If the infinite curve $y=e^{x}, x \geq 0$, is rotated about the $x$-axis, find the area of the resulting surface.

Step 1 ：We are given the infinite curve \(y=e^{x}, x \geq 0\), and we are asked to find the area of the surface resulting from rotating this curve about the x-axis.

Step 2 ：The formula for the surface area of a solid of revolution is: \[ A = 2\pi \int_{a}^{b} y \sqrt{1 + (y')^2} dx \] where \( y' \) is the derivative of \( y \) with respect to \( x \), and \( a \) and \( b \) are the limits of \( x \).

Step 3 ：In this case, \( y = e^x \), \( y' = e^x \), \( a = 0 \), and \( b \) approaches infinity.

Step 4 ：So, we need to calculate the integral: \[ A = 2\pi \int_{0}^{\infty} e^x \sqrt{1 + (e^x)^2} dx \]

Step 5 ：The integral evaluates to infinity, which means the surface area of the resulting solid is infinite. This makes sense because the curve \( y = e^x \) extends to infinity along the x-axis, and rotating it about the x-axis would create a surface with infinite area.

Step 6 ：Final Answer: The area of the resulting surface is \(\boxed{\infty}\).