The following little trick helps me. Interchange the roles of $x$ and $y$. So our integral is now
The change of letters brings us back to the standard setup where the shells method is first introduced, and where we have the most problem-solving practice. We are looking at the curve $y=1-x^2$, a familiar downward-facing parabola. Now it is essential to draw it.
For $x$ going from $0$ to $1$, $1-x^2$ is the (vertical) distance from the point $(x,1-x^2)$ on the curve to the $x$-axis. Draw a thin vertical strip of width "$dx$" down from $(x,1-x^2)$ to the $x$-axis.
And we are rotating about something that would be the $y$-axis if the $6-x$ was $x$. But note that $6-x$ is the distance from the vertical line through $(x,f(x))$ to the vertical line $x=6$. If we rotate our thin strip about the line $x=6$, we will get a shell of thirckness $dx$, radius $6-x$, and height $1-x^2$. "adding up" (integrating) will get us exactly the integral we wrote down.
Now go back to the original integral, by interchanging $x$ and $y$ again, which geometrically means reflecting across the line $y=x$. Or now that we know what is going on, we can see that $x=1-y^2$ is a leftward-opening parabola, and that we are rotating the chunk to the right of the $y$-axis, from $y=0$ to $y=1$, about the line $y=6$.