Problem: Find the surface area of the solid formed by revolving $y = \frac{1}{5}x^5 + \frac{1}{12x^3}$ about the $y$-axis for $x$ between $1$ and $2$.
Question: How do I solve this equation for $x$?
I'd like to use the formula $$ 2\pi\int_{y_1}^{y_2} x(y)\sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy $$ but I can't figure out how to solve for $x$ in order to use it.
Note that $y(x)$ is one-to-one on the $x$-interval, so I know an inverse exists there. I also know that I could implicitly differentiate to find $\frac{dx}{dy}$ if need be; nevertheless, the formula still requires me to know $x(y)$.
Could this be a typo and what was meant is to revolve around the $x$-axis instead? In this case, I have found the solution without a problem.
The area of the solid of revolution of $ y = f(x) $ about the $y$-axis between $x=x_1$ and $x = x_2$ is given by
$ A = 2\pi \displaystyle \int_{y_1}^{y_2} x \ \sqrt{ 1 + \bigg( \dfrac{dx}{dy} \bigg)^2 } dy $
Since $ y = f(x)$ then $dy = f'(x) dx $ and $ \dfrac{dx}{dy} = \dfrac{1}{f'(x)} $
Therefore, the area of revolution is
$ A = 2 \pi \displaystyle \int_{x_1}^{x_2} x f'(x) \sqrt{ 1 + \dfrac{1}{[f'(x)]^2} } dx = 2 \pi \int_{x_1}^{x_2} x \sqrt{ 1 + [f'(x)]^2 } \ dx$