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 curve $r=1-\varphi^2$ where $r>0$.
Explanation of Method
Here the boundary is given directly in polar form, so the "integrate $r$ first" order is immediate. To reverse it we solve $r=1-\varphi^2$ for $\varphi$ in terms of $r$.
- Find the $\varphi$-range where $r=1-\varphi^2>0$.
- Write the direct order using the given polar boundary.
- Solve for $\varphi(r)$ to reverse the order.
Step 1 · Determine the range of $\varphi$
$r>0$ requires $1-\varphi^2>0 \Rightarrow \varphi^2<1 \Rightarrow -1<\varphi<1$ (radians).
Step 2 · Describe the region
The curve $r=1-\varphi^2$ is the outer boundary, and the origin ($r=0$) is the inner boundary. At $\varphi=0$: $r=1$ (maximum); at $\varphi=\pm1$: $r=0$ (curve reaches the origin).
Step 3 · Order 1 — integrate $r$ first, then $\varphi$
$$ \iint_D z\,dx\,dy = \int_{-1}^{1} d\varphi \int_0^{1-\varphi^2} z(r\cos\varphi,r\sin\varphi)\,r\,dr. $$
Step 4 · Order 2 — integrate $\varphi$ first, then $r$
From $r=1-\varphi^2$: $\varphi^2=1-r \Rightarrow \varphi=\pm\sqrt{1-r}$, valid for $0\leqslant r\leqslant1$:
$$ \iint_D z\,dx\,dy = \int_0^{1} dr \int_{-\sqrt{1-r}}^{\sqrt{1-r}} z(r\cos\varphi,r\sin\varphi)\,r\,d\varphi. $$
Additional Notes
- Since the boundary is already given as $r$ as a function of $\varphi$, Order 1 requires no extra algebra — only Order 2 requires inverting the relation.
Result
The limits of integration are: