How to Find Area of a Rotated Surface?

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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?
 
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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].
 
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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