I want to calculate this volume region, using cylindrical coordinates:
$$D=\{y^2+z^2\le5+x^2,4x^2+y^2+z^2\le25\}$$
So, I have a hyperboloid and an ellipsoid.
Is it correct to calculate the volume like this:
$$2\int_{0}^{2\pi}\int_{0}^{3}\int_{2}^{\frac12\sqrt{25-r^2}}rdxdrd\theta + 2\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{\sqrt{5+x^2}}rdrdxd\theta$$
Is there a more efficient method ?
(I don't understand your first integral. The second is o.k.)
Your domain $D$ is a diabolo with the $x$-axis as axis of rotation. It is the intersection of the interior of a one-sheet hyperboloid with an ellipsoid. For the computation we put $y^2+z^2=:\rho^2$. The admissible values for $\rho$ depend on the variable $x$. When $x\geq0$ we have $$0\leq\rho\leq \rho(x):=\sqrt{5+x^2}\quad(0\leq x\leq2),\qquad 0\leq\rho\leq\rho(x):=\sqrt{25-4x^2}\quad\left(2\leq x\leq{5\over2}\right)\ .$$ We now cut up $D$ into "infinitesimal discs" of radius $\rho(x)$ using planes orthogonal to the $x$-axis. These discs have volume $\pi\rho^2(x)\>dx$. It follows that $${\rm vol}(D)=2\pi\int_0^2\rho^2(x)\>dx+2\pi\int_2^{5/2}\rho^2(x)\>dx\ .$$