How the surface area of a cone changes; special relativity

[Hiero]

Homework Statement

A cone has half angle θ₀ and lateral surface area S₀ in the frame in which the cone is at rest. If someone moves at relative speed β=v/c along the cones symmetry axis, what surface area will they see for the cone?

Homework Equations

I believe the lateral surface area of a cone is S=πh²tanθ, where θ is the half angle and h is the height of the cone.
Length contraction is also relevant; h₁=h₀√(1-β²)

The Attempt at a Solution

S₀ = πh₀²tanθ₀
S₁ = πh₁²tanθ₁

If the cone had maximum radius R in the rest frame, it should remain R in the frame moving along the symmetry axis (because R is measured normal to this axis).
So we have tanθ₀=R/h₀, and also tanθ₁=R/h₁, so we see tanθ₁=(h₀/h₁)tanθ₀

Putting all this together (with the length contraction relation) it seems to be that S₁=S₀√(1-β²)

However my book claims S₁=S₀√(1-(βcosθ)²); so where is my mistake?

 

[Orodruin]

You have the wrong expression for the surface area. See https://en.wikipedia.org/wiki/Cone#Surface_area

 

[Hiero]

Ahh. For some reason, when I thought over that formula, I had in my mind that "R=h*sinθ" instead of "R=h*tanθ"
(I even remember thinking this formula over a couple times, and each time I kept incorrectly saying this to myself! I am tired.)

So then the correct formula is actually S=πh²tanθ/cosθ.

So then my answer is off by (needs to be multiplied by) a factor of cosθ₀/cosθ₁

If you take the ratio of these two equations
cosθ₀=1/√(1+(R/h₀)²)
cosθ₁=1/√(1+(R/h₁)²)=1/√(1+(R/h₀)²/(1-β²))
and simplify a fair amount (then use R/h₀=tanθ as well as 1/(1+tan²θ)=cos²θ) then multiply it by my incorrect answer then it indeed becomes the correct answer, so that was the only mistake.

Thank you Orodruin