Find the volume of a solid with unsual axis Find the volume of the solid obtained by rotating the region bounded by the given curves $y=x^6$ and $y=1$ about the specified line which is $y=7$.
Right now I have the integral:
$$
\text{Volume} = 2\pi \int_1 ^7y^{1/6}(7-y) \, \mathrm{d} y.
$$
I am unsure why this is incorrect. Wouldn't I use the shell  method?
Thanks.

I would probably use "washers," though shells work fine. 
If you use shells, the integral should be
$$\int_{y=0}^1 (2\pi)(2y^{1/6})(7-y)\,dy.$$
Using washers, outer radius is $7-x^6$, inner radius is $6$, so we get
$$\int_{x=-1}^1 \pi ((7-x^6)^2 -6^2)\,dx,$$
though I would probably use symmetry, integrate from $x=0$ to $x=1$, and double the result. 
Remark: For either method, a careful picture is necessary. Note that $y=x^6$ kind of looks like $y=x^2$, but is much flatter near the origin. The region is symmetric  about the $y$-axis. The "height" of the cylindrical shell is $2x$, that is, $2y^{1/6}$. The height of the shell at $y$ is $7-y$. 
