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 a circle: $1^{\circ}\ x^2+y^2\leqslant1$; $2^{\circ}\ x^2+y^2\leqslant x$.
Explanation of Method
For Part $1^\circ$ we solve the circle equation for $y$ in terms of $x$ (and vice versa) to get the two boundary curves. For Part $2^\circ$ we first complete the square to see the region is a circle not centered at the origin, then repeat the process about its own center.
- Solve the circle equation for the top/bottom (or left/right) boundary.
- Complete the square for Part $2^\circ$ to find center and radius.
- Write both orders of integration for each part.
Part $1^\circ$ · Circle $x^2+y^2\leqslant1$
Step 1 · Order 1 — integrate $y$ first, then $x$
For $-1\leqslant x\leqslant1$, solving $x^2+y^2=1$ for $y$ gives $y=\pm\sqrt{1-x^2}$:
$$ \iint_D z(x,y)\,dx\,dy = \int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} z(x,y)\,dy\,dx. $$
Step 2 · Order 2 — integrate $x$ first, then $y$
By symmetry, for $-1\leqslant y\leqslant1$, $x=\pm\sqrt{1-y^2}$:
$$ \iint_D z(x,y)\,dx\,dy = \int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} z(x,y)\,dx\,dy. $$
Part $2^\circ$ · Circle $x^2+y^2\leqslant x$
Step 3 · Rewrite in standard form
Completing the square:
$$ x^2-x+y^2\leqslant0 \quad\Longrightarrow\quad \left(x-\tfrac12\right)^2+y^2\leqslant\tfrac14. $$
This is a circle centered at $\left(\frac12,0\right)$ with radius $\frac12$.
Step 4 · Find the range of $x$ and $y$
$$ 0\leqslant x\leqslant1, \qquad -\tfrac12\leqslant y\leqslant\tfrac12. $$
Step 5 · Order 1 — integrate $y$ first, then $x$
Solving $\left(x-\frac12\right)^2+y^2=\frac14$ for $y$: $y^2=\frac14-\left(x-\frac12\right)^2=x-x^2=x(1-x)$, so $y=\pm\sqrt{x-x^2}$:
$$ \iint_D z(x,y)\,dx\,dy = \int_0^1\int_{-\sqrt{x-x^2}}^{\sqrt{x-x^2}} z(x,y)\,dy\,dx. $$
Step 6 · Order 2 — integrate $x$ first, then $y$
Solving for $x$: $x=\frac12\pm\sqrt{\frac14-y^2}$:
$$ \iint_D z(x,y)\,dx\,dy = \int_{-1/2}^{1/2}\int_{\frac12-\sqrt{\frac14-y^2}}^{\frac12+\sqrt{\frac14-y^2}} z(x,y)\,dx\,dy. $$
Summary Table
| Region | $dy\,dx$ | $dx\,dy$ |
|---|---|---|
| $1^\circ$: $x^2+y^2\leqslant1$ | $\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}$ | $\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}$ |
| $2^\circ$: $x^2+y^2\leqslant x$ | $\int_0^1\int_{-\sqrt{x-x^2}}^{\sqrt{x-x^2}}$ | $\int_{-1/2}^{1/2}\int_{\frac12-\sqrt{\frac14-y^2}}^{\frac12+\sqrt{\frac14-y^2}}$ |
Additional Notes
- For Part $1^\circ$, the circle is symmetric about both axes, so both orders are equally simple.
- For Part $2^\circ$, completing the square reveals a circle centered at $\left(\frac12,0\right)$; the boundary formulas differ by which variable is isolated.
Result
The limits of integration are: