By converting to polar coordinates, calculate
By converting to polar coordinates, calculate the following integral: $$ \iint_D \frac{dx\,dy}{(x^2+y^2)\left(1+\sqrt[8]{x^2+y^2}\right)}, $$ where $D$ is the region $x^2-y^2<0$, $1<x^2+y^2<4$.
Explanation of Method
The condition $x^2-y^2<0$ (i.e. $|x|<|y|$) picks out the two opposite "bowtie" wedges of the plane where $y$ dominates $x$; in polar coordinates, $x^2-y^2=r^2\cos2\theta$, so this becomes $\cos2\theta<0$, giving $\theta\in\left(\frac\pi4,\frac{3\pi}4\right)\cup\left(\frac{5\pi}4,\frac{7\pi}4\right)$. Combined with the annulus $1<r<2$, and using the substitution $u=r^{1/4}$ to clear the eighth root, the radial integral reduces to a simple partial-fractions form.
- Convert the region $x^2-y^2<0$ to a polar angle condition.
- Set up the integral using symmetry between the two wedges.
- Integrate with respect to $r$ via $u=r^{1/4}$ and partial fractions, then with respect to $\theta$.
Step 1 · Identify the region
In polar coordinates, $x^2-y^2=r^2\cos2\theta$, so $x^2-y^2<0 \Leftrightarrow \cos2\theta<0$, which holds for $\frac\pi4<\theta<\frac{3\pi}4$ or $\frac{5\pi}4<\theta<\frac{7\pi}4$. Together with $1<x^2+y^2<4 \Leftrightarrow 1<r<2$, the region $D$ consists of two congruent wedge-shaped pieces of the annulus, symmetric to each other.
Step 2 · Convert to polar coordinates
$$ x^2+y^2=r^2, \qquad \sqrt[8]{x^2+y^2}=r^{1/4}, \qquad dx\,dy=r\,dr\,d\theta. $$
By symmetry, the integral over the second wedge equals the integral over the first, so:
$$ I = 2\int_{\pi/4}^{3\pi/4}\int_1^2 \frac{r}{r^2\left(1+r^{1/4}\right)}\,dr\,d\theta = 2\int_{\pi/4}^{3\pi/4}\int_1^2 \frac{dr}{r\left(1+r^{1/4}\right)}\,d\theta. $$
Step 3 · Evaluate the angular integral
$$ \int_{\pi/4}^{3\pi/4}d\theta = \frac{3\pi}{4}-\frac{\pi}{4}=\frac{\pi}{2}. $$
Step 4 · Evaluate the radial integral
Let $u=r^{1/4}$, so $r=u^4$, $dr=4u^3\,du$; when $r=1$, $u=1$, and when $r=2$, $u=2^{1/4}$:
$$ \int_1^2 \frac{dr}{r(1+r^{1/4})} = \int_1^{2^{1/4}} \frac{4u^3\,du}{u^4(1+u)} = 4\int_1^{2^{1/4}} \frac{du}{u(1+u)}. $$
By partial fractions, $\dfrac1{u(1+u)}=\dfrac1u-\dfrac1{1+u}$:
$$ 4\int_1^{2^{1/4}}\left(\frac1u-\frac1{1+u}\right)du = 4\Big[\ln u-\ln(1+u)\Big]_1^{2^{1/4}} = 4\left[\ln\frac{2^{1/4}}{1+2^{1/4}} - \ln\frac12\right]. $$
$$ = 4\left[\ln\frac{2^{1/4}}{1+2^{1/4}}+\ln2\right] = 4\ln\frac{2\cdot2^{1/4}}{1+2^{1/4}} = 4\ln\frac{2^{5/4}}{1+2^{1/4}}. $$
Step 5 · Combine the results
$$ I = 2\cdot\frac\pi2\cdot4\ln\frac{2^{5/4}}{1+2^{1/4}} = 4\pi\ln\frac{2^{5/4}}{1+2^{1/4}} = 4\pi\left[\frac54\ln2-\ln\!\left(1+2^{1/4}\right)\right]. $$
$$ I = \pi\left[5\ln2-4\ln\!\left(1+2^{1/4}\right)\right]. $$
Additional Notes
- Numerically, $I\approx\pi[5(0.6931)-4\ln(2.1892)]\approx1.0417$, confirmed by direct numerical double integration over both wedges of the region.
Result
The solution to the integral is: