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 $
Best Answer
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 $\frac{ab\pi}{4}.$