Calculating Line Integral of (x^3-y^3)dx +(x^3+y^3)dy

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SUMMARY

The line integral of the function (x^3-y^3)dx +(x^3+y^3)dy over the boundary r, defined by the regions x^2+y^2=1 and x^2+y^2=9, yields a result of 120π + 3π/2. The integral over the curve x^2+y^2=1 is calculated as 3π/2, while the double integral for the area between the two curves is 120π. Clarity in the problem statement regarding the direction of the contours is essential for accurate interpretation and solution.

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1.find the line integral of
(x^3-y^3)dx +(x^3+y^3)dy over r, where r is the boundary of the
region limited by x^2+y^2=1 and x^2+y^2=9





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3.
i found that the
line integral over the curve x^2+y^2=1 is 3*Pi/2
and the double integral of the region limited by x^2+y^2=1 and x^2+y^2=9 is 120*Pi
so the answer would be 120*Pi + 3*Pi/2 ?


 
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The question is not very clear. The region between x^2+y^2=1 and x^2+y^2=9 consists of two curves that aren't connected. A proper question would specify the direction for each part of the contour. 120*pi is correct if you make a certain assumption about the direction of each contour. Whoever is giving you these questions should really work on phrasing them better.
 

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