Basic Surface Area of Revolution

Or did you make a mistake in your simplification?In summary, the area of the surface obtained by rotating the curve y=x3, 0≤x≤2 about the x-axis can be found by using the equation SA = ∫2∏y√(1+(dy/dx)2)dx from 0->2. However, a substitution is needed in order to properly solve the integral. By letting u = 1+9x4 and (1/36)du=x3dx, the integral can be simplified to 2∏/36∫√udu. The final answer, after plugging in the value of 2, is a very large number.
  • #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)
 
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  • #2
Saterial said:

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

1. What is the basic surface area of revolution?

The basic surface area of revolution is the measure of the total area of a three-dimensional shape that is formed by rotating a two-dimensional curve or shape around a central axis.

2. How is the basic surface area of revolution calculated?

The basic surface area of revolution is calculated using the formula A = 2πrh, where A is the surface area, π is pi (approximately equal to 3.14), r is the radius of the curve, and h is the height of the curve.

3. What types of shapes can be used to calculate the basic surface area of revolution?

Any two-dimensional shape or curve can be used to calculate the basic surface area of revolution, as long as it is rotated around a central axis to form a three-dimensional shape. Some common examples include circles, parabolas, and ellipses.

4. How is the basic surface area of revolution used in real life?

The basic surface area of revolution has many practical applications in fields such as engineering, architecture, and physics. It is used to calculate the surface area of objects such as cylinders and spheres, which are commonly found in everyday objects and structures.

5. Are there any limitations to using the basic surface area of revolution?

While the basic surface area of revolution is a useful tool for calculating the surface area of many three-dimensional shapes, it does have limitations. It cannot be used for shapes with irregular curves, and it assumes that the shape being rotated is a perfect curve with a uniform radius. In some cases, more advanced mathematical methods may be needed to accurately calculate the surface area.

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