What Is the Correct Surface Area of a Curve Rotated Around the X-axis?

[Saterial]

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]

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 .

 

[Saterial]

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

 

[Millennial]

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