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.


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:

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

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.