Proposed problem

Dear Editor, I would like to propose the following problem

Let 2ax = y^{2}, z = 0 (a > 0) be the equations of a parabola in the (x, y) plane of ℝ^{3}. Let C be the cone of vertex in P = (0, 0, a) and generatrices the segments from P to the points of the parabola. Let S be the sphere of equation x^{2} + y^{2} + z^{2} − az = 0. (1) Calculate the area of the portion of C inside S. (2) Calculate the area of that portion of S inside C.

Results: (1) 3π\sqrt{2}a^{2}/16, (2) π\sqrt{2}a^{2}/4

Proof (1). The parametric equations of the cone are

(x, y, z) ≐ (φ_{1}, φ_{2}, φ_{3})(t, v)

where

(φ_{1}, φ_{2}, φ_{3})(t, v) = (0, 0, a) + t [(\frac{v^{2}}{2a}, v, 0) − (0, 0, a)]

= (t\frac{v^{2}}{2a}, tv, a − ta) , 0 ≤ t ≤ 1, −∞ < v < +∞

The point (x, y, z) of the cone lies inside S if

\frac{t^{2}v^{4}}{4a^{2}} + v^{2}t^{2} + a^{2}(1 − t)^{2} ≤ a^{2}(1 − t), ⟺ t ≤ \frac{a^{2}}{(\frac{v^{2}}{2a} + a)^{2}} < 1

(\frac{∂φ_{1}}{∂t} , \frac{∂φ_{2}}{∂t} , \frac{∂φ_{3}}{∂t} ) ∧ (\frac{∂φ_{1}}{∂v} , \frac{∂φ_{2}}{∂v} , \frac{∂φ_{3}}{∂v} ) = e_{1}at − e_{2}tv − e_{3} \frac{v^{2}t}{2a}

where as usual e_{k} is the k–th versor in R^{3}.

∥(φ_{1}, φ_{2}, φ_{3})_{t} ∧ (φ_{1}, φ_{2}, φ_{3})_{v}∥ = \sqrt{a^{2}t^{2} + t^{2}v^{2} + \frac{v^{4}t^{2}}{4a^{2}}} = \frac{v^{2}t}{2a} + ta

The area of C inside S is given by the integral

∫_{−∞}^{+∞} dv ∫0^{\frac{a^{2}}{(\frac{v^{2}}{2a} +a)^{2}}} dt(\frac{v^{2}}{2a} + a)t = \frac{8a^{7}}{2} ∫_{−∞}^{+∞} dv \frac{1}{(v^{2} + 2a^{2})^{3}} = \frac{3π\sqrt{2}a^{2}}{16}

The last one is a standard integral computable by standard methods (integration by parts and a recursive argument). I prefer to use complex integration and residue’s theory so

4a^{7} ∫_{−∞}^{+∞} dv \frac{1}{(v^{2} + 2a^{2})^{3}} = \frac{4a^{7}\sqrt{2}a}{8a^{6}} ∫_{−∞}^{∞} \frac{dy}{(1 + y^{2})^{3}} =

= \frac{a^{2}}{\sqrt{2}}2πi lim_{v→i} \frac{1}{2} \frac{d^{2}}{dv^{2}} \frac{1}{(v + i)^{3}} = \frac{a^{2}πi}{\sqrt{2}} \frac{12}{(2i)^{5}} = \frac{3π\sqrt{2}a^{2}}{16}

(2). We adopt stereographic parametric equations for the sphere

x ≐ φ_{1}(u, v) = \frac{a^{2}u}{u^{2} + v^{2} + a^{2}} , y ≐ φ_{2}(u, v) = \frac{a^{2}v}{u^{2} + v^{2} + a^{2}} ,

z ≐ φ_{3}(u, v) = \frac{a(u^{2} + v^{2})}{u^{2} + v^{2} + a^{2}}

The point of the sphere (x, y, z) lies inside the cone C if and only if v^{2} ≤ 2au e u ∈ (−∞, +∞).

