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: $1^{\circ}$ Circle $x^2+y^2\leqslant a^2$; $2^{\circ}$ Circular annulus $a^2<x^2+y^2<b^2$.

Explanation of Method

With $x=r\cos\varphi$, $y=r\sin\varphi$, the Jacobian of the transformation is $r$, so $dx\,dy=r\,dr\,d\varphi$. Both regions are centered at the origin, so $\varphi$ simply sweeps the full circle $[0,2\pi]$, while $r$ is bounded by the circle radii.

Derive the Jacobian $r$ for the polar transformation.
Translate each Cartesian inequality into a bound on $r$.
Write both integration orders for each part.

Step 1 · Polar coordinate transformation

$$ x=r\cos\varphi,\qquad y=r\sin\varphi, \qquad \frac{\partial(x,y)}{\partial(r,\varphi)}=\begin{vmatrix}\cos\varphi & -r\sin\varphi\\ \sin\varphi & r\cos\varphi\end{vmatrix}=r, \qquad dx\,dy=r\,dr\,d\varphi. $$

Part $1^\circ$ · Circle $x^2+y^2\leqslant a^2$

Step 2 · Translate to polar bounds

$r^2\leqslant a^2 \Rightarrow 0\leqslant r\leqslant a$, and $\varphi$ sweeps the full circle: $0\leqslant\varphi\leqslant2\pi$.

Step 3 · Both orders

$$ \iint_D z\,dx\,dy = \int_0^{2\pi} d\varphi \int_0^a z(r\cos\varphi,r\sin\varphi)\,r\,dr = \int_0^a dr \int_0^{2\pi} z(r\cos\varphi,r\sin\varphi)\,r\,d\varphi. $$

Part $2^\circ$ · Annulus $a^2<x^2+y^2<b^2$

Step 4 · Translate to polar bounds

$a^2<r^2<b^2 \Rightarrow a<r<b$ (for $a,b>0$), and again $0\leqslant\varphi\leqslant2\pi$.

Step 5 · Both orders

$$ \iint_D z\,dx\,dy = \int_0^{2\pi} d\varphi \int_a^b z(r\cos\varphi,r\sin\varphi)\,r\,dr = \int_a^b dr \int_0^{2\pi} z(r\cos\varphi,r\sin\varphi)\,r\,d\varphi. $$

Additional Notes

Because both regions are centered at the origin, no splitting is needed in polar coordinates — contrast this with Problem 12, where the same annulus required three Cartesian pieces per order.
The factor $r$ from the Jacobian must always accompany $dr\,d\varphi$.

Result

The limits of integration are:

$$ 1^\circ:\ \int_0^{2\pi}\!\int_0^a r\,z\,dr\,d\varphi = \int_0^a\!\int_0^{2\pi} r\,z\,d\varphi\,dr $$ $$ 2^\circ:\ \int_0^{2\pi}\!\int_a^b r\,z\,dr\,d\varphi = \int_a^b\!\int_0^{2\pi} r\,z\,d\varphi\,dr $$