Why is the Volume integration & Surface Area integration of a sphere different?

For both volume & surface area, the sphere is split into many discs and the area or circumference of the discs are summed up in an integral. But the summation process uses $dy$ for volume & $r,dtheta$ (arc-length) for surface area. Why this discrepancy?

Supposing we have a sphere in the $x$–$y$–$z$ plane where you split the sphere into discs along the $y$ axis.. If you visualise the problem from $z$ axis looking down over the $x$–$y$ plane.. The sphere will look like a circle and the disc will be a line segment inside the circle (chord). The length of the line segment will be the diameter of the disc. And the point where the line segment and circle meet – (x,y) can be solved by plugging in the value of y and the x we solve for will then be the radius of the disc.

Now to calculate surface area, we need to sum up the circumference of each disc $ s(x) = 2pi x$ & and for volume, we need to sum up the area of each disc $ v(x) = pi x^2 $

Say, the point $(x,y)$ makes an angle $theta$ with the origin. Then for surface area, we assume for length $r,dtheta$, the disc radius is not changing (across arc length) & we integrate it as: $$int s(x), rdtheta $$

But for volume, instead of using the arc length, we use the diameter $dy$ to integrate it as:
$$int v(x) ,dy$$

Why this discrepancy? In both cases, the number of discs is the same so why should the summation be different?

I tried interchanging the summation process and when i converted everything into polar co-ordinates ($x = r,costheta, y = r,sintheta $) i get an extra $costheta$ since $ dy = rdtheta.costheta$

The same happens to me when i calculate Moment of Inertia for a solid sphere & hollow sphere. Similarly when i calculate gravity for a point outside a solid sphere & hollow sphere.

Can someone please tell me, why we need to change the summation process?? What decides the summation process, why the difference?