Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D (x^2+y^2)\,dx\,dy, $$ where $D$ is the region $x^2+y^2\leqslant1$, $x^2\leqslant y^2$.
Explanation of Method
The condition $x^2\leqslant y^2$ means $|x|\leqslant|y|$, i.e. the region lies within the two "vertical bowtie" wedges around the $y$-axis. In polar coordinates this is $|\cos\theta|\leqslant|\sin\theta|$, giving two symmetric angular ranges of total measure $\pi$ (half of the full circle), each contributing equally to the integral.
- Convert $x^2\leqslant y^2$ to angular bounds in polar coordinates.
- Set up the integral over the two symmetric wedges.
- Integrate with respect to $r$, then $\theta$.
Step 1 · Identify the region
In polar coordinates, $x^2\leqslant y^2 \Leftrightarrow |\cos\theta|\leqslant|\sin\theta|$, which holds for $\theta\in\left[\frac\pi4,\frac{3\pi}4\right]\cup\left[\frac{5\pi}4,\frac{7\pi}4\right]$. Combined with $x^2+y^2\leqslant1 \Leftrightarrow 0\leqslant r\leqslant1$:
Step 2 · Set up the integral
By the symmetry $\theta\to\theta+\pi$, both wedges contribute equally:
$$ I = 2\int_{\pi/4}^{3\pi/4}\int_0^1 r^2\cdot r\,dr\,d\theta. $$
Step 3 · Integrate with respect to $r$
$$ \int_0^1 r^3\,dr = \frac14. $$
Step 4 · Integrate with respect to $\theta$
$$ \int_{\pi/4}^{3\pi/4} d\theta = \frac{3\pi}4-\frac\pi4=\frac\pi2. $$
Step 5 · Combine the results
$$ I = 2\cdot\frac\pi2\cdot\frac14 = \frac\pi4. $$
Additional Notes
- Since the two wedges together cover exactly half the disk's angular range, and the integrand $x^2+y^2=r^2$ depends only on $r$, this integral is exactly half of $\iint_{x^2+y^2\leqslant1}(x^2+y^2)\,dx\,dy=\frac\pi2$ — matching directly.
Result
The solution to the integral is: