Help with change of variables please

  • #31
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
Hi heman,
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. :smile:
 
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  • #32
that i realize Hyper...But why did u write the Cosec For the above triangle..What is the Problem with Sec...Does the eqn of the line y=1 was to be written or how the r is varying ...Pls Help
 
  • #33
heman said:
that i realize Hyper...But why did u write the Cosec For the above triangle..What is the Problem with Sec...Does the eqn of the line y=1 was to be written or how the r is varying ...Pls Help
Hi heman,
If you look at the square, you will see that for the upper triangle, if we measure by the angle the radius makes with the line t=pi/2 (the y-axis), we get the relationship cos(t0) = 1/r. t0 is related to the t of polar coordinates by t0 = pi/2 - t, so we have 1/r = cos(pi/2 - t) = sin(t), so r = 1/sin(t) = csc(t).
 

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