A solid hemisphere of uniform density $\rho$ occupies the region $$x^2+y^2+z^2\le a^2,\qquad z\le0.$$ Find the gravitational potential due to the hemisphere at the point $(0,0,s)$ where $s\gt0$. A uniform rod of density $m$ per unit length lies on the $z$-axis between $(0,0,c)$ and $(0,0,d)$ where $d\gt c\gt0$. Show that the force exerted on the rod by the hemisphere is $$\psi(c)-\psi(d)$$ where $$\psi(\lambda)=\frac{2\pi Gm\rho}{3}\left(\frac{a^3+\lambda^3-\left(a^2+\lambda^2\right)^{3/2}}{\lambda}\right).$$

So to determine the potential at $(0,0,s)$, I evaluated the following integral over the region $R$:

$$\iiint_R\frac{\rho\ G}{\sqrt{x^2+y^2+(z-s)^2}}\,dx\,dy\,dz$$

Using a spherical substitution, the integral becomes:

$$\int_{r=0}^{a}\int_{\theta=\frac{\pi}{2}}^{\pi}\int_{\phi = 0}^{2 \pi}\frac{r^2\sin(\theta)}{\sqrt{r^2+s^2-2rs\cos(\theta)}}d\phi d\theta dr$$

Which evaluates to $$\varphi(s)=\frac{\pi\rho G}{3}\left(\frac{2a^3+3a^2s+s^3-\sqrt{(s^2+a^2)^3}}{s}\right)$$

As for the second part, my understanding is that all we have to do is evaluate


My question is: is there anything wrong with my $\varphi$ or is my method from the second half wrong?



I've got the following:
$\varphi(s)=\frac{2\pi\rho G}{3}\left(\frac{a^3+\frac{3}{2}a^2s+s^3-\sqrt{(s^2+a^2)^3}}{s}\right)$
and a similar approach for the second part, which will give $\psi(c)-\psi(d)$. The $a^2s$ term will be eliminated after subtraction)

My answer might be wrong but maybe check your integration coefficients?


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