(\frac{∂φ_{1}}{∂u} , \frac{∂φ_{2}}{∂u} , \frac{∂φ_{3}}{∂u} ) ∧ (\frac{∂φ_{1}}{∂v} , \frac{∂φ_{2}}{∂v} , \frac{∂φ_{3}}{∂v} ) =

= e_{1} \frac{−2a^{6}u}{(u^{2} + v^{2} + a^{2})^{2}} + e_{2} \frac{2a^{6}v}{(u^{2} + v^{2} + a^{2})^{2}} + e_{3} \frac{a^{4}(a^{2} − u^{2} − v^{2})}{(a^{2} + v^{2} + u^{2})^{3}}

∥(φ_{1}, φ_{2}, φ_{3})_{u} ∧ (φ_{1}, φ_{2}, φ_{3})_{v}∥ = \frac{a^{4}}{(u^{2} + v^{2} + a^{2})^{2}}

The area of S inside C is

a^{4} ∫_{−∞}^{+∞} dv ∫ +∞_{\frac{v^{2}}{2a}} \frac{du}{(u^{2} + v^{2} + a^{2})^{2}} = \frac{πa^{2}\sqrt{2}}{4}

To prove it we introduce polar coordinates in the plane (u, v) u = r cos ϑ, v = r sin ϑ, −π/2 ≤ ϑ ≤ π/2, and v^{2} ≤ 2au becomes r ≤ 2a cos ϑ/(sin ϑ)^{2} so the integral is

a^{4} ∫_{−π/2}^{π/2} dϑ ∫0^{\frac{2a cos ϑ}{sin^{2} ϑ}} \frac{r}{(r^{2} + a^{2})^{2}} dr = a^{4} ∫_{−π/2}^{π/2} dϑ\frac{1}{2} (\frac{1}{a^{2}} − \frac{1}{\frac{4a^{2} cos^{2} ϑ}{sin^{4} ϑ} + a^{2}} )

that is

\frac{a^{2}π}{2} − \frac{a^{2}}{2} ∫_{−π/2}^{π/2} \frac{1}{\frac{4 cos^{2} ϑ}{sin^{4} ϑ} + 1}dϑ

In the integral let’s set ϑ = arctan t and get

\frac{a^{2}π}{2} − \frac{a^{2}}{2} ∫_{−∞}^{+∞} \frac{1}{1 + t^{2}} \frac{t^{4}}{(t^{2} + 2)^{2}} dt

No we use the residue–theorem

−\frac{a^{2}}{2} 2πi (\frac{i^{4}}{2i(−1 + 2)^{2}} + lim_{t→\sqrt{2}i} \frac{d}{dt} \frac{t^{4}(t − \sqrt{2}i)^{2}}{(1 + t^{2})(t − \sqrt{2}i)^{2}(t + \sqrt{2}i)^{2}} ) =

= −\frac{a^{2}}{2} 2πi (\frac{i^{4}}{2i(−1 + 2)^{2}} + \frac{4(\sqrt{2}i)^{3}}{(1 − 2)(2\sqrt{2}i)^{2}} − \frac{2(\sqrt{2}i)^{4}}{(1 − 2)(2\sqrt{2}i)^{3}} − \frac{(\sqrt{2}i)^{4}}{(2\sqrt{2}i)^{2}} \frac{2\sqrt{2}i}{(1 − 2)^{2}} ) =

= −\frac{a^{2}π}{2} + πa^{2}\sqrt{2} − \frac{a^{2}π}{2\sqrt{2}} − a^{2}π\sqrt{2} = −\frac{a^{2}π}{2} + \frac{a^{2}π}{4} \sqrt{2}

whence the final result a^{2} \frac{π}{2} − a^{2} (\frac{π}{2} − \frac{π}{4} \sqrt{2}) = \frac{a^{2}π\sqrt{2}}{4}.

Roma(Italy) 11/09/2016 Best regards Paolo Perfetti

Perfetti Paolo, Dipartimento di Matematica, Università degli studi di Tor Vergata Roma, via della ricerca scientifica, 00133 Roma, Italy – email: perfetti@mat.uniroma2.it