To help you with the integral presented in the image, I'll first need to transcribe it and then solve the second integral for you. Let's start by identifying the expression.

Given integrals:

1. $\int_0^1 \int_0^1 e^{-x^2 - y^2} \, dy \, dx$
2. $2\pi \int_0^\infty e^{-r^2} r \, dr = \pi$

Since you are asking for the solution to the second integral, I'll focus on that.

Solving the Second Integral

The second integral is given as: $\int_0^\infty e^{-r^2} r \, dr$

To solve this, we'll use a standard trick involving polar coordinates and recognizing the Gaussian integral.

1. Convert the integral to a known form:

   This integral can be recognized as part of the Gaussian integral when transformed into polar coordinates. Let's compute the area of the entire Gaussian integral:

   $I = \int_{-\infty}^{\infty} e^{-x^2} dx \cdot \int_{-\infty}^{\infty} e^{-y^2} dy = \left( \int_{-\infty}^{\infty} e^{-x^2} dx \right)^2$

   This is known to be equal to $\sqrt{\pi}$ when evaluated over the entire real line.

2. Use polar coordinates:

   In polar coordinates, $x^2 + y^2 = r^2$ and the differential area element $dx\,dy$ becomes $r \, dr \, d\theta$. Thus:

   $I = \int_0^{2\pi} d\theta \int_0^\infty e^{-r^2} r \, dr$

   This splits into: $I = 2\pi \int_0^\infty e^{-r^2} r \, dr$

3. Solve the radial part:

   The integral $\int_0^\infty e^{-r^2} r \, dr$ is a standard Gaussian integral, which evaluates to:

   $\int_0^\infty e^{-r^2} r \, dr = \frac{1}{2}$

4. Final result:

   $2\pi \times \frac{1}{2} = \pi$

So the final answer is indeed $\pi$, as indicated in the image.

Would you like a more detailed explanation or do you have any questions?

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Related Questions:

1. How do polar coordinates simplify certain types of integrals?
2. What is the Gaussian integral and why is it significant?
3. How does the change of variables affect the limits of integration in double integrals?
4. Can the method used here be applied to integrals involving different functions of $r^2$?
5. What are some common applications of Gaussian integrals in physics?

Tip: When dealing with complicated integrals, consider symmetry and coordinate transformations to simplify the problem.