Volume of the region determined by $x^2 + y^2 + z^2 \leq 10, z \geq 2$ This is question 13 on page 294 of Vector Calculus by Marsden and Tromba.

Find the volume of the region determined by $x^2 + y^2 + z^2 \leq 10, z \geq 2.$

I have attempted it as follows.
The region can be described by using spherical polar coordinates $(r,\theta,\phi)$ and we have $0 \leq r \leq \sqrt{10}$, $0 \leq \theta \leq 2 \pi$.
For the limits for $\phi$, I think that we can find it by saying that $z=r \cos \phi = \sqrt{10} \cos \phi \geq 2$ using that $r = \sqrt{10}$ on the surface. This gives $0 \leq \phi \leq \cos^{-1}(2/\sqrt{10}).$ Hence I get
\begin{align} \iiint_V dV &= \int_0^{2\pi} \int_0^{\cos^{-1}(2/\sqrt{10})}\int_0^{\sqrt{10}} r^2 \sin \phi drd\phi d\theta \\ &= 2\pi \int_0^{\cos^{-1}(2/\sqrt{10})} 10 \dfrac{\sqrt{3}}{3} \sin \phi d\phi \\&= 20\pi \dfrac{\sqrt{10}}{3} - 40\dfrac{\pi}{3}. \end{align}
However, the answer at the back of the book is

$20\pi \dfrac{\sqrt{10}}{3} - 52\dfrac{\pi}{3}$

but I have been unable to identify the mistake. Please, could someone help me? Thank you very much.

From $z=r\cos\phi=2$ you get $r=\frac{2}{\cos\phi}$. The bounds of the first integral which is with respect to $r$ must have been $\frac{2}{\cos\phi}\leq r\leq\sqrt{10}$.
