Why Doesn't the Derivative of Sphere Volume Yield Circle Area?

[Psyguy22]

If you take the derivative of the area of a circle, you get the formula for circumference. When you take the derivative of the volume of the sphere, you do not get the formula for the area of a circle. Why not?
d/dr (4/3pi r^3) =4pi r^2
d/dr (pi r^2)= 2pi r

 

[Mute]

What you actually get when you take the derivative of the volume with respect to radius is the surface area of the ball. Note that there is a technical difference between a ball and a "sphere": a sphere is, strictly speaker, the surface of a ball. It does not include the volume it encloses, whereas a "ball" includes the surface and the volume contained within. Occasionally the forum gets questions about the "volume of a sphere" and someone will answer that it is zero, which is technically correct because the volume of a "sphere", interpreted literally, refers to the volume the surface itself, which is zero, and not the volume contained by the surface; the OP in these cases pretty much always means the volume of the ball and was just unaware of the precise distinction in the terminology. So, just pointing that out.

Anyways, a way to see why it works like this is to consider the following: to build a circle of area $\pi R^2$, you can think of the process of building "shells" of circles of increasing radius, where each shell has an infinitesimal thickness $dr$. By adding more and more shells you are increasing the area of the circle you are building. If you have a circle of radius r, the infinitesimal change in area you get when adding another shell is $dA = 2\pi r dr$ - the circumference of the shell times the thickness.

When you then go and start building a ball in a similar manner, you are not adding shells of circles. Rather, you are adding shells of spheres of thickness $dr$ and surface area $4\pi r^2$. So, the infinitesimal change in volume as you add a shell to a ball of radius r is $dV = 4\pi r^2 dr$ - the surface area times the thickness.

Does that make sense?

 

[lurflurf]

area of circle:circumference::volume of sphere:surface area of sphere
pi r^2:2pi r::4pi r^3/2:4pi r^2
A:A'::V=V'

you get surface area of the sphere

This follows from Stokes theorem

$$\int_\Omega \mathrm {d}\omega = \int_ {\partial \Omega} \omega$$

 

[Psyguy22]

So my old geometry book lied? Isn't a sphere a 3-D shape? Meaning it would have volume? What exactly does the equation 4/3pi r^3 represent then?

 

[pwsnafu]

So my old geometry book lied? Isn't a sphere a 3-D shape? Meaning it would have volume?

Sort of. In natural English sphere is a 3D object. In mathematics it's a 2D object. From Wikipedia:

In mathematics, a careful distinction is made between the sphere (a two-dimensional surface embedded in three-dimensional Euclidean space) and the ball (the interior of the three-dimensional sphere).

Your geometry textbook wasn't being careful.

What exactly does the equation 4/3pi r^3 represent then?

Mute already told you: the term is ball.

 

[Psyguy22]

Ok. Thank you. And sorry for the misuse of words. So just to make sure I understand
4/3 pi r^3 is the volume of a 'ball' and 4 pi r^2 is the surface area of a 'sphere'?

 

[arildno]

Ok. Thank you. And sorry for the misuse of words. So just to make sure I understand
4/3 pi r^3 is the volume of a 'ball' and 4 pi r^2 is the surface area of a 'sphere'?

It is, of course, the surface area of the ball itself, the sphere being its SURROUNDING BOUNDARY.
(Just as the circumference of the DISK constitutes the surrounding circle bounding the disk)

For nice geometrical objects, the surrounding boundary of the object is of 1 dimension lower than the object itself.