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

Question

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 & $r\,d\theta$ (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) = 2\pi 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 $r\,d\theta$, the disc radius is not changing (across arc length) & we integrate it as: $$\int s(x)\, rd\theta $$

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\,cos\theta, y = r\,sin\theta $) i get an extra $cos\theta$ since $ dy = rd\theta.cos\theta$

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?

Answer 1

When you have a ball $B_R:=\bigl\{(x,y,z)\bigm| x^2+y^2+z^2\leq R\bigr\}$ and its boundary $S_R:=\partial B_R= \bigl\{(x,y,z)\bigm| x^2+y^2+z^2= R\bigr\}$ at stake then there are various variables around: Of course $x$, $y$, $z$, and then $r:=\sqrt{x^2+y^2+z^2}$, the geographical longitude $\phi:=\arg(x,y)$, and the geographical latitude $$\theta:=\arg\bigl(\sqrt{x^2+y^2},z\bigr)\quad\in\left[-{\pi\over2},{\pi\over2}\right]\ ,$$ whereby sometimes other normalizations are in place.

Now you are told to compute the volume of $B_R$, or the area of $S_R$. Both tasks involve some integration. This integration can take place in $(x,y,z)$-space, or in the space of spherical coordinates $(r,\phi,\theta)$, and it can also involve "heuristical" arguments, depending on your state of sophistication.