Body bounded with three surfaces

Question

Let B be the body bounded with surfaces:$x^2-y^2=2, x^2-z^2=2, x=2$. Using triple integrals I need to find volume of given body. After projecting this on $OYZ$ plane i get $4$ symmetric parts. And if I take one in the first quadrant, I get bounds $0\le z\le \sqrt{2}, 0\le y\le z$. And I have a problem when it comes to find boundaries for $x$, because we have $3$ surfaces involving $x$. Should I look which one of the first two is closer to plane $x=2$?

Answer 1

For the volume bound between the hyperbolic cylinders and the plane, please note both $y$ and $z$ can be expressed in terms of $x$. So here are your bounds.

$-\sqrt{x^2-2} \leq y \leq \sqrt{x^2-2}$

$-\sqrt{x^2-2} \leq z \leq \sqrt{x^2-2}$

$\sqrt2 \leq x \leq 2$