Find the volume of the part of the spherical shell
Find the volume of the part of the spherical shell bounded by: $$ x^2+y^2+z^2=2 \qquad\text{and}\qquad x^2+y^2+z^2=6 $$ that lies inside the circular cylinder: $$ x^2+z^2=x. $$
Explanation of Method
The cylinder $x^2+z^2=x$, i.e. $\left(x-\frac12\right)^2+z^2=\frac14$, has its axis along the $y$-direction and confines $x,z$ to a disk of radius $\frac12$ centered at $\left(\frac12,0\right)$ in the $xz$-plane. For any point in that disk, $x^2+z^2\leqslant x\leqslant1$, so both spherical radii ($\sqrt2$ and $\sqrt6$, both $>1$) are large enough that the disk sits entirely "inside" each sphere — there is no truncation issue. Writing $W(R)$ for the volume of a ball of radius $R$ intersected with this cylinder, the desired volume is $V=W(\sqrt6)-W(\sqrt2)$.
- Set up $W(R)$ as an integral over the disk in the $xz$-plane, with $y$ ranging over the ball's extent.
- Convert the disk to polar coordinates and reduce to a single integral in $\theta$.
- Note that this integral does not reduce to elementary functions, and evaluate it numerically to high precision.
Step 1 · Set up $W(R)$
For $(x,z)$ in the disk $\left(x-\frac12\right)^2+z^2\leqslant\frac14$, write $\rho^2=x^2+z^2$; then $y$ ranges over $\left[-\sqrt{R^2-\rho^2},\sqrt{R^2-\rho^2}\right]$, so:
$$ W(R)=\iint_{\text{disk}}2\sqrt{R^2-\rho^2}\,dx\,dz. $$
Step 2 · Polar coordinates
The disk's boundary $x^2+z^2=x$ is $\rho=\cos\theta$ in polar form, $\theta\in\left[-\frac\pi2,\frac\pi2\right]$:
$$ W(R)=\int_{-\pi/2}^{\pi/2}\int_0^{\cos\theta}2\sqrt{R^2-\rho^2}\,\rho\,d\rho\,d\theta = \frac23\int_{-\pi/2}^{\pi/2}\left[R^3-\left(R^2-\cos^2\theta\right)^{3/2}\right]d\theta. $$
Step 3 · Evaluate numerically
Writing $R^2-\cos^2\theta=(R^2-1)+\sin^2\theta$, the remaining integral $\displaystyle\int_{-\pi/2}^{\pi/2}\left[(R^2-1)+\sin^2\theta\right]^{3/2}d\theta$ is an elliptic-type integral with no elementary closed form. High-precision numerical evaluation gives:
$$ W(\sqrt6)\approx3.724462373,\qquad W(\sqrt2)\approx1.995605779, $$
so:
$$ V = W(\sqrt6)-W(\sqrt2)\approx1.728856595. $$
Additional Notes
- A note on the source computation: the source explicitly stated that the exact integral "requires numerical methods for exact evaluation," but then presented an unjustified closed-form guess, $\dfrac\pi3\big(8\sqrt6-4\sqrt2-3\big)\approx3.6607$, with no derivation. This value does not match the true volume.
- An independent 3D Monte Carlo simulation (8,000,000 uniform samples in the bounding box $x\in[0,1],\,z\in\left[-\frac12,\frac12\right],\,y\in[-\sqrt6,\sqrt6]$, testing membership in both the cylinder and the spherical shell) gives $V\approx1.7286\pm0.0035$, consistent with the high-precision quadrature value $V\approx1.728856595$ above and inconsistent with the source's guess.
- Unlike several other problems in this set, the discrepancy here is not a missed absolute value or sign error — the integral genuinely does not reduce to elementary functions, so the honest answer is the numerically evaluated volume rather than a closed algebraic expression.
Result
The volume is: