Simple Green's Theorem Problem

[rhm95]

So here is the prompt. Use Green's Theorem to evaluate

. . . . .\(\displaystyle \mbox{Use Green's Theorem to evaluate}\)

. . . . . . . . . .\(\displaystyle \displaystyle \int_C\, \mathbf{F}\, \cdot\, d\mathbf{r}\)

. . . . .\(\displaystyle \mbox{where }\, \mathbf{F}(x,\, y)\, =\, \langle\, y\, \cdot\, \cos(x)\, -\, xy\, \cdot\, \sin(x),\, xy\, +\, x\, \cdot\, \cos(x)\, \rangle \, \)

. . . . .\(\displaystyle \mbox{ and }\, C\, \mbox{ is a tria}\mbox{ngle from }\, (0,\, 0)\, \mbox{ to }\, (0,\, 6)\, \mbox{ to }\, (3,\, 0),\, \mbox{ and back to }\, (0,\, 0).\)

Triangle plotted: found domain is from 0 to 6 on x and from 0 to y = 3 - 0.5x on y.

I'm calling \(\displaystyle \, F\, =\, \langle\, p,\, q\, \rangle \, \dfrac{dp}{dy}\, =\, \cos(x)\, -\, x\, \cdot\, \sin(x)\, \) and \(\displaystyle \, \dfrac{dq}{dy}\, =\, y\, +\, \cos(x)\, -\, x\sin(x)\)

So the integral on C of \(\displaystyle \, \mathbf{F}\, \cdot\, d\mathbf{r}\, \) is equal to double integral from 0 to 3 - 0.5x and 0 to 6 of \(\displaystyle \, \dfrac{dp}{dy}\, -\, \dfrac{dq}{dx}\, \dfrac{dy}{dx}.\)

. . . . .\(\displaystyle \displaystyle \int_0^{3\, -\, 0.5x}\, \int_0^6\, \left(\dfrac{dp}{dy}\, -\, \dfrac{dq}{dx}\right)\, \dfrac{dy}{dx}\)

\(\displaystyle \dfrac{dp}{dy}\, -\, \dfrac{dq}{dx}\, \) comes out to be just y. The double integral of y is just 9. But this is wrong, and I don't know what I did incorrectly.

Thanks in advance for you help.