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?
Best Answer
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,
$\displaystyle \int_0^{4/5} \int_8^{10} 2\pi r \ dr \ dx + \int_{4/5}^{1} \int_8^{8/x} 2\pi r \ dr \ dx$