91Ó°ÊÓ

First, find the intersection points of the given sphere and hyperboloid by setting \(x^2 + y^2 + z^2 = 19\) and \(z^2 - x^2 - y^2 = 1.\) We can solve these equations by setting \(x^2 + y^2 = r^2,\) and substitute \(z^2 - r^2\) for \(z^2 - x^2 - y^2\) from the equation of the hyperboloid. So we have \(r^2 + z^2 = 19,\) and \(z^2 - r^2 = 1.\) Solving the system of equations, we get \(r^2= l^{2}, z=4,\) where \(l=3,\)

We will use spherical coordinates, which are given by the transformations: \(x = \rho\sin\phi\cos\theta,\) \(y = \rho\sin\phi\sin\theta,\) and \(z = \rho\cos\phi.\) The Jacobian of this transformation is \(\rho^2\sin\phi.\) Convert the sphere and hyperboloid equations: Sphere: \((\rho\sin\phi\cos\theta)^2+(\rho\sin\phi\sin\theta)^2+(\rho\cos\phi)^2=19\), which simplifies to \(\rho^2=19\) Hyperboloid: \((\rho\cos\phi)^2-(\rho\sin\phi\cos\theta)^2-(\rho\sin\phi\sin\theta)^2=1,\) which simplifies to \(\cos^2\phi=1+\frac{1}{\rho^2}\).

Now we set up the triple integral over the region bounded by the sphere and hyperboloid. Our limits of integration for \(\rho\) will be from 3 to \(\sqrt{19},\) for \(\theta\) will be from 0 to \(2\pi,\) and for \(\phi\) will be from the arccosine of \(\sqrt{1+\frac{1}{\rho^2}}\) (derived from the hyperboloid equation) to \(\pi\): \(\int_{0}^{2\pi}\int_{3}^{\sqrt{19}}\int_{\arccos\left(\sqrt{1+\frac{1}{\rho^2}}\right)}^{\pi} (\rho^2\sin\phi)d\phi\,d\rho\,d\theta\) Evaluate the integral, we get: \(2\pi\int_{3}^{\sqrt{19}}\rho^2[-\cos\phi]_{\arccos\left(\sqrt{1+\frac{1}{\rho^2}}\right)}^{\pi}d\rho\) \(2\pi\int_{3}^{\sqrt{19}}\rho^2(\cos\arccos\left(\sqrt{1+\frac{1}{\rho^2}}\right)+1)d\rho\) \(2\pi\int_{3}^{\sqrt{19}}\rho^2(\sqrt{1+\frac{1}{\rho^2}}+1)d\rho\) After evaluating the remaining integral, we get the volume of the region as \(V \approx 181.164\) cubic units.