Integrating an exponential function over [itex]|x|+|y| \leq 1[/itex]

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Discussion Overview

The discussion revolves around the integration of the exponential function \( e^{-2(x+y)} \) over a diamond-shaped region defined by \( |x| + |y| \leq 1 \). Participants explore the setup of the double integral and considerations regarding the symmetry of the region.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant notes the symmetry of the diamond region over the x-axis and y-axis but questions the applicability of this symmetry to the integrand \( e^{-2(x+y)} \).
  • Another participant suggests splitting the integral into two parts, indicating that this approach could simplify the integration process.
  • A request for clarification on how to divide the diamond region is made, indicating a need for more detailed explanation on setting up the integrals.
  • A specific method is proposed to divide the diamond along the y-axis, leading to two separate double integrals for the left and right halves of the diamond.

Areas of Agreement / Disagreement

Participants express differing views on the best method to set up the integral, with some suggesting splitting the region while others seek clarification on the specifics of this approach. No consensus is reached on the optimal method for integration.

Contextual Notes

Participants do not fully resolve the mathematical steps involved in setting up the integrals, and there are assumptions regarding the boundaries of integration that remain unclarified.

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OK, I'm new to multi-variable calculus and I got this question in my exercises that asks me to integrate e^{-2(x+y)} over a diamond that is centered around the origin:

\int\int_D e^{-2x-2y} dA

where D=\{ (x,y): |x|+|y| \leq 1 \}

I know that the region I'm integrating over is symmetric over the x-axis and the y-axis, but e^{-x} or e^{-y} are neither odd nor even to use the symmetry that way.

Obviously, the diamond is symmetric over the axes x+y and x-y. Does this help?
 
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Split it into two integrals where one is over one half of the diamond and the other is over the other half!
Then you can write the boundaries of the integrals from D=\{ (x,y): |x|+|y| \leq 1 \} easily.
 
Which haves?

Give me some more details please, I'm not looking for a full solution, only an explanation of how you're setting up the integral.
 
One choice is dividing the diamond in half by the y axis.Then you have two double integrals,one for the left half and the other for the right one.The interval for the left half integral is -(x+1) \ to \ x+1 for integration w.r.t. y and -1 to 0 for integration w.r.t. x.I think you can figure out the intervals for the right half integral by plotting D.
 

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