It almost looks like a cup

let's rotate the shape 90 degrees, so it's easyer to work with.

$y=\ln(x^{3} ) ⇒ e^{\frac{y}{3}} = x$

using $\displaystyle\int_{a}^{b} A(x) dx = \pi\displaystyle\int_{a}^{b} f(x)^{2} dx$

$\pi\displaystyle\int_{0}^{1} e^{\frac{2y}{3}} dy ⇒ \frac{3\pi}{2}\displaystyle\int_{0}^{1} \frac{2}{3}e^{\frac{2y}{3}} dy$

$\frac{3\pi}{2}e^{\frac{2y}{3}}|^{1}_{0} ⇒ \pi\frac{3}{2}(e^{\frac{2}{3}}-1)$

$AB = \frac{3}{2}\frac{2}{3} = \boxed{1}$