Write the limits of integration for the double integral
Write the limits of integration for the double integral $\iint_D z(x,y)\,dx\,dy$ for both possible orders of integration. The region $D$ is an annulus: $4<x^2+y^2<9$.
Explanation of Method
A vertical (or horizontal) line through an annulus can either miss the inner disk entirely, giving one connected segment, or pass through the hole, giving two disconnected segments. We split the $x$-range (and, symmetrically, the $y$-range) into the "single segment" zone and the "two segment" zone.
- Identify the inner circle (radius $2$) and outer circle (radius $3$).
- Classify vertical strips by whether $|x|<2$ (two pieces) or $2\leqslant|x|<3$ (one piece).
- Repeat for horizontal strips by symmetry.
- Note that polar coordinates avoid the splitting entirely.
Step 1 · Identify the region
$D$ lies between the inner circle $x^2+y^2=4$ (radius $2$) and the outer circle $x^2+y^2=9$ (radius $3$), both centered at the origin: $2<r<3$ in polar coordinates.
Step 2 · Determine the range of $x$ and $y$
Overall, $-3\leqslant x\leqslant3$ and $-3\leqslant y\leqslant3$, but not every such pair lies in $D$.
Step 3 · Order 1 — integrate $y$ first, then $x$
A vertical line at $x$ meets the outer circle at $y=\pm\sqrt{9-x^2}$ and the inner circle at $y=\pm\sqrt{4-x^2}$ (when $|x|<2$).
Case $2\leqslant|x|<3$: the line misses the hole entirely — one interval $-\sqrt{9-x^2}<y<\sqrt{9-x^2}$.
Case $|x|<2$: the line passes through the hole — two intervals, $\sqrt{4-x^2}<y<\sqrt{9-x^2}$ and $-\sqrt{9-x^2}<y<-\sqrt{4-x^2}$.
$$ \iint_D z\,dx\,dy = \int_{-3}^{-2}\!\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\! z\,dy\,dx + \int_{2}^{3}\!\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\! z\,dy\,dx + \int_{-2}^{2}\left[\int_{-\sqrt{9-x^2}}^{-\sqrt{4-x^2}} + \int_{\sqrt{4-x^2}}^{\sqrt{9-x^2}}\right] z\,dy\,dx. $$
Step 4 · Order 2 — integrate $x$ first, then $y$
By the same argument with the roles of $x$ and $y$ swapped (the annulus is symmetric about both axes):
$$ \iint_D z\,dx\,dy = \int_{-3}^{-2}\!\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\! z\,dx\,dy + \int_{2}^{3}\!\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\! z\,dx\,dy + \int_{-2}^{2}\left[\int_{-\sqrt{9-y^2}}^{-\sqrt{4-y^2}} + \int_{\sqrt{4-y^2}}^{\sqrt{9-y^2}}\right] z\,dx\,dy. $$
Step 5 · Note on polar coordinates
Polar coordinates avoid the splitting entirely:
$$ \iint_D z(x,y)\,dx\,dy = \int_0^{2\pi}\int_2^3 z(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta. $$
Additional Notes
- An annulus needs multiple Cartesian pieces because a straight line through the region can be interrupted by the hole in the middle.
- This complexity is exactly why polar coordinates — giving the simple bounds $2<r<3$, $0\leqslant\theta\leqslant2\pi$ — are strongly preferred for annular (ring-shaped) regions.
Result
The Cartesian limits (each order split into three pieces) are: