Surface area of a solid of revolution

In summary: I'm not sure what that is.Can someone please clarify what they mean by "fundamentally wrong with our approach"? Could you elaborate on what you think this could be?
  • #1
madgab89
22
0

Homework Statement



Consider now a shape that is obtained by revolving
the “infinitely long” function
f (x) = 1/x , 1 ≤ x < ∞
around the x-axis. Find both the surface area of the resulting object, and the
enclosed volume of it, i.e. the volume of the solid obtained from revolving the
area under the graph of f (x) around the x-axis.

If you do this correctly, you will find an infinite surface area but a finite volume. This would mean that a certain (finite) amount of paint is sufficient to fill up our object, but that this
amount would not be enough to actually paint it. Does this surprising result
mean that there is something fundamentally wrong with our approach? Why
or why not? Comment critically!

Homework Equations


I understand how to calculate the surface area and the volume of a solid of revolution, however it's the second part of the question I don't really undertand, the part about there being something fundamentally wrong with our approach. I was wondering if anyone could give me some hints to point me in the right direction.
 
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  • #2
madgab89 said:

Homework Statement



Consider now a shape that is obtained by revolving
the “infinitely long” function
f (x) = 1/x , 1 ≤ x < ∞
around the x-axis. Find both the surface area of the resulting object, and the
enclosed volume of it, i.e. the volume of the solid obtained from revolving the
area under the graph of f (x) around the x-axis.

If you do this correctly, you will find an infinite surface area but a finite volume. This would mean that a certain (finite) amount of paint is sufficient to fill up our object, but that this
amount would not be enough to actually paint it. Does this surprising result
mean that there is something fundamentally wrong with our approach? Why
or why not? Comment critically!

Homework Equations


I understand how to calculate the surface area and the volume of a solid of revolution, however it's the second part of the question I don't really undertand, the part about there being something fundamentally wrong with our approach. I was wondering if anyone could give me some hints to point me in the right direction.

A good start would be to actually calculate the surface area and volume for this solid. Keep in mind that you are going to have improper integrals for both, since one of the limits of integration is [itex]\infty[/itex].
 
  • #3
Yeah, I did that:

I got pi as the volume of the solid and undefined (infinity) for the surface area. However its the theoretical part that I'm unsure about, where its asking if there's something wrong with the method...
 

1. What is the surface area of a solid of revolution?

The surface area of a solid of revolution is the total area of the curved surface formed by rotating a function or curve around an axis. It is a measure of the outer surface of the three-dimensional shape.

2. How is the surface area of a solid of revolution calculated?

The surface area of a solid of revolution can be calculated using the formula A = 2π∫(f(x)√(1+(f'(x))^2)dx, where f(x) is the function or curve being rotated and f'(x) is its derivative.

3. What is the difference between surface area and volume of a solid of revolution?

Surface area refers to the outer area of a three-dimensional shape, while volume refers to the amount of space inside the shape. The surface area is a two-dimensional measurement, while volume is a three-dimensional measurement.

4. What are some real-life applications of calculating surface area of a solid of revolution?

The surface area of a solid of revolution is commonly used in engineering and architecture to calculate the strength and stability of structures such as bridges and buildings. It is also used in physics and chemistry to determine the surface area of particles and molecules.

5. Are there any limitations to calculating the surface area of a solid of revolution?

One limitation is that the formula for calculating surface area assumes that the shape being rotated is continuous and smooth. It may not accurately calculate the surface area for irregular or discontinuous shapes. Additionally, the formula only applies to solids of revolution and cannot be used for other types of three-dimensional shapes.

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