Can the Simpsons 3/8 Rule be Extended to Calculate Double Integrals?

In summary: In order to use a quadrature approach for a double integral, you will need to be able to parameterize the surface over which the integral is to be evaluated.
  • #1
DivergentSpectrum
149
15
how do i numerically calculate a double integral?
as i understand simpsons 3/8 rule is the optimal method for a single integral, is it still true for double integrals?
if so, how do i extend the 3/8s rule to do a double integral?
 
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  • #2
DivergentSpectrum said:
how do i numerically calculate a double integral?
as i understand simpsons 3/8 rule is the optimal method for a single integral, is it still true for double integrals?
if so, how do i extend the 3/8s rule to do a double integral?

I don't know what you mean by the Simpson's 3/8 rule being 'optimum'. It uses a different number of ordinates than Simpson's first rule (1-4-1), so that an even number of ordinates can be accommodated easily, where Simpson's first rule takes an odd number of ordinates.

Simpson's rules and the trapezoidal rule can be adapted to evaluate double integrals, and they tend to work well for regions which are triangular or rectangular. For more general regions, other methods may be easier to use for such evaluations; there are no hard and fast rules.

This paper illustrates a typical quadrature based on a 2-D Simpson's Rule:

https://www.math.ohiou.edu/courses/math3600/lecture24.pdf
 
  • #3
Only with rectangles? so I am guessing id have to be able to do some kind of change of variables then right?
 
  • #4
DivergentSpectrum said:
Only with rectangles? so I am guessing id have to be able to do some kind of change of variables then right?

Due to the nature of the Simpson's rule setup, a rectangular region will give the most accurate results with the least fuss, because the ordinates must be evenly spaced in a 2-D grid.

If the regions over which you are trying to evaluate a double integral are of a more general shape, perhaps a 2-D Gaussian quadrature method might give you fewer headaches about locating the ordinates in the region where the integrand must be evaluated.
 
  • #5
what I am doing in this case is i am doing a surface integral of a scalar function
using x(u,v),y(u,v),z(u,v) as the parameterized surface
so because this is of the form ∫∫something dudv where umin umax vmin vmax are all constants so its the same as a double integral over a rectangle and it should work right?
 

1. What is a numerical double integral?

A numerical double integral is a mathematical concept used to calculate the area between two curves in a two-dimensional space. It involves taking the sum of infinitesimal rectangles under a given function over a specific region.

2. How is a numerical double integral different from a regular integral?

A regular integral calculates the area under a curve in a one-dimensional space, while a numerical double integral calculates the area between two curves in a two-dimensional space. This means that a numerical double integral involves calculating a volume instead of just an area.

3. What are the different methods for evaluating a numerical double integral?

There are several methods for evaluating a numerical double integral, including the rectangular method, trapezoidal method, and Simpson's method. These methods involve dividing the region into smaller rectangles or trapezoids and using different formulas to calculate the sum of their areas.

4. When would you use a numerical double integral in real life?

Numerical double integrals are used in many fields, such as physics, engineering, and economics. They can be used to calculate the volume of a solid, the work done by a force, or the total revenue of a business.

5. What are the limitations of using numerical double integrals?

One limitation of using numerical double integrals is that they can be time-consuming and require a lot of computational power, especially when dealing with complex equations or large regions. Additionally, numerical double integrals may not always provide an exact solution and may only give an approximation of the true value.

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