Help with change of variables please

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SUMMARY

The discussion centers on evaluating the double integral of the function \(\sqrt{x^2 + y^2}\) over the square region \(R = [0,1] \times [0,1]\) using change of variables. The original poster initially considered using polar coordinates but was advised against it due to the square's geometry. Instead, they successfully transformed the variables by setting \(u = x^2\) and \(v = y^2\), calculating the Jacobian as 1, and integrating over the same limits, resulting in an approximate value of 3.238. However, further insights revealed that the Jacobian was incorrectly computed, leading to a more complex integral.

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  • Understanding of double integrals and their evaluation
  • Familiarity with change of variables in integration
  • Knowledge of Jacobians and their computation
  • Basic concepts of polar coordinates and their application in integration
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  • Study the computation of Jacobians in variable transformations
  • Learn about integrating functions in polar coordinates, especially in non-circular regions
  • Explore the properties of integrals over different geometric shapes
  • Practice evaluating double integrals using various change of variables techniques
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  • #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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