Double Integration for Ellipse $\frac {x^2}{4} + \frac {y^2}{36} =1 $

Question

I have to evaluate the integral for the region covered by $$\frac {x^2}{4} + \frac {y^2}{36} =1 $$ for which I use the change of variables $$x=2u$$ $$y=6v$$ and I get the answer as $$6\pi$$ but isn't the area of ellipse $$\pi ab$$ by which I should get $12\pi$.

Where am I going wrong?

Answer 1

Area of ellipse $= \displaystyle \int_{-2}^{2} \displaystyle \int_{- \sqrt {36-9x^2}}^{\sqrt {36-9x^2}} dxdy$

Using replacement $2u = x, 6v = y$,

$= 12 \displaystyle \int_{-1}^{1} \displaystyle \int_{- \sqrt {1-u^2}}^{\sqrt {1-u^2}} dudv = 24 \displaystyle \int_{-1}^{1} \sqrt {1-u^2} \, du = 24 \displaystyle \int_{-\pi/2}^{\pi/2} \sqrt {1-\sin^2 \theta} \, \cos\theta d\theta$

$= 24 \displaystyle \int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta = 12 \displaystyle \int_{-\pi/2}^{\pi/2} (\cos2\theta + 1) d\theta = 12 \pi$

Answer 2

You probably forgot the jacobian, if you introduce a substitution $x=2u\land y=6v$ the jacobian will be $6\cdot{2}=12$ and your new region will be a circle $x^2+y^2=1$ for which the area is $r^2\pi$, here, $r=1$ , so ,$12\cdot{\pi}$.

Answer 3

I think you're trying to evaluate the following integral: $$\int\int_{D}dxdy$$ where $D$ is the region bounded by your elipse. If you change variables $x = 2u$ and $y = 6v$ as you did, you have: $$\int \int _{D}dxdy = \int \int_{C}|J(u,v)|dudv$$ where, now, $C$ is the region bounded by the unit circle $u^{2}+v^{2}=1$ and $J(u,v)$ is the Jacobian: $$J(u,v) = \mbox{det}\begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \mbox{det}\begin{pmatrix} 2 & 0 \\ 0 & 6 \end{pmatrix} = -12 $$ Thus: $$\int \int _{D}dxdy = \int \int_{C}|J(u,v)|dudv = 12 \int \int _{C}dudv$$ but $\int \int_{C}dudv$ is the area of a circle with radius $r=1$, so $\int \int_{C}dudv = \pi$.