Volume enclosed by a spherical function with constants

Questions
1

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

2

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]…

3

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