Volume of a Sphere with Cylinder in the Middle Question: Let $b$ be a positive constant. Suppose a sphere with radius $2b$, has a cylindrical hole with radius $b$. Find the volume of the sphere using the shell method.
My Attempt:
I tried to think about the general formula for the volume of the sphere, and the cylinder. Knowing that it would have to be $$V = V_{\text{sphere}} - V_{\text{cylinder}}$$
The height of the cylinder would be $h = 2r_{\text{sphere}}$, which will be $h = 4b$. And so, I computed this formula,
\begin{align*}
V &= V_{\text{sphere}} - V_{\text{cylinder}} \\
&= \frac{4}{3}\pi r_{\text{sphere}}^3 - \pi r_{\text{cylinder}}^2h \\
&= \frac{4}{3}\pi (2b)^3 - \pi (b)^2(4b) \\
&= \frac{20\pi b^3}{3}
\end{align*}
If this is the volume of the sphere, I am not sure how I am supposed to show this using the shell method. Any assistance would help! A picture below depicts the problem.

 A: You have note mentioned which axis cylinder is along but let's say it is parallel to $z-$axis. Then in cylindrical coordinates, we have
Equation of sphere: $r^2+z^2 = 4b^2$
Equation of cylindrical hole: $r = b$
To find volume using cylindrical shell method, please take shells along the axis of the cylinder.
At any given radius, $-\sqrt{4b^2 - r^2} \leq z \leq \sqrt{4b^2 - r^2}$
So the height of the shell is $2 \sqrt{4b^2 - r^2}$
Also $b \leq r \leq 2b, \ $ as we are bound by radius of the cylinder and radius of the sphere.
So integration to find volume of the given sphere with cylindrical hole using shell method is,
$\displaystyle \int_b^{2b} 2 \pi r \cdot 2\sqrt{4b^2 - r^2} \ dr $
As far as your calculation without the integration,
at the intersection of cylinder and sphere, we have $z = \pm \sqrt3b$.
So the cylindrical hole is a cylinder of height $2\sqrt3b$ and two spherical caps on top and bottom,
Height of each spherical cap $ = (2-\sqrt3)b$ and radius of its base = $b$.
So, volume of cylinder along with spherical caps
$V_{hole} = 2b\sqrt3 \cdot \pi b^2 + 2 \cdot \frac{\pi b^3}{6} (2-\sqrt3) \big(3 + (2-\sqrt3)^2\big)$
$ = \frac{32 \pi b^3}{3} - 4 \sqrt3 \pi b^3$
So $V_{sphere} - V_{hole} = 4 \sqrt3 \pi b^3$
