Polar coordinates to set up and evaluate double integral

  • Thread starter mikky05v
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  • #1
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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

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

and then my inside would become ∫[itex]^{\pi/2}_{-\pi/2}[/itex] ∫[itex]^{5}_{0}[/itex] (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?
 

Answers and Replies

  • #2
CAF123
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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.
 
  • #3
HallsofIvy
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What about [itex]u= r^2/2[/itex]?

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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