Determine the area of the ellipse $A=\{(x,y) \in \mathbf{R}^² \mid 3x^2+4y^2\leqslant12 \}$

Question

Determine the area of the ellipse $A=\{(x,y) \in \mathbf{R}^² \mid 3x^2+4y^2\leqslant12 \}$

I tried to use polar coordinates here, but couldn't get it to work.

If I have $x=r\cos(\theta)$ and $y=r\sin(\theta)$ I get that $$3r^2\cos^2(\theta)+4r^2\sin^2(\theta)\leqslant 12 \Longrightarrow \frac{r^2\cos^2(\theta)}{4}+\frac{r^2\sin^2(\theta)}{3} \leqslant 1$$

but I cannot get anywhere from here. Similarly, if I don't do the polar conversion I have that $$\frac{x^2}{4}+\frac{y^2}{3} \leqslant 1$$

but this also doens't seem to get me any furhter. What should I do here?

Answer 1

Given I have seen surface integral / line integral questions from you, I am assuming you are aware of Green's theorem. Using parametrization of the ellipse centered at the origin, here is just an alternate method using Green's theorem trick -

$r(t) = (a \cos t, b \sin t)$. Here $a = 2, b = \sqrt3$. So, $\, r(t) = (2 \cos t, \sqrt3 \, \sin t)$

$r'(t) = (- 2 \sin t, \sqrt3 \cos t)$

As per Green's theorem $\int_C P \, dx + Q \, dy = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA \,$ for a simple closed curve.

If we choose a vector field for line integral over our ellipse such that $Q = \frac{x}{2}, P = - \frac{y}{2}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$

So, $A = \int_C P \, dx + Q \, dy = \displaystyle \frac{1}{2} \int_0^{2\pi} x \, dy - y \, dx = \sqrt3\int_0^{2\pi} (sin^2t + cos^2t) \,dt = 2 \sqrt3 \pi$

Answer 2

Hint

Doing the substitution $(x,y)=(2r\cos(\theta ),\sqrt{3}r\sin(\theta ))$ yields $$\text{Area}(A)=\iint_A\,\mathrm d x\,\mathrm d y=\int_0^{2\pi}\int_0^1 \sqrt{12}\,r\,\mathrm d r\,\mathrm d \theta =...$$