Property of a Double Integral involving a limit

In summary, the conversation discusses a statement involving a continuous function on a closed bounded region and a point in the interior of a closed disk. The speaker is unsure about the approach to show the statement, but considers using the mean value theorem for double integrals. The other speaker confirms that this is a good idea, as the left side of the statement is the average value of the function over the disk. The conversation ends with the first speaker thanking the other for their assistance.
  • #1
erogard
62
0
Hi,

I am actually not really concerned about what the whole details are but more whether my approach is correct to show the following statement:

Let [tex]f[/tex] be continuous on a closed bounded region [tex]\Omega[/tex] and let [tex](x_0 ,y_0)[/tex] be a point in the interior of [tex]\D_r[/tex]. Let [tex]D_r[/tex] be the closed disk with center [tex](x_0 ,y_0)[/tex] and radius [tex]r[/tex]. Then

[tex] \displaystyle\lim_{r\to 0}\frac{1}{\pi r^2} \displaystyle\int\displaystyle\int_{D_r} f(x,y)dx dy= f(x_0 , y_0)[/tex]

(see [tex]D_r[/tex] as the region of the entire double integration)
I have thought of considering a "Riemann sum" approach, which seems a little too brutal and complicated to me (although might do well with the limit involved) or using the MVT for double integrals, which invokes the same hypotheses. Does the latter sound like a good idea?
 
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  • #2
Yes, the mean value theorem. In fact, the left side, before the limit, is the "mean" (average) value of f(x,y) over the disk.
 
  • #3
OK, and I almost figured it out. Thank you.
 

1. What is a double integral involving a limit?

A double integral involving a limit is a mathematical concept used in multivariable calculus to determine the area under a curved surface in a specific region. It involves taking the limit of a Riemann sum, which is a series of rectangles that approximate the area.

2. How is a double integral involving a limit calculated?

To calculate a double integral involving a limit, the region of integration is divided into small rectangles, and the function is evaluated at specific points within each rectangle. The sum of these values is then multiplied by the area of each rectangle, and the limit of this sum is taken as the rectangle size approaches zero.

3. What is the purpose of a double integral involving a limit?

The purpose of a double integral involving a limit is to calculate the area under a curved surface in a specific region. It is also used to find the volume of a three-dimensional object and to solve various real-world problems in fields such as physics and engineering.

4. What is the relationship between a double integral involving a limit and a single integral?

A double integral involving a limit is essentially the extension of a single integral from one dimension to two dimensions. While a single integral calculates the area under a curve on a one-dimensional axis, a double integral calculates the volume under a surface in a two-dimensional region.

5. What are some applications of double integrals involving limits?

Double integrals involving limits have numerous applications in real-world problems, including calculating the mass of an object with varying density, determining the center of mass of a two-dimensional object, and finding the average value of a function over a region. They are also used in physics to calculate work and in economics to determine consumer surplus.

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