Volume of a piece of ellipsoid I want to compute the volume of $A:=\{(x,y,z)\in\mathbb{R}^3:\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{25}\leq 1; x\geq\frac{y^2}{9}+\frac{z^2}{25}\}$ (which I think is a piece of ellipsoid) so I set up the following integral: $\iiint_A 1 dxdydz=\int_{x=0}^{x=2}\int_{-3\sqrt{x}}^{3\sqrt{x}}\int_{-\frac{5}{3}\sqrt{9x^2-y^2}}^{\frac{5}{3}\sqrt{9x^2-y^2}}1 dzdydz=30\pi$ which should be equal to the required volume.
Is this correct? (i.e. is the integral I computed equal to the volume of $A$)
Best regards,
lorenzo.

The intersection between the ellipsoid and the paraboloid is given by 
$$
\frac{x^2}{4}+x=1\quad\Rightarrow\quad x=2(\sqrt{2}-1)>0
$$
So the projection of this intersection in the plane $x=0$ is the ellipse
$$
\frac{y^2}{9}+\frac{z^2}{25}=2(\sqrt{2}-1)
$$
Let $D$ be the region bounded by this ellipse:
$$
D=\{(y,z)|\frac{y^2}{9}+\frac{z^2}{25}\le2(\sqrt{2}-1) \}
$$
Your solid is thus defined by
$$
E=\{(x,y,z)|(y,z)\in D,\; \frac{y^2}{9}+\frac{z^2}{25} \le x \le 2\sqrt{1-\frac{y^2}{9}-\frac{z^2}{25}} \}
$$
To integrate this, consider the non distorted version of your problem: suppose the ellipsoid is a sphere with radius $1$ and that the paraboloid has equation $x=y^2+z^2$. In this case, $D$ is a disc with radius $R=\frac{\sqrt{5}-1}{2}$, and the volume equals:
$$
V_0=\int_0^{2\pi}\int_0^R\int_{r^2}^{1-r^2}rdr d\theta = \frac{5\pi(3-\sqrt{5})}{12}
$$
You can adapt this last equation with the coefficients of the initial equations.
