How Do You Estimate a Double Integral Using Riemann Sums?

In summary, the conversation discussed using a Riemann sum to estimate the value of a double integral, with the sample points being the lower right corners of rectangles. The proposed solution involved using a factor of 2 and 1 at the beginning, and the correct setup was confirmed. The solution for the upper left corners of the rectangles was corrected to be 2*1[f(0,-1) + f(0,0) + f(0,1) + f(2,-1) + f(2,0) + f(2,1)], which resulted in an estimated value of -8.
  • #1
AATroop
31
2

Homework Statement


If R = [0,4]x[-1,2], use a Riemann sum with m=2, n=3 to estimate the value of ∫∫(1-xy^2)dA. Take the sample points to be the lower right corners.

Homework Equations


None

The Attempt at a Solution


2*1[f(2,-1) + f(2,0) + f(2,1) + f(4,-1) + f(4,0) + f(4,1)] = some value

Just wondering if the setup is correct, particularly the 2 and the 1 at the beginning.

Edit:

Would the solution for the upper left corners of the rectangles be

2*1[f(0,-1)+ f(0,0)+ f(0,1) +f(2,-1) + f(2,0) + f(2,1)]
 
Last edited:
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  • #2
AATroop said:

Homework Statement


If R = [0,4]x[-1,2], use a Riemann sum with m=2, n=3 to estimate the value of ∫∫(1-xy^2)dA. Take the sample points to be the lower right corners.


Homework Equations


None


The Attempt at a Solution


2*1[f(2,-1) + f(2,0) + f(2,1) + f(4,-1) + f(4,0) + f(4,1)] = some value

Just wondering if the setup is correct, particularly the 2 and the 1 at the beginning.

Yes, that looks OK

Edit:

Would the solution for the upper left corners of the rectangles be

2*1[f(0,-1)+ f(0,0)+ f(0,1) +f(2,-1) + f(2,0) + f(2,1)]

(0,-1) isn't the upper left corner of any rectangle. Did you accidentally do lower left?
 
  • #3
Yes, I think I did do lower left. I changed it and it came out to -8. Sounds close enough to me haha. Thanks a bunch for your help.
 

FAQ: How Do You Estimate a Double Integral Using Riemann Sums?

What is a double Riemann sum?

A double Riemann sum is a method of approximating the area under a two-dimensional function by dividing the region into smaller rectangles and calculating the sum of the areas of those rectangles.

What is the basic problem when using double Riemann sums?

The basic problem when using double Riemann sums is determining the appropriate size and number of rectangles to use in order to accurately approximate the area under the function.

How is the accuracy of a double Riemann sum determined?

The accuracy of a double Riemann sum is determined by the size and number of rectangles used. The smaller the rectangles and the larger the number, the more accurate the approximation will be.

What is the difference between a left and right double Riemann sum?

A left double Riemann sum uses the leftmost point of each rectangle to calculate the area, while a right double Riemann sum uses the rightmost point. This can result in slightly different approximations.

What is the purpose of using double Riemann sums?

The purpose of using double Riemann sums is to approximate the area under a two-dimensional function, which can be used to solve a variety of real-world problems in fields such as physics, engineering, and economics.

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