Change the order of integration: integral from 0 to (2

Change the order of integration: $$ \int_0^{2\pi} dx \int_0^{\sin x} z(x,y)\,dy. $$

Step 1 · Identify the region from the given integral

From $\int_0^{2\pi} dx\int_0^{\sin x} z\,dy$: $0\leqslant x\leqslant2\pi$, $0\leqslant y\leqslant\sin x$. Since $\sin x<0$ for $\pi<x<2\pi$, the constraint $0\leqslant y\leqslant\sin x$ is only meaningful (non-empty) for $0\leqslant x\leqslant\pi$.

Step 2 · Describe the region

The effective region is bounded below by $y=0$, above by $y=\sin x$, between $x=0$ and $x=\pi$ (where $\sin x$ returns to $0$); for $\pi<x\leqslant2\pi$ the integral contributes nothing.

Step 3 · Find the range of $y$

The maximum of $\sin x$ on $[0,\pi]$ is $1$, at $x=\pi/2$, so $0\leqslant y\leqslant1$.

Step 4 · Find the bounds on $x$ for fixed $y$

We need $y\leqslant\sin x$ with $0\leqslant x\leqslant\pi$. On $[0,\pi/2]$, $x=\arcsin(y)$; on $[\pi/2,\pi]$, $x=\pi-\arcsin(y)$. So for fixed $y$, $x$ ranges between these two values:

$$ \arcsin(y) \leqslant x \leqslant \pi-\arcsin(y). $$

Check at $y=1$: both give $x=\pi/2$ (a single point). Check at $y=0$: left $=0$, right $=\pi$ (the full base). ✓

Step 5 · Write the integral with the reversed order

$$ \int_0^1 dy \int_{\arcsin(y)}^{\pi-\arcsin(y)} z(x,y)\,dx. $$

Additional Notes

• The original integral's outer limit $x\in[0,2\pi]$ is misleading at first glance — half of that range (where $\sin x<0$) contributes nothing, since the inner integral would run from $0$ down to a negative number, i.e. is empty (or, by convention, contributes the negative of the region below the axis, which cancels if $z\equiv1$; for a genuine "area" interpretation the effective region is just $0\leqslant x\leqslant\pi$).
• Recognizing the two monotonic branches of $\sin x$ on $[0,\pi]$ is the key step in inverting the boundary.