Basic Surface Area of Revolution

  • Thread starter Saterial
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  • #1
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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)
 

Answers and Replies

  • #2
SammyS
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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:
[itex]\displaystyle \sqrt{1+9x^4}\ne1+3x^2[/itex]​

Notice that [itex]\displaystyle \frac{d}{dx}\ (1+9x^4)=36x^3\,,[/itex] so use a substitution like u = 1 + 9x4 .
 
  • #3
54
0
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
 
  • #4
296
0
No, in fact, it gives the answer. Are you doing some basic arithmetic error?
 

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