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

Mathematics Asked by Vignesh Sk on October 1, 2020

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?

One Answer

When you have a ball $B_R:=bigl{(x,y,z)bigm| x^2+y^2+z^2leq Rbigr}$ and its boundary $S_R:=partial B_R= bigl{(x,y,z)bigm| x^2+y^2+z^2= Rbigr}$ 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 $$theta:=argbigl(sqrt{x^2+y^2},zbigr)quadinleft[-{piover2},{piover2}right] ,$$ 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.

Answered by Christian Blatter on October 1, 2020

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