微积分中计算椭圆面积的几种方法

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

1. Trigonometric Substitutions
   $y=b\sqrt[]{1-\frac{x^2}{a^2}}$
   let $x=a\sin \theta$ then $y=b\cos \theta$ , $dx=a\cos \theta d\theta$
   since $-a<x<a$ so $-\pi /2<\theta <\pi /2$

   $A={\int }_{-a}^{a}2ydx$
   $={\int }_{-\pi /2}^{\pi /2}2ab{\cos }^2\theta d\theta$
   $=ab[\frac{1}{2}\sin 2\theta +\theta {]}_{-\pi /2}^{\pi /2}$
   $=\pi ab$

2. Change of vareables in multiple Integrals and polar coordinates

   let $u=x/a,v=y/b$, we have $x=au,y=bv$ and $u^2+v^2=1$

   The Jacobian of the transformation from (x, y) to (u, v) is
   $\frac{\partial (x,y)}{\partial (u,v)}=\left|\begin{array}{cc}\frac{\partial (x)}{\partial (u)}& \frac{\partial (x)}{\partial (v)}\\ \frac{\partial (y)}{\partial (u)}& \frac{\partial (y)}{\partial (v)}\end{array}\right|=\left|\begin{array}{cc}a& 0\\ 0& b\end{array}\right|=ab$

   $A={\iint }_{R}dA={\iint }_S|\frac{\partial (x,y)}{\partial (u,v)}|dudv={\iint }_{S}abdudv$

   let $u=r\cos \theta$ and $v=r\sin \theta$,

   The Jacobian of the transformation from (u, v) to (r, θ) is
   $\frac{\partial (u,v)}{\partial (r,\theta )}=\left|\begin{array}{cc}\frac{\partial (u)}{\partial (r)}& \frac{\partial (u)}{\partial (\theta )}\\ \frac{\partial (v)}{\partial (r)}& \frac{\partial (v)}{\partial (\theta )}\end{array}\right|=\left|\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right|=r$

   then
   $A={\iint }_{S}abdudv=={\iint }_{S}ab|\frac{\partial (u,v)}{\partial (r,\theta )}|drd\theta ={\int }_0^{2\pi }{\int }_0^{1}abrdrd\theta =\pi ab$

3. Green’s Theorem

   The Green’s Theorem gives the following formulas for the area of D:
   $A={\oint }_{C}xdy={\oint }_{C}ydx=\frac{1}{2}{\oint }_{C}xdy-ydx$

   The ellipse has parametric equations $x=a\cos t$ and $y=b\sin t$, where $0<t<2\pi$. Using the third formula in Equation, we have
   $A=\frac{1}{2}{\oint }_{C}xdy-ydx$
   $=\frac{1}{2}{\int }_0^{2\pi }(a\cos t)(b\cos t)dt-(b\sin t)(-a\sin t)dt$
   $=\frac{ab}{2}{\int }_0^{2\pi }dt=\pi ab$