Evaluating Double Integrals of Odd and Even Functions on a Disk

In summary, the problem is to evaluate the integral of h(x,y) over a disk D centered at the origin, where h(x,y) equals the product of an even function and an odd function.
  • #1
ChiefKeeper92
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0

Homework Statement



Suppose f : ℝ→ℝ and g : ℝ →ℝ are continuous. Suppose that f is odd and g is even. Define h(x,y) : f(x)*g(y).
Let D be a disk centered at the origin in the plane. What is

∫∫h(x,y)dA?
D


The Attempt at a Solution


I know there's probably a trick to it. Is it 0 because h becomes odd over a disk that is symmetrical to the origin?
 
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  • #2
Yes, that's basically it. Set it up as a dx*dy integral with limits if you want to show it explicitly.
 
  • #3
ChiefKeeper92 said:
#Error

The problem posted was to evaluate the integral of h(x,y) over a disk D centered on the origin, where h(x,y)=f(x)g(y), f is an even function, g is odd.
 

FAQ: Evaluating Double Integrals of Odd and Even Functions on a Disk

1) What is the concept of "evaluating double integrals of odd and even functions on a disk"?

Evaluating double integrals of odd and even functions on a disk is a mathematical process used to find the area under a curve on a disk-shaped region. This concept is commonly used in calculus and is important in the study of multivariable calculus.

2) What is an odd function?

An odd function is a mathematical function that satisfies the property f(-x)=-f(x). This means that when the input is multiplied by -1, the output is equal to the negative of the original output. In terms of graph, an odd function has symmetry about the origin.

3) What is an even function?

An even function is a mathematical function that satisfies the property f(-x)=f(x). This means that when the input is multiplied by -1, the output remains the same. In terms of graph, an even function has symmetry about the y-axis.

4) How is a double integral of an odd function on a disk evaluated?

To evaluate a double integral of an odd function on a disk, the disk-shaped region is divided into small rectangles and the function is evaluated at the midpoint of each rectangle. The sum of these values is then multiplied by the area of each rectangle and the result is the approximate value of the double integral. The approximation becomes more accurate as the number of rectangles increases.

5) What is the difference between evaluating a double integral of an odd function and an even function on a disk?

The main difference is in the symmetry of the region. For an odd function, the region is symmetric about the origin, while for an even function, the region is symmetric about the y-axis. This means that the limits of integration for an odd function will be from -r to r (where r is the radius of the disk), while for an even function, the limits will be from 0 to r. Additionally, the value of the double integral for an odd function will always be 0, while for an even function, it will be twice the value of the integral of half the region.

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