Change the order of integration: integral from -1 to 1 dx
Change the order of integration: $$ \int_{-1}^{1} dx \int_0^{\sqrt{1-x^2}} z(x,y)\,dy. $$
Explanation of Method
The upper bound $y=\sqrt{1-x^2}$ is the upper unit semicircle, and the lower bound $y=0$ is the $x$-axis, so the region is the upper half-disk. We swap the order by solving $x^2+y^2=1$ for $x$ in terms of $y$.
- Identify the region as the upper half-disk of radius $1$.
- Find the $y$-range.
- Solve for the left/right boundaries in terms of $y$.
Step 1 · Identify the region from the given integral
From $\int_{-1}^{1} dx\int_0^{\sqrt{1-x^2}} z\,dy$: $-1\leqslant x\leqslant1$, $0\leqslant y\leqslant\sqrt{1-x^2}$.
Step 2 · Analyze the curve $y=\sqrt{1-x^2}$
Squaring gives $x^2+y^2=1$ — the upper unit semicircle (since $y\geqslant0$).
Step 3 · Describe the region
$D$ is the upper half-disk: bounded below by $y=0$, above by $y=\sqrt{1-x^2}$, left by $x=-1$, right by $x=1$.
Step 4 · Find the range of $y$
$$ 0\leqslant y\leqslant1. $$
Step 5 · Find the bounds on $x$ for fixed $y$
From $x^2+y^2=1$: $x=\pm\sqrt{1-y^2}$.
Step 6 · Write the integral with the reversed order
$$ \int_0^1 dy \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} z(x,y)\,dx. $$
Additional Notes
- This is the simplest possible order-swap: the region is symmetric about the $y$-axis, so the reversed limits are simply $\pm\sqrt{1-y^2}$ with no splitting needed.
Result
The order-reversed integral is: