# Integration/Differentiation of Formulas

Periapsis
So I am new to calculus, and have been playing around with integration and differentiation. I have learned that integrating and deriving formulas, you get other formulas that can be used, like integrating surface area of a sphere to get volume. I was wondering why this is, and what limits you from continuing to integrate on and on. For example, you can integrate the surface area of a sphere, but integrating the volume of a sphere doesnt get you anything.

Staff Emeritus
Gold Member
2021 Award
For the sphere - ball relationship, the key is that if you slice up a ball into spheres of constant radius, then each spherical slice has volume (surface area)*(delta r) and adding all of those up gives the volume of the ball. To integrate the ball's volume and get something useful you need to have a four dimensional shape which can be sliced up into a bunch of three dimensional balls.

Homework Helper
Gold Member
I was slicing up a 4 dimensional orange with my high-tech 4 dimensional knife and accidentally cut 4 fingers at once. 4D is tricky. Periapsis
@Office_Shredder
So to integrate anything, you need a theoretical higher exponent type dimension? Then why can you take a two dimensional surface area and integrate it to find the volume? V = ∫ f(r) 4πr2 dr right? Integrating that would result in a cubed radius?

Staff Emeritus
Gold Member
2021 Award
theoretical higher exponent type dimension?

I have no idea what these words in succession mean.

Then why can you take a two dimensional surface area and integrate it to find the volume? V = ∫ f(r) 4πr2 dr right? Integrating that would result in a cubed radius?

You can take a two dimensional surface area and integrate it to find the volume, in the sense that you can integrate the surface area of a sphere to get the volume of a ball. I don't know what your function f(r) is supposed to be here - if it's completely arbitrary, then you're just integrating an arbitrary function.

Periapsis
So lets take position for example, a function of time. s(t). Why is it that when you take the derivative of position, you get velocity, and from there, acceleration, jerk, and jounce? Why does it start with a function of time?

Staff Emeritus
Gold Member
2021 Award
What are your definitions of velocity and acceleration and jerk and jounce? Velocity is typically defined as the instantaneous change in position as a function of time, which is the definition of the derivative. Acceleration is the instantaneous change in velocity as a function of time, which is again a derivative by definition.

If you have a different definition then we can work through how one is a derivative of another.

Periapsis
I suppose what i'm really trying to ask, why is position, a function of time. and why is velocity, a function of position? why is acceleration a function of velocity? and why is jerk a function of acceleration?

Staff Emeritus
Gold Member
2021 Award
Acceleration is not a function of velocity, it's also a function of time (in the typical physics setup). Jerk is a function of time as well. If s(t) is position, v(t) velocity, a(t) acceleration and j(t) jerk, we have
s'(t) = v(t)
s''(t) = v'(t) = a(t)
s'''(t) = v''(t) = a'(t) = j(t)

Periapsis
Okay, now I understand that, thanks. But now what about using that same principal to solve for anything? What is circumference a function of? I learned the integral of the circumference, is the area. But what is the beginning formula?