Make Aristotle Proud

Calculus Level 5

Suppose we are given the circle x 2 + ( y − 1 ) 2 = 1 x^2+(y-1)^2=1 and we mark the point on it that resides at the origin. Then we decide to roll it on the x x -axis (in the positive x x -direction) one full rotation (with no slipping). The path of the point we marked traces a cycloid. Then we take the cycloid and rotate it about the x x -axis to form a surface of revolution. Then we take this surface of revolution and place it on a flat surface so that its two points of greatest curvature remain at the same height above the flat surface. Slice the surface of revolution with an imaginary plane that crosses through its two points of greatest curvature and the point where the surface of revolution touches the flat surface to mark a curve that is the intersection of the plane with the underside of the surface of revolution. Finally, roll this surface of revolution so its two points of greatest curvature remain at the same height, and the very bottom does not slip as it rolls. The marked curve traces a new surface. The surface area of this surface can be given by:

∫ 0 2 π ∫ 0 2 π ( a − cos ⁡ u ) 4 sin ⁡ 2 v + ( a − cos ⁡ u ) 2 ( b − ( a − cos ⁡ u ) cos ⁡ v ) 2 + [ ( a − cos ⁡ u ) sin ⁡ u sin ⁡ 2 v − sin ⁡ u cos ⁡ v ( b − ( a − cos ⁡ u ) cos ⁡ v ) ] 2 d u d v . \displaystyle \int_0^{2\pi}\int_0^{2\pi} \sqrt{(a-\cos{u})^4\sin^2v+(a-\cos{u})^2(b-(a-\cos{u})\cos{v})^2+[(a-\cos{u})\sin{u}\sin^2{v}-\sin{u}\cos{v}(b-(a-\cos{u})\cos{v})]^2}\,du\, dv.

Enter your answer as a + b a+b .


The answer is 3.

James Wilson by James Wilson , 28, USA

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1 solution

James Wilson
Jan 21, 2021

With the help of my understanding of Aristotles's wheel paradox, I determined that the equations for the surface generated by the rolling surface of revolution are:

x = 2 v − ( 1 − cos ⁡ u ) sin ⁡ v x=2v-(1-\cos{u})\sin{v}

y = 2 − ( 1 − cos ⁡ u ) cos ⁡ v y=2-(1-\cos{u})\cos{v}

z = u − sin ⁡ u . z=u-\sin{u}.

Furthermore,

∂ r ∂ u = ⟨ − sin ⁡ u sin ⁡ v , − sin ⁡ u cos ⁡ v , 1 − cos ⁡ u ⟩ , \frac{\partial r}{\partial u}=\langle-\sin{u}\sin{v},-\sin{u}\cos{v},1-\cos{u}\rangle,

∂ r ∂ v = ⟨ 2 − ( 1 − cos ⁡ u ) cos ⁡ v , ( 1 − cos ⁡ u ) sin ⁡ v , 0 ⟩ . \frac{\partial r}{\partial v}=\langle 2-(1-\cos{u})\cos{v},(1-\cos{u})\sin{v},0 \rangle.

∂ r ∂ u × ∂ r ∂ v = ⟨ − ( 1 − cos ⁡ u ) 2 sin ⁡ v , ( 1 − cos ⁡ u ) ( 2 − ( 1 − cos ⁡ u ) cos ⁡ v ) , − ( 1 − cos ⁡ u ) sin ⁡ u sin ⁡ 2 v + sin ⁡ u cos ⁡ v ( 2 − ( 1 − cos ⁡ u ) cos ⁡ v ) ⟩ . \frac{\partial r}{\partial u}\times\frac{\partial r}{\partial v}=\langle -(1-\cos{u})^2\sin{v},(1-\cos{u})(2-(1-\cos{u})\cos{v}),-(1-\cos{u})\sin{u}\sin^2{v}+\sin{u}\cos{v}(2-(1-\cos{u})\cos{v}) \rangle.

The surface area is equal to ∫ 0 2 π ∫ 0 2 π ∣ ∂ r ∂ u × ∂ r ∂ v ∣ d u d v . \int_0^{2\pi}\int_0^{2\pi}|\frac{\partial r}{\partial u}\times\frac{\partial r}{\partial v}|dudv.

In the end, you get a = 1 a=1 and b = 2 b=2 .

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