Find curve integral:

$\displaystyle \int\limits_{\Gamma}(\tan x+2y)dx+(3x+4\tan^5y)dy$

where $\displaystyle \Gamma$ is a paralelogram with vertices at (0,0), (1,2), (1,3), (0,1)

solved

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- May 16th 2011, 03:03 PMzehIntegral, paralelogram
Find curve integral:

$\displaystyle \int\limits_{\Gamma}(\tan x+2y)dx+(3x+4\tan^5y)dy$

where $\displaystyle \Gamma$ is a paralelogram with vertices at (0,0), (1,2), (1,3), (0,1)

solved - May 17th 2011, 05:39 AMHallsofIvy
I'm glad you were able to solve it. For future reference, few people are going to help you with a problem where you have showed no work of your own at all.

I assume you used

$\displaystyle \int_C f(x,y)dx+ g(x,y)dy= \int\int_R \frac{\partial f}{\partial y}- \frac{\partial g}{\partial y} dxdy$

where C is the closed curve bounding the region R. Here the integrand on the right is a constant so the integral is that constant times the area of the parallelogram.