hypermorphism
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Hi heman,heman said:That was quite impressive Hyper...But does it really matters to calculate the other one too...
It seems obvious that both of them are equal and if u are calculating the above one with cosec
You're right. Since the surface {(r, t, r)} (polar/cylindrical coords) is symmetric with respect to t, it's sufficient to compute the integral above one of the triangles and multiply by 2. For surfaces without symmetry, we don't inherit that convenience, though.