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
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
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.