Polar coordinates to set up and evaluate double integral

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SUMMARY

The discussion focuses on evaluating the double integral of the function f(x,y) = e^(-(x²+y²)/2) over the region defined by x²+y²≤25 and x≥0 using polar coordinates. The correct setup involves the integral ∫^{π/2}_{-π/2} ∫^{5}_{0} (e^(-r²/2) r) dr dθ. Participants confirm that substitution is necessary for integration, specifically using u = r², and clarify that limits of integration can be adjusted accordingly without explicitly changing them during substitution.

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mikky05v
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Homework Statement



Use polar coordinates to set up and evaluate the double integral f(x,y) = e-(x2+y2)/2 R: x2+y2≤25, x≥0


The Attempt at a Solution



First I just want to make sure I'm understanding this

my double integral would be

∫^{\pi/2}_{-\pi/2} because x≥0 ∫^{5}_{0} because my radius is 5 (e-(x2+y2)/2) r dr dθ

and then my inside would become ∫^{\pi/2}_{-\pi/2} ∫^{5}_{0} (e-r2/2) r dr dθ

can anyone confirm for me that this is correct and give me a brief break down on integrating.

obviously I would use substitution because I have r er2 but the -1/2 is throwing me a bit when it comes to the substitution.

Also how would i go about changing the limits while I'm substituting.
u= r2
du = 2r dr
isn't there something I have to do with my limits of integration that involves my u and du?
 
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mikky05v said:
u= r2
du = 2r dr
isn't there something I have to do with my limits of integration that involves my u and du?
Consider another u substitution. You don't have to explicitly change the bounds - you can just call them u1 and u2 midcalculation and then sub back in the r dependence at the end.
 
What about u= r^2/2?

You can either, as CAF123 says, do the integration and then change back to r, or you can just replace the "r" limits with the corresponding "u" limits. When r= 0, what is u? When r= 5, what is u?
 

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