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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

2. rsst123

@ganeshie8

3. rsst123

I understand what to do if there is one circle but not two

4. anonymous

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

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

Greens Theorem.

7. anonymous

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

Better yet, take @IrishBoy123's advice :)

9. rsst123

how would i use greens theorem given two circles? would i just evaluate them separately then add them?

10. IrishBoy123

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

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

@IrishBoy123

13. IrishBoy123

$$1 ≤ r ≤ 2$$ $$0 ≤ \theta ≤ \pi/2$$ 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

Ok i see now so the radius would just be 1 to 2 awesome Thanks!

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