Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D e^{-x^2-y^2}\,dx\,dy, $$ where $D$ is the disk $x^2+y^2\leqslant a^2$.
Explanation of Method
In polar coordinates, $x^2+y^2=r^2$, so the integrand becomes $e^{-r^2}$ and the extra factor of $r$ from the area element $dA=r\,dr\,d\theta$ makes the radial integral directly antidifferentiable — this is the classical trick that makes the Gaussian integral tractable over a disk (unlike over the whole plane in Cartesian form).
- Convert to polar coordinates.
- Integrate with respect to $r$ using the substitution $u=r^2$.
- Integrate the constant result with respect to $\theta$.
Step 1 · Convert to polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$, the disk becomes $0\leqslant r\leqslant a$, $0\leqslant\theta\leqslant2\pi$, and $x^2+y^2=r^2$:
$$ I = \int_0^{2\pi}\int_0^a e^{-r^2}\cdot r\,dr\,d\theta. $$
Step 2 · Integrate with respect to $r$
With $u=r^2$, $du=2r\,dr$:
$$ \int_0^a e^{-r^2}r\,dr = \frac12\int_0^{a^2} e^{-u}\,du = \frac12\left[-e^{-u}\right]_0^{a^2} = \frac12\left(1-e^{-a^2}\right). $$
Step 3 · Integrate with respect to $\theta$
$$ I = \int_0^{2\pi} \frac12\left(1-e^{-a^2}\right)d\theta = 2\pi\cdot\frac12\left(1-e^{-a^2}\right) = \pi\left(1-e^{-a^2}\right). $$
Additional Notes
- As $a\to\infty$, $I\to\pi$, which is the value of the classical Gaussian integral $\iint_{\mathbb R^2} e^{-x^2-y^2}\,dx\,dy=\pi$ — consistent with $\left(\int_{-\infty}^\infty e^{-x^2}\,dx\right)^2=\pi$.
Result
The solution to the integral is: