Area of ellipse using double integral

Question

I am trying to find the area of a quadrant of an ellipse by double integrating polar coordinates but the answer I'm getting is incorrect.

ellipse : $ x^2/a^2 + y^2/b^2 =1 $

Any point on ellipse : $ ( a\cos(\theta), b\sin(\theta)) $

At $ \theta$, taking $ d\theta $ segment, Thus $ r^2 = a^2\cos^2(\theta) + b^2\sin^2(\theta) $ [Using pythagoras theorem] $$ Area = \int_{0}^{\pi/2} \int_{0}^{\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)}} rdrd\theta $$

$$ = 1/2 \int_{0}^{\pi/2} r^2 \Big|_{0}^{\sqrt{a^2\cos^2(\theta)+b^2\sin^2(\theta)}} d\theta $$

$$ = 1/2 \int_{0}^{\pi/2} (a^2\cos^2(\theta)+b^2\sin^2(\theta)) d\theta $$ $$ = 1/2 \int_{0}^{\pi/2} ((a^2 - b^2)\cos^2(\theta)+b^2) d\theta $$ $$ = 1/4 \int_{0}^{\pi/2} (a^2 - b^2)(1+ \cos(2\theta)) d\theta +2b^2 d\theta $$ I am getting $$ \pi/8 (a^2 + b^2).$$ But the correct answer is $ \pi ab/4 $

Answer 1

Your mistake is to believe that $\theta$ is the polar argument. It is not, because

$$\tan\phi=\frac yx=\frac ba\tan \theta\ne\tan\theta.$$

You can fix by taking the differential

$$(\tan^2\phi+1)\,d\phi=\frac ba(\tan^2\theta+1)\,d\theta$$

and substituting

$$\left(\frac{b^2}{a^2}\tan^2\theta+1\right)\,d\phi=\frac ba(\tan^2\theta+1)\,d\theta$$

so that the integral becomes

$$\frac12\int_0^{\pi/2}\frac{\dfrac ba(\tan^2\theta+1)}{\dfrac{b^2}{a^2}\tan^2\theta+1}(a^2\cos^2\theta+b^2\sin^2\theta)\,d\theta \\=\frac{ab}2\int_0^{\pi/2}\frac{\sin^2\theta+\cos^2\theta}{b^2\sin^2\theta+a^2\cos^2\theta}(a^2\cos^2\theta+b^2\sin^2\theta)\,d\theta.$$

Answer 2

Set $x=ar\cos \theta, y=br\sin \theta.$ The Jacobian is $abr$ and we compute the area $$\mathcal {A}= \int_0^{\pi/2} \int_0^1 abr \;d r \;d\theta,$$ which is $\frac{ab\pi}{4}.$

Answer 3

The given ellipse is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.To transform it into polar coordinates,substitute $(x,y)=(r\cos\theta,r\sin\theta)$ to get $r=\dfrac{ab}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}$.

Take an elementary area $rdrd\theta$ inside the ellipse.

Then the area of the ellipse in the first quadrant is the sum of all such elementary areas

$\displaystyle\int_{\theta=0}^{\frac{\pi}{2}}\int_{r=0}^{\dfrac{ab}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}} r \,dr \, d\theta$

$=\dfrac12\displaystyle\int_{\theta=0}^{\frac{\pi}{2}}\dfrac{a^2b^2}{b^2\cos^2\theta + a^2\sin^2\theta} d\theta$

$=\dfrac12\displaystyle\int_{\theta=0}^{\frac{\pi}{2}}\dfrac{a^2b^2\sec^2\theta}{b^2+ a^2\tan^2\theta} d\theta$

$=\dfrac{b^2}{2}\displaystyle\int_{\theta=0}^{\frac{\pi}{2}}\dfrac{\sec^2\theta}{\dfrac{b^2}{a^2}+ \tan^2\theta} d\theta$

$=\left.\dfrac{b^2}{2}\dfrac{a}{b}\tan^{-1}\left(\dfrac{a\tan\theta}{b}\right)\right|_{0}^{\frac{\pi}{2}}$

$=\boxed{\boxed{\dfrac{\pi ab}{4}}}$

Answer 4

You can also use green's theorem. Set x=acos($\theta$), y=bsin($\theta$). Now we can set up the intergal. $A=\frac{1}{2}\int_{c} xdy-ydx. \\ dy=bcos(\theta), dx=-asin(\theta) \\ A=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}abcos^2(\theta)+absin^2(\theta)\text{ }d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}ab \text{ }d\theta=\frac{\pi ab}{4}. \\ $