Double Integral: Error in Calculation?

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The discussion revolves around a discrepancy in the results of a double integral calculation, where the user obtained 312.5 using Wolfram Alpha, while the book states the answer is 0. The problem involves integrating over a triangle's surface or the pyramid formed by its planes. Correct integration techniques, including using the curl of a vector field and normalizing the surface area element, are emphasized as crucial for obtaining the correct result. The user is encouraged to check their work against these methods, which consistently yield an answer of zero. Proper application of the mathematical principles is essential to resolve the error in calculation.
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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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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