Hopefully this explanation makes sense, as its tricky to convey what I mean exactly and there is almost certainly terminology that I should be using I'm unaware of!

My question relates to something I intuitively suspect to be true, but am not sure how to go about expressing mathematically as its a bit beyond where my mathematical education stopped!

I'm interested in working out what the apparent 2D area of a 3D cylinder is, when viewed from a specific point, when this 3D cylinder is rotated around a fixed point (such that the ends of the cylinder would trace out a sphere). To my mind, this is akin to asking what the area of the shadow cast by the object would be when illuminated from one direction (without any 'magnification effects' of the shadow over distance). Since this is tricky to explain, I've drawn a diagram of what I perceive to be the 'solution'.

enter image description here

In the most extreme cases, starting from the cylinder being perpendicular to the observer the 'area' of the cylinder should be simply $h x d$ where $h$ is the height, and $d$ is the diameter, in this case $h_p$, which I'm calling the 'projected height', is equal to $h$ if we ignore any magnification. If its rotated 90˚, then $h_p$ becomes the same as $d$, and the area is then $\pi r^2$, if $r = d/2$.

When the cylinder is somewhere in between 0˚ and 90˚, its $h_p$ and therefore area will be some reduced amount. This will also apply to all 360˚ (and it should be enough to consider just one hemisphere since the situation is symmetric).

So tl;dr

What is the general solution/how would one go about calculating the average 'observed' area for a cylinder that rotates to trace out every point on a sphere?

The real motivation for this is to understand how cylindrical nanoparticle sizes might be 'misreported' when sized by light scattering (which typically uses spherical ones).

Hopefully that makes sense, and I'd just ask for simple explanations where possible as I'm no mathematician!


1 Answer 1


The average projection area of a convex 3D body is equal to $1/4$ of its surface Area. You can read more about it here (in particular, the proof for 3D case is given in section 6).

  • $\begingroup$ Perfect! Thank you - I knew this must be known and reasonably trivial. $\endgroup$
    – Joe Healey
    Commented Oct 22, 2019 at 11:01
  • $\begingroup$ Would it be consistent to calculate the average dimension for $h_p$ in this case - presumably rearranging $ A_s = 2 \pi rh + 2 \pi r^2$ for $h$ should be equivalent? $\endgroup$
    – Joe Healey
    Commented Oct 22, 2019 at 11:10
  • $\begingroup$ If you integrate over semicirlce and not hemisphere, you need to account for a non-uniform probability density $\endgroup$ Commented Oct 22, 2019 at 11:16
  • $\begingroup$ Indeed, this would presumably be normally distributed, peaking toward the centre, but would become maximal where the distance from the centre point is equal to or less than the diameter, since this would be covered at all times. $\endgroup$
    – Joe Healey
    Commented Oct 22, 2019 at 11:33

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