I am trying to derive the results for the gravitational field for a solid sphere and a spherical shell. In both cases the sphere's radius is $R$ and integration is performed with the origin of the $x$ axis located a the sphere/shell center. For the solid sphere a differential disc is integrated from while for the spherical shell a differential ring is used.
Disc
The differential volume of our differential disc of radius $r$ is written $$dV = 2 \pi r^2 dx$$ where the origin of the $x$ passes through the sphere's center. Integrating this from $x = -R$ to $x = R$ will give the volume of a sphere $$\int dV = \int_{x=-R}^{x=+R} 2 \pi r^2 dx$$ The differential disc's radius $r$ equals $\sqrt{R^2-x^2}$ so this becomes $$V = \int_{x=-R}^{x=+R} 2 \pi (R^2 - x^2) dx$$ Performing this integration we get the expected result $4 \pi R^3/3$.
Ring
The differential surface area of a disc of radius $r$ is written $$dA = \textrm{perimeter} \times \textrm{width} = [2\pi \sin(\theta) R] \times [R d\theta]$$ where $\theta$ is the angle formed between the $x$-axis and sphere radius connected to the top of the ring. Integrating this from $\theta = 0$ to $\theta = \pi$ gives the surface area of a sphere $$\int dA = \int_{\theta = 0}^{\theta = \pi} 2\pi R^2 sin(\theta) d\theta$$ Performing this integration we get the expected result $4\pi R^2$.
Statement
I understand how each of the above work and are derived. What I do not understand is why in the disc case we can integrate over $x$ but for the ring we must integrate over $\theta$.
Question
Why can we use the volume of a cylinder to get $dV$ for the differential disc BUT we can NOT use the surface area of a cylinder to get $dA$ for the differential ring?
That is, we can NOT mimic what we used to find $dV$ and write $$dA = 2 \pi r \times dx$$ replace $r = \sqrt{R^2 - x^2}$ and then integrate from $x = -R$ to $x = +R$? Doing that integration does NOT give $4 \pi R^2$. Am I just missing something obvious?