In the double integral double integral over D z(x,y) dx
In the double integral $\iint_D z(x,y)\,dx\,dy$, convert to polar coordinates $\varphi$ and $r$ and write the limits of integration for both possible orders. The region $D$ is a rectangle with vertices $O(0,0)$, $A(1,0)$, $B(1,1)$, $C(0,1)$.
Explanation of Method
In polar coordinates a straight edge $x=1$ becomes $r=\sec\varphi$ and an edge $y=1$ becomes $r=\csc\varphi$. Because a ray from the origin hits a different edge depending on its angle relative to the diagonal $OB$ ($\varphi=\pi/4$), the region must be split into two angular pieces (or, for the reversed order, two radial pieces).
- Locate the vertices in polar form.
- Split the angular range at the diagonal $\varphi=\pi/4$.
- Convert each edge equation to $r(\varphi)$.
- Reverse the order, splitting radially at $r=1$ instead.
Step 1 · Identify the region
$D=[0,1]\times[0,1]$, the unit square.
Step 2 · Determine the polar bounds
Vertices in polar form: $O(0,0)$: $r=0$; $A(1,0)$: $r=1,\varphi=0$; $B(1,1)$: $r=\sqrt2,\varphi=\pi/4$; $C(0,1)$: $r=1,\varphi=\pi/2$. The angle sweeps the first quadrant, $0\leqslant\varphi\leqslant\pi/2$.
For $0\leqslant\varphi\leqslant\pi/4$, a ray hits the right edge $x=1$ first: $r\cos\varphi=1\Rightarrow r=\sec\varphi$.
For $\pi/4\leqslant\varphi\leqslant\pi/2$, a ray hits the top edge $y=1$ first: $r\sin\varphi=1\Rightarrow r=\csc\varphi$.
Step 3 · Order 1 — integrate $r$ first, then $\varphi$
$$ \iint_D z\,dx\,dy = \int_0^{\pi/4} d\varphi \int_0^{\sec\varphi} r\,z\,dr \ + \ \int_{\pi/4}^{\pi/2} d\varphi \int_0^{\csc\varphi} r\,z\,dr. $$
Step 4 · Order 2 — integrate $\varphi$ first, then $r$
For $0\leqslant r\leqslant1$: the circular arc of radius $r$ lies entirely inside the square, so $\varphi$ sweeps the full first quadrant, $0\leqslant\varphi\leqslant\pi/2$.
For $1<r\leqslant\sqrt2$: the arc pokes outside the square except between where it crosses $x=1$ (at $\varphi=\arccos(1/r)$) and $y=1$ (at $\varphi=\arcsin(1/r)$):
$$ \iint_D z\,dx\,dy = \int_0^{1} dr \int_0^{\pi/2} r\,z\,d\varphi \ + \ \int_1^{\sqrt2} dr \int_{\arccos(1/r)}^{\arcsin(1/r)} r\,z\,d\varphi. $$
Additional Notes
- Straight edges of a Cartesian region become $r=\sec\varphi$ or $r=\csc\varphi$ type curves in polar coordinates — useful to recognize whenever a rectangular boundary appears in a polar setup.
- The split point $\varphi=\pi/4$ (equivalently $r=1$ radially) is exactly the square's diagonal.
Result
The limits of integration are: