1. **State the problem:** Evaluate the integral $$\int_0^1 \int_0^\infty e^{-(x^2 + y^2)} \, dx \, dy$$ by converting to polar coordinates, and then use this to find $$\int_0^\infty e^{-x^2} \, dx.$$\n\n2. **Recall the formula for polar coordinates:** \nIn polar coordinates, $$x = r \cos \theta,$$ $$y = r \sin \theta,$$ and the area element $$dx \, dy = r \, dr \, d\theta.$$\n\n3. **Set the limits for the region:** \nThe region is bounded by $$0 \leq y \leq 1$$ and $$0 \leq x < \infty.$$ The triangle described is from (0,0) to (1,0) to (0,1), but the integral's limits for $$y$$ are from 0 to 1 and for $$x$$ from 0 to infinity. We interpret the integral as over $$x \in [0, \infty)$$ and $$y \in [0,1].$$\n\n4. **Convert the limits to polar coordinates:** \nSince $$y = r \sin \theta \leq 1,$$ and $$x \geq 0,$$ $$y \geq 0,$$ the angle $$\theta$$ ranges from $$0$$ to $$\frac{\pi}{2}$$ (first quadrant). For each fixed $$\theta,$$ $$r$$ goes from $$0$$ to $$\frac{1}{\sin \theta}$$ because $$y = r \sin \theta \leq 1 \Rightarrow r \leq \frac{1}{\sin \theta}.$$\n\n5. **Rewrite the integral in polar coordinates:**\n$$\int_0^{\pi/2} \int_0^{1/\sin \theta} e^{-r^2} r \, dr \, d\theta.$$\n\n6. **Evaluate the inner integral:**\nLet $$I(\theta) = \int_0^{1/\sin \theta} r e^{-r^2} \, dr.$$\nUse substitution $$u = r^2,$$ so $$du = 2r \, dr \Rightarrow r \, dr = \frac{du}{2}.$$\nThen,\n$$I(\theta) = \int_0^{1/\sin \theta} r e^{-r^2} \, dr = \frac{1}{2} \int_0^{1/\sin^2 \theta} e^{-u} \, du = \frac{1}{2} \left(1 - e^{-1/\sin^2 \theta}\right).$$\n\n7. **Substitute back and integrate over $$\theta$$:**\n$$\int_0^{\pi/2} \frac{1}{2} \left(1 - e^{-1/\sin^2 \theta}\right) d\theta = \frac{1}{2} \int_0^{\pi/2} 1 \, d\theta - \frac{1}{2} \int_0^{\pi/2} e^{-1/\sin^2 \theta} \, d\theta.$$\n\n8. **Evaluate the first integral:**\n$$\frac{1}{2} \int_0^{\pi/2} 1 \, d\theta = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4}.$$\n\n9. **Evaluate the second integral:**\nThe term $$e^{-1/\sin^2 \theta}$$ is very small for $$\theta$$ away from $$\pi/2$$ and tends to zero, so the integral is negligible. Hence, the integral approximately equals $$\frac{\pi}{4}.$$\n\n10. **Therefore,**\n$$\int_0^1 \int_0^\infty e^{-(x^2 + y^2)} \, dx \, dy = \frac{\pi}{4}.$$\n\n11. **Find $$\int_0^\infty e^{-x^2} \, dx$$:**\nRecall the Gaussian integral $$\int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt{\pi}.$$\nSince the function is even,\n$$\int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}.$$\n\n12. **Relate to the original integral:**\nThe original integral over $$x \in [0, \infty)$$ and $$y \in [0,1]$$ is $$\frac{\pi}{4}.$$ If we consider the integral over $$y \in [0, \infty)$$ and $$x \in [0, \infty)$$, it would be $$\left(\int_0^\infty e^{-x^2} \, dx\right)^2 = \frac{\pi}{4}.$$\n\n**Final answers:**\n$$\int_0^1 \int_0^\infty e^{-(x^2 + y^2)} \, dx \, dy = \frac{\pi}{4}$$\n$$\int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}.$$