Help with change of variables please

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The discussion revolves around evaluating a double integral using change of variables, specifically the integral of √(x² + y²) over the region [0,1] x [0,1]. The original poster initially considered using polar coordinates but was advised against it due to the square region not being suitable for such a transformation. Instead, they successfully applied a change of variables by setting u = x² and v = y², calculating the Jacobian, and determining the limits of integration remained [0,1] x [0,1]. The resulting integral was computed to yield approximately 3.238, but concerns were raised about the accuracy of the Jacobian calculation and whether polar coordinates could be effectively utilized. Ultimately, the discussion highlighted the complexities of integrating over non-circular regions and the importance of correctly applying the Jacobian in transformations.
  • #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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