How to Find Area of a Rotated Surface?

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Homework Help Overview

The discussion revolves around finding the area of a surface generated by rotating the function \( y = \frac{x^3}{2} + \frac{1}{6x} \) around the x-axis within the domain \( \frac{1}{2} \leq x \leq 1 \). Participants explore the appropriate formulas for calculating this area.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial formula for the area of a solid of rotation and question its correctness. There is an exploration of the correct formula for surface area, leading to the introduction of a formula involving \( 2\pi \) and the derivative of the function.

Discussion Status

The discussion has progressed with participants clarifying the correct formula for surface area and attempting to set up the integral accordingly. Some participants express uncertainty about the correctness of their calculations, while others provide reassurance regarding the conceptual understanding.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the extent of guidance provided. There is an acknowledgment of potential errors in detail while maintaining a focus on conceptual understanding.

stunner5000pt
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Homework Statement


Find the exact value when [itex]y = \frac{x^3}{2} + \frac{1}{6x}[/itex] in domain [itex]\frac{1}{2} \leq x \leq 1[/itex] is rotated around the X axis


Homework Equations


Area is given by
[tex]A = \int \pi \left(f(x)\right)^2[/tex]


The Attempt at a Solution


If the above formula is correct, then I should be integrating the following

[tex]A = \int_{\frac{1}{2}}^{1} ( \frac{x^3}{2} + \frac{1}{6x})^2 dx[/tex]

Is this set up alone correct?
 
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I think the formula you are using is for the volume of the solid of rotation: you can cut it up into thin slices of surface area [itex]\pi r^2 = \pi f(x)^2[/itex] and "thickness" dx.

Do you have a formula with a [itex]2\pi[/itex] and an f'? :)
 
CompuChip said:
I think the formula you are using is for the volume of the solid of rotation: you can cut it up into thin slices of surface area [itex]\pi r^2 = \pi f(x)^2[/itex] and "thickness" dx.

Do you have a formula with a [itex]2\pi[/itex] and an f'? :)

Hmm i think I am mistaken...
I was trying to look this up and foudn the following formula:

[tex]A = \int_{a}^{b} 2 \pi f(x) \sqrt{1 + (f'(x))^2} dx[/tex]

is this the fomula I need to be using?
 
Last edited:
stunner5000pt said:
Hmm i think I am mistaken...
I was trying to look this up and foudn the following formula:

[tex]A = \int_{a}^{b} 2 \pi f(x) \sqrt{1 + (f'(x))^2} dx[/tex]

is this the fomula I need to be using?

If it's surface area you want, then that looks right.
 
Dick said:
If it's surface area you want, then that looks right.

Ok that's perfect...

Now just to clarify, the integral becomes:

[tex]2\pi \int_{\frac{1}{2}}^{1} \left( \frac{3x^4+1}{6x^2} \right) \sqrt{1+\left(\frac{9x^4-1}{6x^2}\right)^2} dx[/tex]


So I was a bit lazy so I plugged into wolfram alpha and got the answer [itex]445\pi/768[/itex]

is this correct? is there anything that I missed to consider?
 
stunner5000pt said:
Ok that's perfect...

Now just to clarify, the integral becomes:

[tex]2\pi \int_{\frac{1}{2}}^{1} \left( \frac{3x^4+1}{6x^2} \right) \sqrt{1+\left(\frac{9x^4-1}{6x^2}\right)^2} dx[/tex]So I was a bit lazy so I plugged into wolfram alpha and got the answer [itex]445\pi/768[/itex]

is this correct? is there anything that I missed to consider?

Conceptually you aren't missing anything. In detail, it might be wrong. [Edit: my mistake, it is correct].
 
Last edited:
Dick said:
Conceptually you aren't missing anything. In detail, it might be wrong. [Edit: my mistake, it is correct].

Thank you :)
Been stressing about these questions for quite some time now
 

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