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

Question

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.

Answer 1

What you have computed is the volume of a larger region, which is the region that you have been given plus the cone whose vertex is located at the origin and whose base is the the intersection of the solid sphere centered at the origin with radius $\sqrt{10}$ and the plane $z=2$.

I suggest the use of cylindrical coordinates:\begin{align}\int_0^{2\pi}\int_2^{\sqrt{10}}\int_0^{\sqrt{10-z^2}}r\,\mathrm dr\,\mathrm dz\,\mathrm d\theta&=2\pi\int_2^{\sqrt{10}}5-\frac{z^2}2\,\mathrm dz\\&=20\pi\frac{\sqrt{10}}3-52\frac\pi3.\end{align}

Answer 2

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 $\frac{2}{\cos\phi}\leq r\leq\sqrt{10}$.