Double Integral: Error in Calculation?

In summary, there is a discrepancy between the calculated answer and the answer in the book for a double integral problem. The problem involves a line integral around a triangle and can also be solved by integrating over the surface of a pyramid. The correct method involves using the expression for ## \int \nabla \times \vec{F} \cdot \hat{n} \, dS ## and correctly normalizing the unit vector. Both methods result in an answer of zero.
  • #1
fonseh
529
2

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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  • #2
fonseh said:
which part of my working is wrong

That depends. What is the actual problem statement?
 
  • #3
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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Related to Double Integral: Error in Calculation?

What is a double integral?

A double integral is a mathematical concept used to calculate the area under a surface in a two-dimensional space. It is represented by the symbol ∬ and is a type of definite integral.

What is the purpose of using a double integral?

The purpose of using a double integral is to calculate the volume or area of a three-dimensional shape or surface. It is also used to find the average value of a function in a two-dimensional space.

What is the error in calculation for a double integral?

The error in calculation for a double integral is the difference between the exact value of the integral and the estimated value obtained through numerical methods. This error can occur due to rounding errors, approximation methods, or incorrect input values.

What are some common sources of error in double integral calculations?

Some common sources of error in double integral calculations include improper bounds of integration, incorrect input values, incorrect application of integration rules, and using numerical methods with low precision.

How can errors in double integral calculations be reduced?

To reduce errors in double integral calculations, one can use more precise numerical methods, double check input values, and carefully apply integration rules. It is also helpful to break down the integral into smaller parts and check the calculations at each step.

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