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rsst123

  • one year ago

**Will Medal** Calc 3 Can someone help me go through this problem please

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  1. rsst123
    • one year ago
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  2. rsst123
    • one year ago
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    @ganeshie8

  3. rsst123
    • one year ago
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    I understand what to do if there is one circle but not two

  4. anonymous
    • one year ago
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    Have you tried parameterizing? First break up the whole path into sub-paths. (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.

  5. anonymous
    • one year ago
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    Whoops, forgot my labels:|dw:1436727081690:dw| So you have \[\begin{align*}I&=\int_C(4+e^{\cos x})\,dx+(\sin y+3x^2)\,dy\\\\ &=\int_{C_1}(\cdots)+\cdots+\int_{C_4}(\cdots) \end{align*}\]

  6. IrishBoy123
    • one year ago
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    Greens Theorem.

  7. anonymous
    • one year ago
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    Define \(C_1\) 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

  8. anonymous
    • one year ago
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    Better yet, take @IrishBoy123's advice :)

  9. rsst123
    • one year ago
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    how would i use greens theorem given two circles? would i just evaluate them separately then add them?

  10. IrishBoy123
    • one year ago
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    Green 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.

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

  12. rsst123
    • one year ago
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    @IrishBoy123

  13. IrishBoy123
    • one year ago
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    \(1 ≤ r ≤ 2\) \(0 ≤ \theta ≤ \pi/2\) 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

  14. rsst123
    • one year ago
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    Ok i see now so the radius would just be 1 to 2 awesome Thanks!

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