Working backwards from cylindrical shells I have the volume of a solid, and I'm supposed to find the shape of the solid:
$$\int_0^1 2\pi(6-y)(1-y^2) \mathrm dy $$
This resembles a volume by cylindrical shells, but I'm not sure how to work backwards. It looks like 6-y is an adjusted rotation (i.e. not rotated around the x-axis). However, it's -y and not positive y, so I'm not sure how to find the original shape.
 A: You're right - it seems funny. For the shell method, we're used to $\int 2 \pi r(x) f(x) \mathrm{d}x$. It's sort of like that here, and you're exactly right that we should think of it  through the shell method.
I recommend two options. Either draw a picture - for example, note that at $y = 0$, the 'radius' of the solid is 6 and the 'height' is 1. At $y = 1$, the radius is 5 and the height is 0. From the equation, we see that a linear change in radius is a quadratic change in height - so if you draw it well, it will become apparent.
Or, do a change of variable to make it seem more familiar. In particular, I recommend $x = (6 - 6)$. Don't forget the chain rule! The resulting integral will be in the exact same shell method as normal.
A: The following little trick helps me.  Interchange the roles of $x$ and $y$. So our integral is now
$$\int_0^1 2\pi(6-x)(1-x^2)\,dx.$$
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 $y=6$. 
