# Basic Surface Area of Revolution

## Homework Statement

Find the area of the surface obtained by rotating the curve y=x3, 0≤x≤2 about the x-axis.

## Homework Equations

\begin{equation*}
SA = \int_{0}^{2} 2 \pi y L
\end{equation*}

## The Attempt at a Solution

SORRY, I don't know how to use LaTeX yet.

∫2∏y√(1+(dy/dx)2)dx from 0->2
=∫2∏y√(1+(3x2)2)dx
=2∏∫x3√(1+9x4)dx
=2∏∫x3(1+3x2)dx
=2∏∫x3+3x5dx
=2∏[x4/4 + x6/2] 0->2
=plug in 2
=72∏

I don't see where I went wrong. The answer is ∏/27*(145√(145)-1)

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Find the area of the surface obtained by rotating the curve y=x3, 0≤x≤2 about the x-axis.

## Homework Equations

\begin{equation*}
SA = \int_{0}^{2} 2 \pi y L
\end{equation*}

## The Attempt at a Solution

SORRY, I don't know how to use LaTeX yet.

∫2∏y√(1+(dy/dx)2)dx from 0->2
=∫2∏y√(1+(3x2)2)dx
=2∏∫x3√(1+9x4)dx
=2∏∫x3(1+3x2)dx
=2∏∫x3+3x5dx
=2∏[x4/4 + x6/2] 0->2
=plug in 2
=72∏

I don't see where I went wrong. The answer is ∏/27*(145√(145)-1)
Review basic algebra:
$\displaystyle \sqrt{1+9x^4}\ne1+3x^2$​

Notice that $\displaystyle \frac{d}{dx}\ (1+9x^4)=36x^3\,,$ so use a substitution like u = 1 + 9x4 .

Trying it with that, I now have:

∫2∏y√(1+(dy/dx)2)dx from 0->2
=∫2∏y√(1+(3x2)2)dx
=2∏∫x3√(1+9x4)dx

let u = 1+9x4, (1/36)du=x3dx

=2∏/36∫√udu
=2∏/36[2u^3/2/3] 0->2
=2∏/36[2(1+9x4)3/2/3] 0->2
=plug in 2
=some ridiculously huge number compared to answer

No, in fact, it gives the answer. Are you doing some basic arithmetic error?