Answer
$S\approx258.85$
Work Step by Step
To find the surface area of $y=\frac{1}{3}x^{3}$ revolved around the x-axis from $[0,3]$:
$\frac{dy}{dx}=x^{2}$
Then, use the formula for surface area of a revolution, where 'r' will equal the function:
$S=2\pi\int_{a}^{b}r\sqrt (1+(\frac{dy}{dx})^{2})dx$
$S=2\pi\int_{0}^{3}(\frac{1}{3}x^{3})(\sqrt (1+(x^{2})^{2})dx$
$S=\frac{2\pi}{3}\int_{0}^{3}(x^{3})(\sqrt (1+x^{4})dx$
let $u=1+x^{4}$
$du=4x^{3}dx$
$\frac{1}{4}du=x^{3}dx$
$u(3)=82$
$u(0)=1$
$S=\frac{2\pi}{3}(\frac{1}{4})\int_{1}^{82}\sqrt u du$
$S=\frac{\pi}{6}[\frac{2}{3}u^{\frac{3}{2}}]_{1}^{82}$
$S=\frac{\pi}{9}[82^{\frac{2}{3}}-1^{\frac{2}{3}}]$
$S=\frac{\pi}{9}[82\sqrt 82-1]$
$S\approx258.85$
