Use the shell method to find the volume of the solid obtained by rotating the region bounded by the given curves $$y=\sqrt{x}$$  $$y=0$$ $$x=1$$ about x=-1
Does this set up look alright:
$$V =  \int_0^1 2 \pi (x+1)(\sqrt{x}) dx$$

Your setup is correct.
Note the formula for computing the volume via the shell method about the y-axis is:
$$\int_{a}^{b}2\pi x(f(x)-g(x))dx$$
You've correctly identified the circumference of the circle:
$$2\pi r= 2\pi (x+1)$$
Furthmore, since
$$f(x)=\sqrt{x}$$
and 
$$g(x)=0$$
The following equation (which you listed with the correct limits of integration) is correct:
$$V =  \int_0^1 2 \pi (x+1)(\sqrt{x}) dx$$
