Solution to Problem #70

Ignacio Larrosa Caņestro found the surface area to be 17π/12 = 4.450589... Here is his solution:

i) Volume

V = 8*Int(1, (x, y, z), A)

where A = {(x, y, z) in R^3 | 0 ≤ x^(2/3) + y^(2/3) + z^(2/3) ≤ 1}

Let

(r, s, t) = (x^(2/3), y^(2/3), z^(2/3))

So (x, y, z) = (r^(3/2), s^(3/2), t^(3/2))

The Jacobian is (27/8)(r*s*t)^(1/2), and

V = 8*(27/8)*Int(r^(1/2)*s^(1/2)*t^(1/2), (r, s, t), T)

where T = {(r, s, t) in R^3 | 0 ≤ r, s, t ≤ 1}

We have

V = 27*Int(Int(Int(r^(1/2)*s^(1/2)*t^(1/2), t, 0, 1- r - s), s, 0, 1 - r),
r, 0, 1)

  = 27*Int(r^(1/2)*Int(s^(1/2)*Int(t^(1/2), t, 0, 1- r - s), s, 0, 1 - r),
r, 0, 1)

  = 18*Int(r^(1/2)*Int(s^(1/2)*(1 - r - s)^(3/2), s, 0, 1 - r), r, 0, 1)

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For I = Int(s^(1/2)*(1 - r - s)^(3/2), s, 0, 1 - r), we have

I = Int(s^2*((1 - r - s)/s)^(3/2), s, 0, 1 - r)

Let

u^2 = (1 - r)/s - 1  ===> s = (1 - r)/(1 + u^2)

ds = -2u(1 - r)/(1 + u^2)^2

I = -2*(1 - r)^3*Int(u^4/(1+u^2)^4, u, inf, 0)

  = 2*(1 - r)^3*Int(u^4/(1+u^2)^4, u, 0, inf)

Let  u = tan(v), du = (1/cos^2(v))dv,

I = 2*(1 - r)^3*Int(cos^2(v)*sin^4(v), v, 0, pi/2)  (int. by parts)

 = 2*(1 - r)^3*(1/5)*Int(sin^6(v), v, 0, pi/2)

 = 2*(1 - r)^3*(1/5)*(5/6)*Int(sin^4(v), v, 0, pi/2)

 = 2*(1 - r)^3*(1/5)*(5/6)*(3/4)*Int(sin^2(v), v, 0, pi/2)

 = 2*(1 - r)^3*(1/5)*(5/6)*(3/4)*(1/2)*Int(1, v, 0, pi/2)

 = 2*(1 - r)^3*(1/5)*(5/6)*(3/4)*(1/2)*(pi/2) = (1 - r)^3*pi/16

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V  = (9pi/8)*Int(r^(1/2)*(1 - r)^3, r, 0, 1) = 4pi/35 ~= 0.359039...

(Incidentally, the volume in the positive octant is V/8 = pi/70,
coincidence with the problem number?)

The result can be obtained more quickly using Eulerian functions Beta and
Gamma and/or Dirichlet formula.

ii)

The surface equation is

z = (1 - x^(2/3) - y^(2/3))^3/2

Then

S = 8*Int((1 + (dz/dx)^2 + (dz/dy)^2)^(1/2), (x, y), A)

 =  8*Int(((x^(2/3) + y^(2/3) - x^(4/3) - x^(2/3)*y^(2/3) -
y^(4/3))/(x^(2/3)*y^(2/3)))^(1/2), (x, y), A)

where A = {(x, y) in R^2 | 0 ≤ x^(2/3) + y^(2/3) ≤ 1}

Let (r, s) = (x^(2/3), y^(2/3))


So (x, y) = (r^(3/2), s^(3/2))

The Jacobian is (9/4)(r*s)^(1/2), and

S = 8*(9/4)*Int(Int((r + s - r^2 - r*s - s^2)^(1/2), (r, s), T)

where T = {(r, s) in R^2 | 0 ≤ r, st ≤ 1}. We have

S = 18*Int(Int((r(1 - r) + s(1 - r) - s^2), s, 0, 1 - r), s, 0, 1)

 = 18*Int((1 - r)(3r + 1)*atan((r(1 - r))^(1/2)/(2r))/4 + (1 - r)(r(1 -
r))^(1/2)/2, r, 0, 1)

 = 17pi/12  ~= 4.450589592