jmc

Rotating the triangle about one of its legs produces a cone with radius 2 and height 2:

The base of the cone is a circle with radius 2, which has area $2^2\pi =4\pi$.

When unrolled, the curved lateral area of the cone becomes a flat sector of a circle: The sector's radius is the cone's slant height, which, by the Pythagorean theorem, is $$\sqrt{2^2+2^2}=2\sqrt{2}.$$The sector's arc length is the cone's base perimeter, which is $$2(\pi )(2)=4\pi .$$The circle's circumference is $$2(\pi )(2\sqrt{2})=4\sqrt{2}\pi ,$$so the ratio of the sector's area to the circle's area is $\frac{4\pi }{4\sqrt{2}\pi }=\frac{1}{\sqrt{2}}$. The circle's area is $$(2\sqrt{2}{)}^2\pi =8\pi ,$$so the sector's area is $$\frac{1}{\sqrt{2}}\cdot 8\pi =4\sqrt{2}\pi .$$Summing the lateral area and the base area gives a total surface area of $4\sqrt{2}\pi +4\pi$, so its total surface area is $\boxed{4\sqrt{2}+4}$ times $\pi$.