Gravitational potential of a disc [closed]

Question

The question says

Find the potential at the center of a disc whose surface area density varies as $$σ = σ_0(1+\cosθ)r $$ where theta is the angle made by the radius with the horizontal and $r$ is the distance of the point from the center

My textbook says is to first integrate -$$-Gσ_0(1+\cosθ)r\mathrm{d}r\mathrm{d}θ$$ with respect to dr, then integrate the result with respect to d0. I have understand the integration process, I wanted to know the physical meaning between integrating like suppose for finding for a ring we choose a small portion dm then we find for the complete ring but here at every point the mass density varies, how exactly the integration works. Can this process be physically interpreted the one we did for ring?

Answer 1

In this setup, mass density of the disc is a function of two variables, namely $\theta$ and $r$.

Hence, double integration is required to solve for the gravitational potential of the disc at the centre.

We first consider an differential ring of thickness $dr$ and then a differential patch of length $rd\theta$ on this ring. This area of this patch is $dA$.

While doing double integration, we first integrate w.r.t that variable whose element is considered last, here, $\theta$. While integrating w.r.t $\theta$, $r$ is considered to be constant.

Here, we are basically finding the gravitational potential at the centre due to a differential ring by summing up the contribution of each small patch on the ring.

Next, we integrate w.r.t $r$.

Here, we sum up the contribution of each differential ring to find the gravitational potential due to the whole disc.

Note:

For this particular integral, the result will come out to be same if we first integrate w.r.t $r$ and then w.r.t $\theta$. However, it is more intuitive to first integrate w.r.t $\theta$.