Change of integration order of a double integral

Question

Which is the integral equivalent to $$\int_{0}^{\pi}\int_{0}^{\sin x}f(x, y)dydx$$ to make a change in the order of integration?

Answer 1

Draw a picture. We are integrating $f(x,y)$ over the "first" complete arch of the sine curve. We assume that $f(x,y)$ is "well-behaved" enough that the interchange is valid.

For any value of $y$, the variable $x$ travels from $\arcsin y$ to $\pi-\arcsin y$. And then $y$ travels from $0$ to $1$. Thus our integral can be rewritten as $$\int_0^1 \left(\int_{\arcsin y}^{\pi-\arcsin y} f(x,y)\,dx\right)\,dy.$$

Answer 2

$\large\mbox{No pictures !!!. Use Theta function.}$ \begin{align} &\color{#ff0000}{\large\int_{0}^{\pi}\int_{0}^{\sin\left(x\right)}{\rm f} \left(x, y\right)\,{\rm d}y\,{\rm d}x} = \int_{0}^{\pi}\int_{0}^{1}{\rm f} \left(x, y\right)\Theta\left(\sin\left(x\right) - y\right)\,{\rm d}y\,{\rm d}x \\[3mm]&= \int_{0}^{1}\int_{0}^{\pi}{\rm f}\left(x, y\right) \Theta\left(\sin\left(x\right) - y\right)\,{\rm d}x\,{\rm d}y \\[3mm]&= \int_{0}^{1}\left[% \int_{0}^{\pi/2}{\rm f}\left(x, y\right) \Theta\left(\sin\left(x\right) - y\right)\,{\rm d}x + \int_{0}^{\pi/2}{\rm f}\left(\pi - x, y\right) \Theta\left(\sin\left(x\right) - y\right)\,{\rm d}x \right]\,{\rm d}y \\[3mm]&= \color{#ff0000}{\large\int_{0}^{1}\left\{% \int_{\arcsin\left(y\right)}^{\pi/2} \left[{\rm f}\left(x, y\right) + {\rm f}\left(\pi - x, y\right)\right] \,{\rm d}x \right\}\,{\rm d}y} \end{align}