Volume of solid of rotation - can you always get same answer using shell or disk-washer? The question is to find the volume of revolution of the region bonded by $x=0$ and $y=10$ and $y=8$ and $xy=8$ when rotated about the $x$ axis.
Using the Shell method is almost trivial:
$V=2 \pi \int y (8/y)dy $ with bounds from $8$ to $10$.  The y variables cancel and the answer is $32 \pi$.
I'm having difficulty doing this with the disk-washer method.
Is it always possible to achieve the correct answer using both methods?
 A: 
For disk method, you are considering radius along $y$ axis and that is not defined by the same curves throughout $8 \leq y \leq 10$.
$xy = 8 \implies x = 1 $ when $y = 8$.
$xy = 8 \implies x = \frac{8}{10} $ when $y = 10$.
For $0 \leq x \leq \frac{4}{5}$, the radius of revolution is defined by lines $y = 8$ and $y = 10$.
For $\frac{4}{5} \leq x \leq 1$, the radius of revolution has lower bound of $y = 8$ and upper bound of $y = \frac{8}{x}$ which is hyperbola curve.
So volume of revolution using disk method,
$\displaystyle \int_0^{4/5} \int_8^{10} 2\pi r \ dr \ dx + \int_{4/5}^{1} \int_8^{8/x} 2\pi r \ dr \ dx$
A: 
Is it always possible to achieve the correct answer using both methods?

It depends on what you mean by "achieve."
Both methods should always give the same answer--when the integral can be evaluated explicitly by both methods.
However, it may be that one antiderivative may not have an elementary expression, or that the limits of integration cannot be expressed by elementary antiderivatives, and so the volume requires the use of special functions to calculate (or approximate) if the "wrong" method is selected.
One simple example would be the area produced by revolving the region $0 \leq x \leq 1, x^5 + x \leq y \leq 2$ about the $y$-axis:

Even writing down the integrand to evaluate this volume via disk integration requires Bring radicals, which are not elementary functions: $$\text{Volume } = \pi \int_0^2 BR(y)^2 dy.$$ It's debatable at best whether computing the volume of this solid via the disk method is "achievable."
Yet if we set this up using shell integration instead, any second-semester high-school calculus student can evaluate the resulting volume integral easily: $$\text{Volume } = 2\pi \int_0^1 x (2 - x^5 - x) dx = 2\pi [x^2 - \frac{x^7}{7} - \frac{x^3}{3}]_0^1 = \frac{22 \pi}{21}.$$ Perhaps there's some clever substitution which could be used to evaluate the disk integral above, in which case we would get the same answer of $\frac{22 \pi}{21}$.
But I'm pretty sure that any such substitution would in fact end up being a thinly (or not-so-thinly) disguised version of the shell integral, regardless.
