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**dmbocci** 3.Evaluate: Integral of region C Mdx + Ndy for $\displaystyle M=e^x\sin y, N=e^x\cos y, C: x=t-\sin t, y=1-\cos t$ from (0,0) to (2pi,0).

Looks like unit circle. I don't know how to set it up.

Thanks

$\displaystyle \displaystyle C$ is a contour, so I suspect it is a contour integral that's wanted.

The contour, $\displaystyle \displaystyle C$, is a single arch of a cycloid.

Find $\displaystyle \displaystyle dx$ and $\displaystyle \displaystyle dy$ in terms of $\displaystyle \displaystyle dt$ from the contour equations.

Take $\displaystyle \displaystyle x$ and $\displaystyle \displaystyle y$ from the contour equations and plug them into the equations for $\displaystyle \displaystyle M$ and $\displaystyle \displaystyle N$. The resulting integral looks like it would be messy.

Alternative: Use Green's Theorem:

$\displaystyle \displaystyle \oint_{C_1} Mdx + Ndy=\underset{R}{\int\int}\ \left({{\partial M}\over{\partial x}}- {{\partial N}\over{\partial y}}\right)\ dA$, where $\displaystyle \displaystyle C_1$ is a closed path around region, $\displaystyle \displaystyle R$.

Break $\displaystyle \displaystyle C_1$ up into path $\displaystyle \displaystyle C$ as defined above, and path $\displaystyle \displaystyle C_0$, which goes from $\displaystyle \displaystyle (2\pi,\ 0)$ to $\displaystyle \displaystyle (0,\ 0)$ along the $\displaystyle \displaystyle x$-axis. $\displaystyle \displaystyle dy = 0$ along $\displaystyle \displaystyle C_0$. Also, $\displaystyle \displaystyle M=0$ along $\displaystyle \displaystyle C_0$. So the integral along $\displaystyle \displaystyle C_0$ contributes nothing to the integral around $\displaystyle \displaystyle C_0$.

Therefore, $\displaystyle \displaystyle \oint_{C_1} Mdx + Ndy=\int_{C} Mdx + Ndy$.

Now, compute: $\displaystyle \displaystyle \underset{R}{\int\int}\ \left({{\partial M}\over{\partial x}}- {{\partial N}\over{\partial y}}\right)\ dy\ dx$.

This will also take some work. Good Luck.