Find the area of the surface generated by revolving the curve about the indicated axis.
-$y = x ^ { 3 } / 9,0 \leq x \leq 2$; $x$-axis
A) $\frac { 98 } { 81 } \pi$
B) $\frac { 256 } { 27 } \pi$
C) $98 \pi$
D) $\frac { 1163 } { 2187 } \pi$

Question

Quizplus · Accepted Answer

The surface area $A$ generated by revolving the curve $y = \frac{x^3}{9}$ about the x-axis from $x = 0$ to $x = 2$ can be found using the formula $A = 2\pi \int_{a}^{b} y \sqrt{1 + (dy/dx)^2} dx$. Here, $dy/dx = x^2/3$, so $1 + (dy/dx)^2 = 1 + (x^2/3)^2$. Integrating this from 0 to 2 and simplifying gives $\frac{98}{81}\pi$.

Find the Area of the Surface Generated by Revolving the Curve