Assuming that phi and r are polar coordinates, change
Assuming that $\varphi$ and $r$ are polar coordinates, change the order of integration: $$ \int_0^{\pi/2} d\varphi \int_0^{a\sqrt{\sin2\varphi}} z(\varphi,r)\,dr \qquad (a>0). $$
Explanation of Method
We locate the maximum of $r=a\sqrt{\sin2\varphi}$ by calculus (or by recognizing $\sin2\varphi$ peaks at $\varphi=\pi/4$), then invert the relation. Since $a\sqrt{\sin2\varphi}$ rises from $0$ to $a$ and back to $0$ as $\varphi$ runs over $[0,\pi/2]$, each fixed $r<a$ corresponds to two $\varphi$ values — one on the rising branch, one on the falling branch.
- Maximize $r=a\sqrt{\sin2\varphi}$ over $[0,\pi/2]$.
- Invert $r=a\sqrt{\sin2\varphi}$ to get both branches.
- Write and verify the reversed integral.
Step 1 · Identify the region
From $\int_0^{\pi/2} d\varphi\int_0^{a\sqrt{\sin2\varphi}} z\,dr$: $0\leqslant\varphi\leqslant\frac\pi2$, $0\leqslant r\leqslant a\sqrt{\sin2\varphi}$.
Step 2 · Determine the range of $r$
Differentiating, $\frac{d}{d\varphi}\left(a\sqrt{\sin2\varphi}\right)=\frac{a\cos2\varphi}{\sqrt{\sin2\varphi}}=0$ gives $\cos2\varphi=0\Rightarrow\varphi=\frac\pi4$, where $\sin2\varphi=1$, so $r_{\max}=a$. Hence $0\leqslant r\leqslant a$.
Step 3 · Find bounds on $\varphi$ for fixed $r$
From $r=a\sqrt{\sin2\varphi}$: $\sin2\varphi=\frac{r^2}{a^2}$, so $2\varphi=\arcsin\!\left(\frac{r^2}{a^2}\right)$ or $2\varphi=\pi-\arcsin\!\left(\frac{r^2}{a^2}\right)$. Since $a\sqrt{\sin2\varphi}$ rises on $[0,\pi/4]$ and falls on $[\pi/4,\pi/2]$, a fixed $r\leqslant a$ is reached once on each branch:
$$ \frac12\arcsin\!\left(\frac{r^2}{a^2}\right) \leqslant \varphi \leqslant \frac{\pi}{2}-\frac12\arcsin\!\left(\frac{r^2}{a^2}\right). $$
Step 4 · Write the integral with the reversed order
$$ \int_0^a dr \int_{\frac12\arcsin(r^2/a^2)}^{\frac{\pi}{2}-\frac12\arcsin(r^2/a^2)} z(\varphi,r)\,d\varphi. $$
Step 5 · Verification
At $r=0$: $\varphi$ ranges from $\frac12\arcsin(0)=0$ to $\frac\pi2-0=\frac\pi2$. ✓ At $r=a$: $\varphi$ ranges from $\frac12\arcsin(1)=\frac\pi4$ to $\frac\pi2-\frac\pi4=\frac\pi4$ (a single point). ✓
Additional Notes
- Whenever the outer polar boundary is not monotonic in $\varphi$, reversing the order produces two symmetric branches, as here.
Result
The order-reversed integral is: