Register to reply 
Formula involving double integral over a disc? 
Share this thread: 
#1
Apr2612, 08:42 AM

P: 1

Problem:
Can anyone help me out with the following problem: I am given a uniformly continuous function : [itex]g:\mathbb{R}^{2}\rightarrow [0,\infty )[/itex] such that the following condition is satisfied: [tex]\sup_{r> 0}\iint_{{x^{2}+y^{2}\leq r^{2}}}g(x,y)dxdy< \infty [/tex] The question is to prove that:[tex]\lim_{ (x,y) \to \infty}g(x,y)=0[/tex] I tried to use polar coordinates instead of rectangular ones, but it didn't work out. Any help? 


#2
Apr2612, 06:01 PM

P: 169

First, note that the limit intuiteively means that if you go really far from the origin, g becomes almost zero.
if you first suppose by contradiction the limit does not hold and [itex]g(x,y)=g(x(r,\theta),y(r,\theta))[/itex], then there must be some constant angle theta along which [itex]g_{\theta}(x(r),y(r))[/itex] has nonzero limit, right? If you try to use polar coordinates, you get [itex]\iint r g(x(r,\theta),y(r,\theta)), \theta\in[0,2\pi], r\in[0,\infty)[/itex] Because g is always positive and uniformly continous, I expect this is larger than the integral over the radiant with theta such that there is a nonzero limit (admittedly, this needs to worked out a bit), which exists as assumed by contradiction. So then you just need something like [itex]\int_0^\infty f(x) dx < \infty, f(x)>0 \iff f(x)\rightarrow0[/itex], which is easier than what you had first. Hope that helps a bit! 


Register to reply 
Related Discussions  
Question involving resistivity using the formula?  Introductory Physics Homework  6  
Hollow Metal disc loading  the needed thickness of the disc?  General Engineering  1  
Using polar coord. to change double integral into single integral involving only r.  Calculus & Beyond Homework  5  
How to solve an equation involving an integral? integral of f(x) = 1?  Calculus & Beyond Homework  1  
Property of a Double Integral involving a limit  Calculus & Beyond Homework  2 