Polar coordinates to set up and evaluate double integral

mikky05v
Messages
53
Reaction score
0

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?
 
Physics news on Phys.org
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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top