Register to reply

Formula involving double integral over a disc?

by b.krom
Tags: analysis, calculus
Share this thread:
b.krom
#1
Apr26-12, 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?
Phys.Org News Partner Science news on Phys.org
Bees able to spot which flowers offer best rewards before landing
Classic Lewis Carroll character inspires new ecological model
When cooperation counts: Researchers find sperm benefit from grouping together in mice
jacobrhcp
#2
Apr26-12, 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 co-ord. 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