I'm adding this answer because the other answers so far answer different questions (either a different region, or a different method) than what was originally asked. I'm also going to try to explain my work in more depth to make it easier to follow along.
Note that the cone $z = \sqrt{(x^2 + y^2)}$ can be represented in spherical coordinates as $\phi = \frac{\pi}{4}$. You already calculated that $z = r \sin \phi$, and we know from the spherical coordinate transform that $z = r \cos \phi$. Combining these two equations, we get $\sin \phi = \cos \phi$ or $tan \phi = 1$, which means $\phi = \frac{\pi}{4}$. Intuitively/geometrically, you can also imagine that the cone covers the full range from $\theta = 0$ to $\theta = 2 \pi$ of rotation around the z-axis, meaning $\theta$ can be any value, and the radius can be any value as well (the cone is unbounded above), so the only parameter needed to define the cone is $\phi$.
To set up the region of integration, we know that to calculate the volume of the sphere, we would have the radius ranging from 0 to 1, with $\theta$, the angle of rotation around the z-axis, ranging from 0 to $2 \pi$, and $\phi$, the angle with the z-axis, ranging from 0 to $\pi$. In this case however, we want to exclude the region of the cone, which is the region from $\phi = 0$ to $\phi = \frac{\pi}{4}$. Thus we want to include $\phi$ only over the range from $\frac{\pi}{4}$ to $\pi$.
Thus, we have the below integral:
$$\int_{0}^{1} \int_{0}^{2 \pi} \int_{\frac{\pi}{4}}^{\pi} r^2 sin(\phi) d\phi d\theta dr$$
where $r^2 sin(\phi)$ comes from the change of base formula, and can be calculated as the absolute value of the determinant of the derivative matrix of $x$, $y$, $z$ with respect to $r$, $\theta$, and $\phi$.
The integral can be evaluated in any order, probably $\theta$ then $r$ then $\phi$ to get $2 \pi \frac{1}{3} (1 + \frac{1}{\sqrt{2}}) \approx 3.58$.
A quick intuitive check confirms that this answer makes sense: the volume of the sphere is given by $\frac{4}{3}\pi r^3$ which in this case is just $\frac{4}{3}\pi \approx 4.19 $, and we're taking out a small fraction of the top half of the sphere.
In spherical coordinates, you are looking for the volume of the following region:
$$
E=\{(\rho,\theta,\phi)\;|\;0\le \rho\le 4\cos\phi,0\le \theta \le2\pi, \frac{\pi}{4} \le \phi\le\frac{\pi}{2}\}
$$
So your volume equals
$$
V(E)=\iiint_E \rho^2\sin\phi\; d\rho d\theta d\phi
$$
Best Answer
First find where the cone intersects the inner sphere:
$$x^2+y^2=1-z^2$$
$$z^2=3-3z^2$$
$$z^2=\frac{3}{4}$$
This means that the radius of the boundary between the cone and inner sphere is $\frac{1}{2}$, and the radius of the boundary between the cone and the radius of the boundary for the outer sphere can be found to be $\frac{\sqrt{33}}{2}$, using the same process. Converting the sphere equations into cylindrical coordinates and using as bounds $0 \le \theta < 2\pi$ and $\frac{1}{2}\le r < \frac{\sqrt{33}}{2}$, the volume integral is as follows:
$$V=\int_0^{2\pi}\int_{\frac{1}{2}}^{\frac{\sqrt{33}}{2}}(\sqrt{9-r^2}-\sqrt{1-r^2})rdr d\theta$$
A similar procedure can be followed to find the integral in rectangular and spherical coordinates.