Change the order of integration: integral from 0 to (2
Change the order of integration: $$ \int_0^{2\pi} dx \int_0^{\sin x} z(x,y)\,dy. $$
Explanation of Method
Since $0\leqslant y\leqslant\sin x$ is only a non-empty range where $\sin x\geqslant0$, the effective region only uses $0\leqslant x\leqslant\pi$. To swap the order we invert $y=\sin x$ on that interval, which has two branches (increasing on $[0,\pi/2]$, decreasing on $[\pi/2,\pi]$) both expressed with $\arcsin$.
- Restrict to where $\sin x\geqslant0$, i.e. $x\in[0,\pi]$.
- Find the maximum height of the region.
- Invert $y=\sin x$ on each monotonic branch.
- Write the reversed integral using $\arcsin(y)$ for both boundaries.
Step 1 · Identify the region
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 condition $0\leqslant y\leqslant\sin x$ only holds where $\sin x\geqslant0$, i.e. for $0\leqslant x\leqslant\pi$.
Step 2 · Determine the range of $y$
The maximum of $\sin x$ on $[0,\pi]$ is $\sin(\pi/2)=1$, so $0\leqslant y\leqslant1$.
Step 3 · Find bounds on $x$ for fixed $y$
For fixed $y\in[0,1]$, we need $y\leqslant\sin x$ with $0\leqslant x\leqslant\pi$. On $[0,\pi/2]$, $x=\arcsin(y)$ (sine increasing); on $[\pi/2,\pi]$, $x=\pi-\arcsin(y)$ (sine decreasing). For $\sin x\geqslant y$:
$$ \arcsin(y) \leqslant x \leqslant \pi-\arcsin(y). $$
Step 4 · Write the integral with the reversed order
$$ \int_0^1 dy \int_{\arcsin(y)}^{\pi-\arcsin(y)} z(x,y)\,dx. $$
Additional Notes
- Half of the stated $x$-range $[0,2\pi]$ contributes nothing, since $\sin x<0$ there makes the inner integral's bounds empty (or reversed) in the area sense.
- Recognizing the two monotonic branches of $\sin x$ on $[0,\pi]$ is the key step in inverting the boundary.
Result
The order-reversed integral is: