Change the order of integration: integral from -2 to ² dx
Change the order of integration: $$ \int_{-2}^{2} dx \int_{-\frac{1}{\sqrt2}\sqrt{4-x^2}}^{\frac{1}{\sqrt2}\sqrt{4-x^2}} z(x,y)\,dy. $$
Explanation of Method
Squaring the boundary curves reveals an ellipse. Once we know the semi-axes, swapping the order is the same computation as for a circle, just rescaled.
- Square the boundary curves to find the ellipse equation.
- Read off the semi-major and semi-minor axes.
- Solve the ellipse equation for $x$ in terms of $y$.
Step 1 · Identify the region from the given integral
From $\int_{-2}^{2} dx\int_{-\frac{1}{\sqrt2}\sqrt{4-x^2}}^{\frac{1}{\sqrt2}\sqrt{4-x^2}} z\,dy$: $-2\leqslant x\leqslant2$, $-\frac{1}{\sqrt2}\sqrt{4-x^2}\leqslant y\leqslant\frac{1}{\sqrt2}\sqrt{4-x^2}$.
Step 2 · Analyze the boundary curves
$y=\pm\frac{1}{\sqrt2}\sqrt{4-x^2}$; squaring gives $y^2=\frac{4-x^2}{2}$, so
$$ x^2+2y^2=4 \quad\Longrightarrow\quad \frac{x^2}{4}+\frac{y^2}{2}=1, $$
an ellipse with semi-major axis $a=2$ along $x$ and semi-minor axis $b=\sqrt2$ along $y$.
Step 3 · Describe the region
$D$ is the interior of the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$.
Step 4 · Find the range of $y$
$$ -\sqrt2 \leqslant y \leqslant \sqrt2. $$
Step 5 · Find the bounds on $x$ for fixed $y$
From $\frac{x^2}{4}+\frac{y^2}{2}=1$: $x^2=4-2y^2=2(2-y^2)$, so
$$ x=\pm\sqrt2\sqrt{2-y^2}. $$
Step 6 · Write the integral with the reversed order
$$ \int_{-\sqrt2}^{\sqrt2} dy \int_{-\sqrt2\sqrt{2-y^2}}^{\sqrt2\sqrt{2-y^2}} z(x,y)\,dx. $$
Additional Notes
- The original $x$-bounds already had the "$\frac{1}{\sqrt2}$" scale factor baked in; recognizing the ellipse form early avoids algebra mistakes when solving for the other variable.
Result
The order-reversed integral is: