Assuming that phi and r are polar coordinates, change
Assuming that $\varphi$ and $r$ are polar coordinates, change the order of integration: $$ \int_0^{a} d\varphi \int_0^{\varphi} z(\varphi,r)\,dr \qquad (0<a<\pi). $$
Explanation of Method
Here the outer boundary $r=\varphi$ is linear (a spiral arm), so both the direct and reversed limits are simple linear inequalities — no trigonometric inversion is needed.
- Note that $r\leqslant\varphi\leqslant a$ forces $r\leqslant a$.
- Swap the roles of $r$ and $\varphi$ directly from the linear inequality.
- Verify at the endpoints $r=0$ and $r=a$.
Step 1 · Identify the region
From $\int_0^{a} d\varphi\int_0^{\varphi} z\,dr$: $0\leqslant\varphi\leqslant a$, $0\leqslant r\leqslant\varphi$.
Step 2 · Determine the range of $r$
Since $r\leqslant\varphi\leqslant a$, we get $0\leqslant r\leqslant a$ (maximum $r=a$ occurs at $\varphi=a$).
Step 3 · Find bounds on $\varphi$ for fixed $r$
From $r\leqslant\varphi$ and $\varphi\leqslant a$: $r\leqslant\varphi\leqslant a$.
Step 4 · Write the integral with the reversed order
$$ \int_0^{a} dr \int_r^{a} z(\varphi,r)\,d\varphi. $$
Step 5 · Verification
At $r=0$: $\varphi$ ranges over $[0,a]$. ✓ At $r=a$: $\varphi$ ranges from $a$ to $a$ (a single point). ✓ At $r=a/2$: $\varphi$ ranges from $a/2$ to $a$. ✓
Additional Notes
- Swapping limits when the boundary curve $r(\varphi)$ is linear (or, symmetrically, $\varphi(r)$ is linear) is the simplest possible case of order reversal — the region is just a triangle in the $(\varphi,r)$-plane.
Result
The order-reversed integral is: