I am trying to find the volume of the solid that is the region bounded by $y=x^2+1,y=0,x=0,x=1$ rotated about the $x$-axis by the disc method.
What I'm struggling with is the radius. It seems to me that the radius should just be 1 since the vertical shape never gets any skinnier. $$A(y) = 1\pi\ dx$$ And my limits would be 2 to 0 (max value). Just wondering if this is the right direction. Thanks for the help!
I think you're confusing $x$ with $y$ here. First, since the disks are perpendicular to the $x$-axis, the area function should depend on $x$ not $y$. The radius of each disk is the distance from the $x$-axis (which is $y=0$) up to the curve $y=x^2+1$. So the radius is $x^2+1 - 0$. So write instead
$$A(x) = \pi (x^2+1)^2 \; dx.$$
Finally, your limits are in the $x$ direction, not the $y$-direction, so they're given to you in the problem statement. $x$ goes from $0$ to $1$.