Proving the Volume of a Sphere (Cylindrical Co-ordinates feat. triple integrals)

We are always taught that the volume of a sphere is 4/3πr^3 at school. But how can this be proved? One of the ways is to use cylindrical co-ordinates and integrate over suitable ranges in each of this co-ordinate system’s dimensions.

But before we can do this, what are cylindrical co-ordinates?

Cylindrical co-ordinates is similar to polar co-ordinates, and in many respects it is it’s 3D equivalent. Co-ordinates are given in the form (r,θ,z):

  • r is the distance from the origin in the x-y plane of the point we are describing (similar to the modulus of a complex number)
  • θ is the angle r makes with the positive x axis
  • z is the height above the x-y plane where the point lies

In essence it is the polar co-ordinate, with the height above the x-y plane added.

Image result for cylindrical coordinates

Cylindrical co-ordinates


Once we have established what this co-ordinate system is all about we are going to integrate in all three of the dimensions to get a volume. We need to set bounds for this triple integral, as this will define the solid that we will get the volume of.

We are going to be finding the volume of a Sphere with radius “a”. We are not using “r” to not confuse it with the r part of our co-ordinate system. So what we be the limits to our integrals? :

  • r will always be between 0 and a to be within the sphere so:  0 ≤ r ≤ a
  • θ will be between 0 and 2π so: 0 ≤ θ ≤ 2π
  • For z it is slightly less straight forwards. In the x-z plane the sphere will be a circle of radius “a” therfore:

We therefore have the limits for our triple integral and can go on to evaluate it.

First we integrate “r” with respect to z. This will of course give us “rz”. When we evaluate this between the limits we previously set we get the following, now we have a double integral left to evaluate:

Next we have to evaluate the the integral with respect to “r”. It is clear that a u-substitution will make this integral feasible:

*I know that the derivative is infact -2r, but by switching my limit (a^2 should be at the bottom) to the top, I can cancel out the negative from the derivative*

The final step is the most simple. Evaluating this integral with respect to θ between 2π and 0 is equivalent to just multiplying it by 2π, as the 0 term will be equal to 0:

Thanks to PlanckTime user @fematika for suggesting this post. I really enjoyed learning about this co-ordinate system and doing this challenge. You can find his forum post here. 

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