Integral with elliptical coordinates Good evening everyone. I'm doing an integral of which I know the result but it comes out different to me.
Anyone able to tell me where am I wrong?
The result is $\frac{64}{27} \sqrt{3} \pi$.
The starting integral was a triple integral. Passing in elliptical coordinates I found:
$x = \frac 23 + \frac 43\rho cos\theta; y=\frac {2\sqrt{3}}{3}\rho sen \theta $
with  $\rho  \in [0,1), \theta \in [0, 2\pi).$
$\int \int (-3x^2+4x+4-4y^2) dx dy  = \int_0^{2\pi} \int_0^1 [-3(\frac 49+ \frac {16}{9} \rho^2cos^2\theta+\frac {16}{9} \rho\cos\theta)+\frac 83 + \frac {16}{3}\rho\cos\theta+4-\frac {16}{3}\rho^2sen^2\theta] d\rho d\theta=   \int_0^{2\pi} \int_0^1 [- \frac 43 - \frac {16}{3}\rho^2cos^2\theta -\frac {16}{3}\rho cos\theta + \frac 83 + \frac {16}{3}\rho cos \theta +4 - \frac {16}{3}\rho^2sen^2\theta] d\rho d\theta =\int_0^{2\pi} \int_0^1  [\frac {16}{3}-\frac {16}{3}\rho^2] d\rho d\theta = \frac {64\pi}{9}$
The starting integral is:
$\int_\Omega (2z) dxdydx$
$\Omega = (x,y,z) \in R^3 | 2 \sqrt {x^2+y^2}<z<x+2$

In short, you are missing the Jacobian of transformation in your integral. If you plug that in, rest of your work is correct.
If the projection of $z = x + 2$, intersected by the paraboloid surface, in xy-plane is $E$,
$x + 2 \geq 2 \sqrt{ (x^2 + y^2)} \implies 3x^2 - 4x + 4 y^2 \le 4$
$ \displaystyle E: ~\left(x - \frac 23\right)^2 + \frac 43 y^2 \leq \frac {16}9$
We use substitution $ \displaystyle x = \frac 23 + \frac 43 \rho \cos\theta~; ~y=\frac {2\sqrt{3}}{3}\rho \sin \theta$
$$0 \leq \rho \leq 1, 0 \leq \theta \leq 2\pi$$
$ \displaystyle |J| = \frac{ 8 \rho}{3 \sqrt3}$
Also note that,
$\begin {aligned} 
-3x^2+4x+4-4y^2 &= -3 \left(x^2 - \frac{4x}{3} + \frac{4y^2}{3} - \frac 43\right) \\
&= \frac{16}3 - 3 \left[\left(x - \frac{2}{3}\right)^2 + \frac{4y^2}{3}\right] \\
&= \frac{16}3 (1 - \rho^2) \\
\end {aligned}$
$ \begin {aligned} 
\iiint_{\Omega} 2 z ~dV &=  \iint_E [(x + 2)^2 - 4 (x^2 + y^2)] ~dx ~ dy \\
&=  \iint_E (-3x^2+4x+4-4y^2) ~dx ~ dy \\
&= \int_0^{2\pi} \int_0^1  \frac {16}{3} (1 -\rho^2) ~|J| ~ d\rho ~ d\theta \\
&= \frac{128}{9 \sqrt3} \int_0^{2\pi} \int_0^1 (\rho - \rho^3) ~d\rho ~d\theta = \frac{64 \sqrt 3 \pi}{27} \\
\end {aligned}$
