Find the volume of solid defined by the following inequalities :
$$z \geq 3x^2+2y^2, \ \ 3x^2+2y^2+5z^2 \le 1$$
We have an ellipse, which the semi-axis are $\sqrt{\frac{z}{2}}$ and $\sqrt{\frac{z}{3}}$, so the area is $A(z)=\frac{\pi z}{\sqrt{6}}$. So now how can I find the volume bounded by the equations ? I didn't see anything so far.
The first inequality describes a region above a paraboloid; the second inequality is the interior of an ellipsoid. But because a scaling transformation on both inequalities $$u = x \sqrt{3}, \quad v = y \sqrt{2}, \quad w = z \sqrt{5}$$ gives $$w \ge \sqrt{5}(u^2 + v^2), \quad u^2 + v^2 + w^2 \le 1,$$ and the latter is now a sphere, it is straightforward to consider the volume in $(u,v,w)$-space as a solid of revolution described by the inequalities $$r^2 \sqrt{5} \le w \le \sqrt{1 - r^2},$$ where $r^2 = u^2 + v^2$. The intersection point where $r^2 \sqrt{5} = \sqrt{1-r^2}$ is given by $$r = c = \sqrt{\frac{\sqrt{21}-1}{10}},$$ so the resulting volume by cylindrical shells is $$V^* = 2\pi \int_{r=0}^c r\left(\sqrt{1-r^2} - r^2 \sqrt{5}\right) \, dr.$$ Since the scaling transformation also scaled the original volume by a factor of $\sqrt{3}\sqrt{2}\sqrt{5}$, the actual volume is $$V = V^* /\sqrt{30}.$$