS=\iint_Bdx~dy$

By Green's theorem, we know

$\iint_B(M_x-N_y)~dx~dy=\int_E(N~dx+M~dy)$

Thus, in order to apply Green's theorem to our problem we need $M,N$ so that

$M_x-N_y=1$

We try the obvious; say

$M={\frac {bx}{b+a}},~~~~N={-\frac {ay}{b+a}}$

Then

$M_x={\frac {b}{b+a}},~~~~N_y={-\frac {a}{b+a}},~~~~~M_x-N_y=1$

and hence, by Green, we have

$\begin{align*}A&=\iint_B(M_x-N_y)~dx~dy\\ &=\int_E(N~dx+M~dy)\\ &=\int_E{-\frac{ay~dx}{a+b}}+{\frac{bx~dy}{a+b}}\end{align*}$

Since $E=\{(a\cos t, b\sin t):t\in [0,\pi/4]\}$ , we get $dx=-a\sin t~dt$ and $dy=b\cos t~dt$. Therefore,

$\begin{align*}A&=\int_E{-\frac{ay~dx}{a+b}}+{\frac{bx~dy}{a+b}}\\ &={\frac 1{a+b}}\int_0^{\pi/2}(a^2b\sin^2t+ab^2\cos^2t)~dt\\ &={\frac 1{2(a+b)}}\int_0^{\pi/2}(a^2b(1-\cos2t)+ab^2(1+\cos2t))~dt\\ &={\frac {(a^2b+ab^2)}{2(a+b)}}\int_0^{\pi/2}dt+{\frac {(-a^2b+ab^2)}{2(a+b)}}\int_0^{\pi/2}\cos2t~dt\\ &={\frac {\pi ab}{4}}+{\frac {(-a^2b+ab^2)}{4(a+b)}}\sin2t\mid_0^{\pi/2}\\ &={\frac {\pi ab}{4}}\end{align*}$

Therefore, area of the ellipse is

$\begin{align*}4A&={\frac {4\pi ab}{4}}\\ &=\pi ab\end{align*}$