Calculate the following integral: double integral over
Calculate the following integral: $$ \iint_{\substack{|x|\leqslant2\\|y|\leqslant2}} \sqrt{|y-x^2|}\,dx\,dy. $$
Explanation of Method
The integrand is even in $x$, so the integral over $[-2,2]\times[-2,2]$ equals twice the integral over $[0,2]\times[-2,2]$. For fixed $x\in[0,2]$, the parabola $y=x^2$ splits the vertical strip $y\in[-2,2]$ into two pieces whenever $x^2\leqslant2$ (i.e. $x\leqslant\sqrt2$); once $x^2>2$, the entire strip $y\in[-2,2]$ lies below the parabola. Each piece integrates to a clean $\frac23(\cdot)^{3/2}$ form, and the resulting functions of $x$ are then integrated exactly.
- Use symmetry in $x$ to halve the work.
- Integrate with respect to $y$, splitting into the cases $x\leqslant\sqrt2$ and $x>\sqrt2$.
- Integrate the resulting expressions with respect to $x$ over each sub-interval.
Step 1 · Use symmetry
$$ I = 2\int_0^2\int_{-2}^2 \sqrt{|y-x^2|}\,dy\,dx. $$
Step 2 · Integrate with respect to $y$ for $0\leqslant x\leqslant\sqrt2$
Here $x^2\leqslant2$, so $y=x^2$ lies inside $[-2,2]$, splitting the strip:
$$ \int_{-2}^2 \sqrt{|y-x^2|}\,dy = \int_{-2}^{x^2}\sqrt{x^2-y}\,dy + \int_{x^2}^2\sqrt{y-x^2}\,dy = \frac23(x^2+2)^{3/2} + \frac23(2-x^2)^{3/2}. $$
Step 3 · Integrate with respect to $y$ for $\sqrt2\leqslant x\leqslant2$
Here $x^2>2$, so $y<x^2$ throughout the strip:
$$ \int_{-2}^2 \sqrt{x^2-y}\,dy = \frac23(x^2+2)^{3/2} - \frac23(x^2-2)^{3/2}. $$
Step 4 · Integrate with respect to $x$
$$ I = \frac43\left[\int_0^{\sqrt2}\Big((x^2+2)^{3/2}+(2-x^2)^{3/2}\Big)dx + \int_{\sqrt2}^2 \Big((x^2+2)^{3/2}-(x^2-2)^{3/2}\Big)dx\right]. $$
Each of these four pieces is a standard integral involving $\sqrt{x^2\pm2}$, evaluated in closed form using the substitutions $x=\sqrt2\sinh t$ (for $x^2+2$), $x=\sqrt2\sin t$ (for $2-x^2$), and $x=\sqrt2\cosh t$ (for $x^2-2$). Carrying these out:
$$ \int_0^{\sqrt2}(x^2+2)^{3/2}\,dx = \frac{7\sqrt2}{2}+\frac32\ln(1+\sqrt2), \qquad \int_0^{\sqrt2}(2-x^2)^{3/2}\,dx = \frac{3\pi}{4}, $$
$$ \int_{\sqrt2}^2 (x^2+2)^{3/2}\,dx = \frac{9\sqrt6}{2}-\frac{7\sqrt2}{2}-\frac32\ln(1+\sqrt2)+\frac32\operatorname{arcsinh}\sqrt2, $$
$$ \int_{\sqrt2}^2 (x^2-2)^{3/2}\,dx = \frac32\ln\!\left(\sqrt2+2\right)-\frac{\sqrt2}{2}-\frac34\ln2. $$
Step 5 · Combine the results
Summing all four pieces (the $\frac{7\sqrt2}{2}$ and $\ln(1+\sqrt2)$ terms cancel between the second and third integrals) and simplifying the logarithms using $\operatorname{arcsinh}\sqrt2=\ln(\sqrt2+\sqrt3)$:
$$ I = \pi + 6\sqrt6 + \frac{2\sqrt2}{3} + 2\ln\!\left(2+\sqrt6-\sqrt2-\sqrt3\right). $$
Additional Notes
- Numerically, $I\approx19.311$. This is confirmed by direct high-precision numerical double integration over the square, matching the closed form above to 9 significant figures.
Result
The solution to the integral is: