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ok so here is the question:

evaluate $\oint \left(3y^2+e^{\cos x}\,dx\right) + \left(\sin y+5x^2\,dy\right)$, where $C$ is the boundary of the region in the upper half-plane ($y\ge 0$)between the circle $x^2+y^2=a^2$ and $x^2+y^2=b^2$ where $0\lt a\lt b$

note: the region can be expressed in polar coordinates $0\le\theta\le\pi$ and $a\le r \le b$ good time for Greens theorem?

I dont know if it is the right answer because it is negative.

first I took $\partial$$F_2$/$dx$ and got $10x$ then $\partial F_1/\partial y$ and got $6y$

I converted $\iint_D (10x-6y)\,dA$ to $\int_0^\pi\!\int_a^b(10r\cos\theta-6r\sin\theta)r\,drd\theta$

then when I integrated I got $-4b^3+4a^3$

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  • $\begingroup$ Its e^cosx sorry i didnt know how to write the command $\endgroup$ – Harvind May 18 '13 at 4:12
  • $\begingroup$ I've edited the question to fix the formatting. You can view the commands I used by right-clicking the formatted math and selecting "View math as"->"TeX commands" $\endgroup$ – Alex Becker May 18 '13 at 4:19
  • $\begingroup$ sorry man i was editing it too so i didnt think it worked! @AlexBecker $\endgroup$ – Harvind May 18 '13 at 4:23
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You used correctly the Green's Theorem for that line integral and your last result is also correct. In fact, we have $$\iint_D(Q_x-P_y)dA=\iint_D (10x-6y)$$ in which $P=3y^2+e^{\cos x},~~Q=\sin y+5x^2$

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  • $\begingroup$ Nice confirmation and elaboration! $\endgroup$ – Namaste May 19 '13 at 0:13

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