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Volume enclosed by a spherical function with constants

Posted 1 year ago

Hi,

I want to determine the volume covered by a spherical function with three constants (a, b, c) given as-

f = a* Cos[Theta]^3 + b* Cos[Theta] Sin[Theta]^2 - c* (Cos[Phi] - 3 Cos[3 Phi]) Sin[Theta]^3

So I tried the Integration -

Integrate[Sin[Theta]*f, {Theta, 0, Pi}, {Phi, 0, 2 Pi}]

Which gives the o/p to be zero, suggesting that the function is symmetric on positive and negative axes irrespective of the values of the three constants. Since I am only interested in determining the volume enclosed by the function, I also tried integrating the absolute value of it, but due to the presence of three constants, it did not succeed.
Would appreciate any suggestion to Integrate this function. thanks

POSTED BY: S G
2 Replies
Posted 1 year ago

Thanks. Any suggestion to find the enclosed volume, other than the Integration which I tried.

POSTED BY: S G

Hi, actually, this is function expressed in polar coordinates, and "f" only gives the radius (which can be negative). Defining

f[Theta_, Phi_] := 
 a*Cos[Theta]^3 + b*Cos[Theta] Sin[Theta]^2 - 
  c*(Cos[Phi] - 3 Cos[3 Phi]) Sin[Theta]^3

I find that

Integrate[Sin[Theta]*f[Theta, Phi], {Theta, 0, Pi}, {Phi, 0, 2 Pi}]

is null. But this is not the right way to measure the volume enclosed in the surface (3 lobes), which is certainly not zero: see

FullSimplify[Sin[Theta]*f[Theta, Phi], Trig -> True]
Print[SphericalPlot3D[% /. {a -> 1, b -> 1, c -> 1}, {Theta, 0, 
   Pi}, {Phi, 0, 2 Pi}, 
  PlotStyle -> Directive[Cyan, Opacity[0.7], Specularity[White, 10]], 
  Mesh -> None, PlotPoints -> 50, AspectRatio -> Automatic, 
  ImageSize -> Large]]

So, this is a bit more complicated than merely integrating Sin[Theta]*f[Theta, Phi]...

POSTED BY: Claude Mante
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