Find the surface area of the surface of revolution generated by revolving the graph y=x3, 0x7 around the x-axis.
The formula for the area of a surface of revolution about the $\displaystyle x$-axis is
$\displaystyle
\begin{array}{rcl}
\mbox{surface area}&=&\int_{x_1}^{x_2}2\pi y\,ds\\
&=&2\pi\int_{x_1}^{x_2}y\sqrt{1+\left(\frac{dy}{dx }\right)^2}\,dx.
\end{array}
$
We get this from the formula for the area of a cone frustum.
In our case, the integral becomes
$\displaystyle 2\pi \int_0^7 x^3\sqrt{1+(3x^2)^2}\,dx.$