Cylinder Average Distance

Calculus Level 3

Consider the following cylinder:

x = cos ⁡ θ y = sin ⁡ θ − 1 ≤ z ≤ + 1 0 ≤ θ ≤ 2 π x = \cos \theta \\ y = \sin \theta \\ -1 \leq z \leq +1 \\ 0 \leq \theta \leq 2 \pi

What is the average distance from a point on the cylinder to the point ( x , y , z ) = ( 1 , 0 , 0 ) (x,y,z) = (1,0,0) ?

Note: Take a surface-area-weighted average


The answer is 1.426.

Steven Chase by Steven Chase , 37, USA

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1 solution

Using the Fact that on a right circular cylindrical surface of radius a a , The surface area element in cylindrical coordinates is given by d S = a d z d θ dS = adzd\theta So averaging over the top half of the cylinder(since it is symmetric about the z-axis ) we have

1 2 π ∫ 0 2 π ∫ 0 1 4 sin ⁡ 2 ( θ 2 ) + z 2 d z d θ = 1.426 \displaystyle \frac{1}{2\pi}\int_{0}^{2\pi}\int_{0}^{1} \sqrt{4\sin^{2}(\frac{\theta}{2}) + z^{2}}\, dzd\theta = 1.426

I do not know how to calculate this integral in terms of known antiderivatives so I used Desmos to calculate it.

I tried using circular strips of constant radius and thickness d z dz to find the surface area average but I ran into symmetry problems. If anyone has a solution to this integral or even a separate approach which leads to a doable integral by hand then please post it.

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