Find the mass of a circular annulus (ring) if at each
Find the mass of a circular annulus (ring) if at each point the surface density is inversely proportional to the square of the distance from the center of the ring.
Explanation of Method
With density $\mu=k/r^2$, the factor of $r$ from the polar area element $dA=r\,dr\,d\theta$ exactly cancels the $r^2$ in the denominator, leaving an elementary $1/r$ radial integral over the annulus $a\le r\le b$.
- Set up the density function and mass integral.
- Convert to polar coordinates.
- Evaluate the resulting logarithmic integral.
Step 1 · Density and mass integral
With inner radius $a$ and outer radius $b$: $\mu(x,y)=\dfrac k{x^2+y^2}=\dfrac k{r^2}$, and $m=\displaystyle\iint_D\mu\,dA$ over the annulus $D:a\le r\le b$.
Step 2 · Polar coordinates
$$ m = \int_0^{2\pi}\int_a^b\frac k{r^2}\cdot r\,dr\,d\theta = \int_0^{2\pi}\int_a^b\frac kr\,dr\,d\theta. $$
Step 3 · Evaluate
$$ m = 2\pi k\big[\ln r\big]_a^b = 2\pi k\ln\frac ba. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically (e.g. $a=1.2,\,b=3.5,\,k=2.3$): direct polar double integration gives $m\approx15.4693$, matching $2\pi k\ln(b/a)$ exactly.
Result
The mass of the annulus is: