Find the area of the surface of revolution $y=\frac{1}{3}x^3$ from $x=0$ to $3$, about the y-axis.
$$S=\int_{0}^{9} 2 \pi x \sqrt{1+(\d{x}{y}})^2\,dy$$
$$=2\pi\int_{0}^{9} x \sqrt{1+\frac{1}{x^4}}\,dy$$
$$=2\pi\int_{0}^{9} \frac{1}{x}\sqrt{x^4+1}\,dy$$
$$=2\pi\int_{0}^{9}...