# 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?

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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