The problem is to evaluate the double integral $\int_0^\infty \int_0^\infty e^{-(x^2 + y^2)} \, dx \, dy$ using a change of variables.

Solution:

The integral involves the expression $x^2 + y^2$, which suggests that converting to polar coordinates ($r, \theta$) is appropriate. The steps are as follows:

1. Switch to polar coordinates:

   • In polar coordinates, $x = r\cos\theta$ and $y = r\sin\theta$, so: $x^2 + y^2 = r^2.$
   • The Jacobian for the transformation is $r$, so the area element $dx\,dy$ becomes $r\,dr\,d\theta$.

2. Set up the integral in polar form:

   • The region $x, y \in [0, \infty)$ corresponds to $r \in [0, \infty)$ and $\theta \in [0, \frac{\pi}{2}]$.
   • Thus, the integral becomes: = \int_0^{\pi/2} \int_0^\infty e^{-r^2} \cdot r \, dr \, d\theta.$$

3. Evaluate the $r$-integral:

   • The inner integral over $r$ is: $\int_0^\infty r e^{-r^2} \, dr.$ Substituting $u = r^2$, so $du = 2r\,dr$, the limits of integration become $u \in [0, \infty)$: $\int_0^\infty r e^{-r^2} \, dr = \frac{1}{2} \int_0^\infty e^{-u} \, du.$ The integral of $e^{-u}$ over $[0, \infty)$ is 1, so: $\int_0^\infty r e^{-r^2} \, dr = \frac{1}{2}.$

4. Evaluate the $\theta$-integral:

   • The remaining integral is: $\int_0^{\pi/2} \frac{1}{2} \, d\theta = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}.$

Final Answer:

$\int_0^\infty \int_0^\infty e^{-(x^2 + y^2)} \, dx \, dy = \frac{\pi}{4}.$

Would you like me to clarify any step or explore further variations? Here are five related questions:

1. How does the Jacobian factor affect the change of variables in integrals?
2. What other integrals are simplified by switching to polar coordinates?
3. Why does the substitution $u = r^2$ work in this scenario?
4. How can we generalize this result to other Gaussian integrals?
5. What is the physical significance of Gaussian integrals in probability theory?

Tip: When encountering $\sqrt{x^2 + y^2}$ or $x^2 + y^2$, polar coordinates often simplify the problem significantly.