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 bounded by the lines $x=2$, $y=x$, and $y=x\sqrt3$.

Step 1 · Identify the region

Bounded by the vertical line $x=2$, and the two rays $y=x$ (angle $\pi/4$) and $y=x\sqrt3$ (angle $\pi/3$).

Step 2 · Convert lines to polar form

$y=x$: $\tan\varphi=1\Rightarrow\varphi=\pi/4$. $y=x\sqrt3$: $\tan\varphi=\sqrt3\Rightarrow\varphi=\pi/3$. $x=2$: $r\cos\varphi=2\Rightarrow r=2\sec\varphi$.

Step 3 · Describe the region in polar coordinates

A sector between the rays $\varphi=\pi/4$ and $\varphi=\pi/3$, bounded outward by $r=2\sec\varphi$.

Step 4 · Order 1 — integrate $r$ first, then $\varphi$

$$ \iint_D z\,dx\,dy = \int_{\pi/4}^{\pi/3} d\varphi \int_0^{2\sec\varphi} z(r\cos\varphi,r\sin\varphi)\,r\,dr. $$

Step 5 · Order 2 — integrate $\varphi$ first, then $r$

The largest $r$ occurs at the smaller angle $\varphi=\pi/4$: $r_{\max}=2\sec(\pi/4)=2\sqrt2$.

For $0<r\leqslant2$: the whole angular sector $\pi/4\leqslant\varphi\leqslant\pi/3$ lies inside $x\leqslant2$.
For $2<r\leqslant2\sqrt2$: only the part beyond where the circle of radius $r$ crosses $x=2$ (at $\varphi=\arccos(2/r)$) up to $\varphi=\pi/3$ is inside the region:

$$ \iint_D z\,dx\,dy = \int_0^{2} dr \int_{\pi/4}^{\pi/3} r\,z\,d\varphi \ + \ \int_2^{2\sqrt2} dr \int_{\arccos(2/r)}^{\pi/3} r\,z\,d\varphi. $$

Additional Notes

• Any line through the origin corresponds to a fixed $\varphi$ in polar coordinates — the two given lines fix the sector's angular width, $\pi/3-\pi/4=\pi/12$.
• The radial split at $r=2$ is where the vertical line $x=2$ first "enters" the sector, namely at $\varphi=\pi/4$.