# 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 & $$r\,d\theta$$ (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?

• Sep 29, 2020 at 5:44

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.