# The gravitational potential experienced by a point on the $z$-axis due to a hemisphere

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

$$m\int_{c}^{d}\varphi^{'}(s)\,ds=m\varphi(d)-m\varphi(c)$$

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

Thanks!

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