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For the double integral:$$\int^{2\pi}_0\int^1_{\cos(x)}f(x,y)dydx$$

I want to sketch the region of integration and then obtain an equivalent double integral with the order of integration reversed.

My region is simply bound by the upper limit of $y=1$ and lower of $y = \cos(x)$. Lower of $x = 0$ and upper of $x = 2\pi$.

Is this the correct region? If so, how will i have my new double integral set as if i use horizontal strips i go from the $\cos(x)$ line back to the $\cos(x)$ line. from observation ( i most likely am wrong) i took $x$ to go from (lower ) $x = 1-x^3$ to (upper) $x=??$ This is proving to be a bit difficult for me.

Using horizontal strips my $y$ goes from $y=-1$ to $y=1$, which would be my outside integral.

Any help would with sketching this region/ obtaining an equivalent double integral with the order of integration reversed would be greatly appreciated.

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You have to use the $\arccos(y)$ function. Remember, the domain of the $\arccos(y)$ function is limited to $-\pi\le{y}\le\pi$, so this will only cover the left half of the region. For the right have, you will have to use a second function: $-\arccos(y)+2\pi$. Your region will therefore become:

$$\begin{align} \arccos(y)\le &x <-\arccos(y)+2\pi\\ -1\le&y<1 \end{align}$$

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  • $\begingroup$ Please replace $\arccos(y)+2\pi$ by $2\pi-\arccos(x)$. $\endgroup$ – Did Aug 20 '13 at 7:49
  • $\begingroup$ @Did Done. Sorry for the confusion. $\endgroup$ – Ataraxia Aug 20 '13 at 7:53

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