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rsst123
 one year ago
**Will Medal**
Calc 3
Can someone help me go through this problem please
rsst123
 one year ago
**Will Medal** Calc 3 Can someone help me go through this problem please

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rsst123
 one year ago
Best ResponseYou've already chosen the best response.0I understand what to do if there is one circle but not two

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Have you tried parameterizing? First break up the whole path into subpaths. (see sketch) dw:1436726978652:dw Let \(x=2\cos t\) and \(y=2\sin t\) for the larger circle (radius 2) and \(x=\cos t\) and \(y=\sin t\) for the smaller circle (radius 1). The paths along the axes are straightforward (literally), so you can set \(x=0\) for the vertical line and \(y=0\) for the horizontal one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whoops, forgot my labels:dw:1436727081690:dw So you have \[\begin{align*}I&=\int_C(4+e^{\cos x})\,dx+(\sin y+3x^2)\,dy\\\\ &=\int_{C_1}(\cdots)+\cdots+\int_{C_4}(\cdots) \end{align*}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Define \(C_1\) by \(\begin{cases}x=2\cos t\\y=2\sin t\end{cases}\), so that \(\begin{cases}dx=2\sin t\\dy=2\cos t\end{cases}\), valid over \(0\le t\le\dfrac{\pi}{2}\). The first integral is then \[\int_{C_1}(\cdots)=\int_{0}^{\pi/2}\left[(4+e^{\cos (2\cos t)})(2\sin t)+(\sin (2\sin t)+12\cos^2t)(2\cos t)\right]\,dt\] Bit of an awkward integral, though, good luck computing it :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Better yet, take @IrishBoy123's advice :)

rsst123
 one year ago
Best ResponseYou've already chosen the best response.0how would i use greens theorem given two circles? would i just evaluate them separately then add them?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2Green looks something like: \(\int \vec F \bullet \vec dr = \int\int curl \vec F \ dA\). [my latex skills aren't up to the detail] so have you thought about integrating the curl of the field over the area, which is as per @SithsAndGiggles sketch it is a closed loop. curl F = 6x that's a double integral in polar.

rsst123
 one year ago
Best ResponseYou've already chosen the best response.0ok so using greens theorem i would get something like this converting to cylindrical but what would my radius be 1 or 2? \[6\int\limits_{0}^{\pi/2}\int\limits_{0}^{?}rcos \theta r drd \theta \]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\(1 ≤ r ≤ 2\) \(0 ≤ \theta ≤ \pi/2\) cylindrical? I was seeing \(\int \int 6x \ dx \ dy\) and converting that to polar. and then \(x = r cos \theta \) so it's: \(\int \int 6 \ r \ cos \theta \ r \ dr \ d\theta \) which is where you come out too IOW: i think we agree on this ... with the limits above cool

rsst123
 one year ago
Best ResponseYou've already chosen the best response.0Ok i see now so the radius would just be 1 to 2 awesome Thanks!
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