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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.

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    With a change of variable , the problem boils down to finding the volume of the intersection between a sphere and a paraboloid. That is simple through the disk method (every $x$-section is a circle). – 2017-01-24
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    I think your answer is incorrect. Is $30\pi$ given to you in the question? – 2017-01-24

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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.