Triple integral, how to bound the region

Question

Given a region, which is bounded below by z=0, and enclosed by x^2+y^2+4z^2=a^2, where a is positive. I want to know how to bound the region. Which coordinates should I use, cylindrical or spherical? Is there some technics to do this situation?

Answer 1

If the surface was $x^2+y^2+z^2 = a^2$, you'd want to use spherical coordinates. Since the surface is $x^2+y^2+(2z)^2 = a^2$ instead, a slight modification yields bounds which are constants. Set $$x = \rho \cos\theta \sin \phi, \quad y = \rho \sin\theta \sin\phi, \quad z = \dfrac{1}{2}\rho \cos\phi.$$

The bounds are $\rho \in [0,a]$, $\theta \in [0,2\pi]$, and $\phi \in [0,\tfrac{\pi}{2}]$ (since $z \ge 0$), and the Jacobian of the transformation (if you need it) is $\tfrac{1}{2}\rho^2\sin\phi$.