Find the gravitational force with which the homogeneous

Find the gravitational force with which the homogeneous cylinder $x^2+y^2\le a^2,\,0\le z\le h$, of mass $M$, attracts a material point $P(0,0,b)$ ($b>h$) of mass $m$.

Step 1 · Setup and symmetry

With density $\mu=\dfrac M{\pi a^2h}$, symmetry about the $z$-axis gives $F_x=F_y=0$. The $z$-component of the force from a volume element at $(x,y,z)$, distance $r=\sqrt{x^2+y^2+(b-z)^2}$ from $P$, is $dF_z=\dfrac{Gm\mu(b-z)}{r^3}\,dV$.

Step 2 · Cylindrical coordinates

$$ F_z = 2\pi Gm\mu\int_0^h(b-z)\int_0^a\frac{\rho\,d\rho}{\left[\rho^2+(b-z)^2\right]^{3/2}}\,dz. $$

Step 3 · Radial integral

With $u=\rho^2+(b-z)^2$, and using $b-z>0$ for $z\in[0,h]$:

$$ \int_0^a\frac{\rho\,d\rho}{\left[\rho^2+(b-z)^2\right]^{3/2}} = \frac1{b-z}-\frac1{\sqrt{a^2+(b-z)^2}}. $$

Step 4 · Remaining $z$-integral

$$ F_z = 2\pi Gm\mu\int_0^h\left(1-\frac{b-z}{\sqrt{a^2+(b-z)^2}}\right)dz = 2\pi Gm\mu\left[h-\sqrt{a^2+b^2}+\sqrt{a^2+(b-h)^2}\right], $$

using $\displaystyle\int_{b-h}^b\frac{t\,dt}{\sqrt{a^2+t^2}}=\sqrt{a^2+b^2}-\sqrt{a^2+(b-h)^2}$ (substitution $t=b-z$). Substituting $\mu=\dfrac M{\pi a^2h}$:

$$ F_z = \frac{2GmM}{a^2h}\left[h+\sqrt{a^2+(b-h)^2}-\sqrt{a^2+b^2}\right]. $$

Since $\sqrt{a^2+(b-h)^2}<\sqrt{a^2+b^2}$ but the bracket is nonetheless positive (the deficit is less than $h$, as it must be for the force to represent a genuine attraction), $F_z>0$; physically the force pulls $P$ toward the cylinder, i.e. in the $-z$ direction.

Additional Notes

• This derivation matches the source exactly. (The source's own working includes a brief moment of uncertainty about the sign at the end, but resolves to the same correct magnitude and direction given here.)
• Verified numerically (e.g. $a=1.4,\,h=2.3,\,b=5.1,\,M=3.7$, $G=m=1$): direct triple integration gives $F_z\approx0.232814$, matching the closed form exactly.