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Given integral $\int \limits_0^{\pi}dx \int \limits_{0}^{2\sin x} f(x,y)dy$. Change the order of integration.

I've drawn the graph of original integral it is just one piece of $2\sin x$ function squeezed between $[0,2]$ on x-axis.

Then I'm trying to flip the axes, and I thought it's similar to rotate the graph on $90$ degree. Than I have to do some sort of manipulations to express $y$ in terms of $x$ I guess. I thought I simply can solve $2\sin x = y \Rightarrow x = \arcsin \frac{y}{2}$ and do the same thing for lower bound, but it is 0, so I can't express $x$ in terms $y$.

So now I'm confused. I don't really understand what sort of manipulations I have to perform to find the right bounds.

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What about this:

$$\int_0^{2}d y \int_{\arcsin{\frac{y}{2}}}^{\pi -\arcsin {\frac{y}{2}}} f(x,y)dx$$

enter image description here

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  • $\begingroup$ Can you please explain how did you derived the bounds for $x$ and $y$? $\endgroup$ – False Promise Dec 20 '17 at 2:54
  • $\begingroup$ @FalsePromise I've add a sketch to better explain the derivation. $\endgroup$ – gimusi Dec 20 '17 at 21:04

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