Formula involving double integral over a disc?

In summary, the problem is to prove that for a uniformly continuous function g from $\mathbb{R}^{2}$ to [0,$\infty$], if the integral of g over a circle with radius r tends to infinity, then the limit of g as the point (x,y) goes to infinity is 0. The conversation suggests using polar coordinates and a proof by contradiction to show this result.
  • #1
b.krom
1
0
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?
 
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  • #2
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!
 
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What is a double integral over a disc?

A double integral over a disc is a mathematical formula used to calculate the volume or surface area of a solid or a curved surface, respectively, by integrating a function over a two-dimensional region in the shape of a disc.

What is the formula for a double integral over a disc?

The formula for a double integral over a disc is ∬D f(x,y) dA, where D represents the region in the shape of a disc, f(x,y) is the function being integrated, and dA represents the infinitesimal area element.

What is the difference between a single integral and a double integral over a disc?

A single integral calculates the area under a curve on a one-dimensional interval, while a double integral calculates the volume or surface area of a two-dimensional region, such as a disc, by integrating a function over that region.

What are the limits of integration for a double integral over a disc?

The limits of integration for a double integral over a disc depend on the specific region being integrated over. In general, the limits will be the equations that define the boundary of the disc.

What are some applications of a double integral over a disc?

A double integral over a disc has various applications in mathematics, physics, engineering, and other fields. It can be used to calculate the volume of a solid of revolution, the surface area of a curved surface, the center of mass of a disc-shaped object, and more.

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