I have tried to integrate the expression


where t is in [0, π], as a triple integral to get volume. The same function is visualized as shown in the figure below. I know that the surface is enclosed in a box of dimension 0.5 x 0.5 x 2.0, so the box's volume is 0.50 (Unit^3),

Spherical plot of the function

However, when I integrate the expression with various methods, I have got,

V = 0.837758 (Unit^3 !)

Integration of the function in spherical coordinate


  (1/2)*(1/2)*(3*Cos[t]*Cos[t] - 1)*(3*Cos[t]*Cos[t] - 1)*r*r*Sin[t], 
  {r, 0, 1}, {t, 0, Pi}, {p, 0, 2*Pi}]

= 0.837758

My point is: It can be clearly seen that the surface is well encapsulated in the box, so its volume is expected < 0.50 Unit^3, but I'm not getting that result.

So I am confused here. Can anyone help me to figure what went wrong?


1 Answer 1


(1/2)*(1/2)*(3*Cos[t]*Cos[t] - 1)*(3*Cos[t]*Cos[t] - 1) is not the density, it is the maximal radius of the body. So you evaluate the wrong integral. Try this:

 r r Sin[t], 
 {t, 0, Pi}, 
 {p, 0, 2*Pi}, 
 {r, 0, 1/4 (3 Cos[t] Cos[t] - 1)^2}


  • $\begingroup$ Many Thanks it agrees with stochastic int. methods..... $\endgroup$
    – numeric
    Commented Nov 28, 2018 at 6:46
  • $\begingroup$ You're welcome. $\endgroup$ Commented Nov 28, 2018 at 6:54

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