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Evaluate the line integral

$3xy^2dx+8x^3dy$ where $C$ is the boundary of the region between the circles $x^2+y^2=1$ and $x^2+y^2=64$ having positive orientation.

Okay so I first looked at the first circle. I took $x=\cos t$ and $y=\sin t$. $dx=-\sin t dt$ and $dy=\cos t dt$

I looked at the second circle. I took $x=8\cos t$ and $y =8\sin t$. Then $dx = -8\sin t dt$ and $dy = 8\cos t dt$.

I looked up how to solve these kinds of problem and so I solved it two ways. First I plugged it into $A = (1/2)$ the integral of $xdy + ydx$. I then subtracted $A2-A1$ but I got $63\pi$. Lon capa said this was incorrect. Then I plugged it into the $f(x,y)$ equation that the problem gave me. However lon capa said that was incorrect as well... Please can someone tell me what I'm doing wrong?

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    $\begingroup$ Who is Lon Capa? $\endgroup$ – Git Gud Apr 20 '14 at 19:43
  • $\begingroup$ Its a math site where you do your homework.... It gives you a problem and you have 8 tried to get the correct answer $\endgroup$ – Ayoshna Apr 20 '14 at 19:46
  • $\begingroup$ I see. $\endgroup$ – Git Gud Apr 20 '14 at 19:47
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Green's theorem relates the line integral of a vector field around a closed path to the area contained within the closed path. Using your data as input we have:

\begin{align} \text{Green's Theorem in the plane : }\quad\oint_C\vec F\cdot d\vec r = \iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA \end{align}

\begin{align} \vec F = P\hat i+Q\hat j = 3xy^2\:\hat i + 8x^3\:\hat j \end{align}

where $D$ is the region in the $xy-$plane bounded by $C$. Traverse the boundary once in a counterclockwise fashion and choosing the same parameterization you have chosen we have:

\begin{align} & Q_x = \frac{\partial}{\partial x}\left(8x^3\right) = 24x^2\qquad P_y = \frac{\partial}{\partial y}\left(3xy^2\right) = 6xy\\[2mm] & Q_x-P_y = 6x\left(4x-y\right) \end{align}

And then the integral over the annulus:

\begin{align} &6\int_0^{2\pi}\int_1^8\left(4r^2\cos^2t-r\sin t\right)rdrdt=6\int_0^{2\pi}\left(\left .r^4\right|_1^8\cos^2t-\left .\frac{1}{3}r^3\right|_1^8\sin t\right)dt=\\ & 6\int_0^{2\pi}\left(4095\cos^2t-\frac{511}{3}\sin t\right)dt= 6\int_0^{2\pi}\left(\frac{4095}{2}\left(1+\cos 2t\right)-\frac{511}{3}\sin t\right)dt=\\ & \left .6\left(\frac{4095}{2}\left(t+\frac{\sin 2t}{2}\right)+\frac{511}{3}\cos t\right)\right|_0^{2\pi} = \left .12285t\right|_0^{2\pi} = 24570\pi \end{align}

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