Double Integral: Error in Calculation?

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SUMMARY

The discussion centers on the calculation of a double integral involving a triangular region defined by vertices at (5,0,0), (0,5,0), and (0,0,5). The user initially obtained an answer of 312.5 using Wolfram Alpha, while the correct answer is 0, as confirmed by multiple contributors. The key to resolving the discrepancy lies in the proper application of the curl of the vector field, specifically using the formula ∫ ∇ × F · n̂ dS and ensuring the normal vector is normalized correctly. Contributors emphasized the importance of evaluating the integral over the surface of the pyramid formed by the triangle.

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  • Understanding of double integrals and their applications in vector calculus.
  • Familiarity with the concept of curl in vector fields.
  • Knowledge of surface integrals and normal vectors.
  • Proficiency in using mathematical software like Wolfram Alpha for verification.
NEXT STEPS
  • Study the properties of the curl of a vector field and its physical significance.
  • Learn how to compute surface integrals using the formula ∫ ∇ × F · n̂ dS.
  • Practice normalizing vectors in three-dimensional space for integration purposes.
  • Explore the geometric interpretation of double integrals over triangular regions and their corresponding surfaces.
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Students and professionals in mathematics, physics, and engineering who are working with vector calculus, particularly those dealing with double integrals and surface integrals in three-dimensional space.

fonseh
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Homework Statement


I have calculate my double integral using wolfram alpha , but i get the ans = 312.5 , but according to the book , the ans is = 0 , which part of my working is wrong

Homework Equations

The Attempt at a Solution


Or is it z =0 , ? i have tried z = 0 , but still didnt get the ans = 0 , Which part is wrong?[/B]
 

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fonseh said:
which part of my working is wrong

That depends. What is the actual problem statement?
 
I can see what the problem statement seems to be=the line integral around the triangle with vertices at (5,0,0), (0,5,0), and (0,0,5). Instead of trying to integrate across the plane of this triangle, you can also integrate over the surface consisting of the three triangle plane surfaces making up a pyramid. i.e. dydz for the x-component of the curl F at x=0; dxdz for the y-component at y=0; and dxdy for the z component at z=0. I did get zero for an answer when I summed these. I will be glad to check your work to see if you get the same answer I did for each of these. These 2-D integrals are not difficult to evaluate=their limits of integration just take a couple of minutes to compute. ## \\ ## Editing: I also solved it the way you are attempting, but you need to use ## \int \nabla \times \vec{F} \cdot \hat{n} \, dS ##. Your ## \hat{n} ## needs a ##1/\sqrt{3} ## to normalize it which will cancel the ## cos(\gamma) ## factor of ## cos(\gamma)dS=dxdy ## so that ## dS=\frac{dxdy}{cos(\gamma)} ##. (## cos(\gamma)=\hat{n} \cdot \hat{k}=1/\sqrt{3} ##. ). You correctly used ## z=5-x-y ##, but your expression that you integrate will be much simpler if you correctly use ## \nabla \times \vec{F} ## instead of ## \vec{F} ##. And yes, I did get zero for an answer this way as well.
 
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