Integrate f(x,y) Over the Set A

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Homework Help Overview

The problem involves integrating the function f(x,y) = e^{-(x^2 + y^2)} over a specified region A in the first quadrant defined by the conditions x > 0, y > 0, and x^2 + y^2 < a. The discussion centers around the application of polar coordinates and the change of variables theorem.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply a polar coordinate transformation to simplify the integration process. Some participants question the necessity of mentioning diffeomorphisms in the context of the problem, while others emphasize the importance of the change of variables theorem.

Discussion Status

The discussion is ongoing, with some participants expressing agreement with the original poster's approach. However, there are differing opinions on the relevance of certain mathematical concepts, indicating a mix of interpretations regarding the problem's setup.

Contextual Notes

Participants are navigating the implications of using the change of variables theorem and the conditions under which it applies, particularly in relation to the properties of the transformation function.

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Homework Statement



Integrate f(x,y) = e^{-(x^2 + y^2)} over the set A = \{ (x,y): x &gt; 0, y &gt; 0, x^2 + y^2 &lt; a \}



Homework Equations





The Attempt at a Solution



The polar coordinate transformation g(r,\theta) = (r cos\theta, r sin\theta) is a diffeomorphism from A to the set B = \{(r,\theta): 0 &lt; r &lt; a, 0 &lt; \theta &lt; \pi / 2 \}, so I can use the change of variables theorem.

So \int_A e^{-(x^2 + y^2)} = \int_B re^{-r^2} = \int_0^a \int_0^{\pi/2} re^{-r^2} d \theta dr = (\pi / 2) \int_0^a re^{-r^2} dr.

Let u = -r^2. Then -du/2 = rdr, so we have (\pi / 2) \int_0^a re^{-r^2} dr = (- \pi / 4) \int_0^{-a^2} e^u du = (- \pi / 4)(e^{-a^2} - 1).

I get the same answer if I reverse the order of integration. Is this answer correct?
 
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It all looks fine to me.
 
All good buddy. Why even mention diffeomorphisms? This be calculus, there be dragons there.
 
Because as far as I know, the change of variables theorem requires g to be a C^r diffeomorphism.
 

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