Find the area of the surface generated when the given curve is revolved about the​ x-axis.
yequalsx cubed plus StartFraction 1 Over 12 x EndFraction
​, for one half less than or equals x less than or equals 1
Question content area bottom
Part 1
The area of the surface is 
  
enter your response here square units.
​(Type an exact​ answer, using pi as​ needed.)

To find the area of the surface generated by revolving a curve \( y = f(x) \) about the x-axis over an interval \([a, b]\), we use the surface area formula:

\[
A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
\]

Given:
\[
y = x^3 + \frac{1}{12x}
\]
over the interval \(\frac{1}{2} \leq x \leq 1\).

### Step 1: Compute the derivative \(\frac{dy}{dx}\)
\[
\frac{dy}{dx} = \frac{d}{dx}\left(x^3 + \frac{1}{12x}\right)
\]
\[
\frac{dy}{dx} = 3x^2 - \frac{1}{12x^2}
\]

### Step 2: Plug into the surface area formula
The surface area \(A\) is:

\[
A = 2\pi \int_{\frac{1}{2}}^{1} \left( x^3 + \frac{1}{12x} \right) \sqrt{1 + \left(3x^2 - \frac{1}{12x^2}\right)^2} \, dx
\]

### Step 3: Simplify the expression under the square root
First, compute:
\[
\left(3x^2 - \frac{1}{12x^2}\right)^2 = 9x^4 - 2\cdot 3x^2 \cdot \frac{1}{12x^2} + \left(\frac{1}{12x^2}\right)^2
\]
\[
= 9x^4 - \frac{1}{2} + \frac{1}{144x^4}
\]
So:
\[
1 + \left(3x^2 - \frac{1}{12x^2}\right)^2 = 1 + 9x^4 - \frac{1}{2} + \frac{1}{144x^4} = 9x^4 + \frac{1}{2} + \frac{1}{144x^4}
\]

### Step 4: Evaluate the integral
The integral can be complex and generally needs advanced integration techniques or numerical methods for an exact solution. However, we can represent the exact surface area in terms of an integral as follows:

\[
A = 2\pi \int_{\frac{1}{2}}^{1} \left( x^3 + \frac{1}{12x} \right) \sqrt{9x^4 + \frac{1}{2} + \frac{1}{144x^4}} \, dx
\]

This expression is the exact area of the surface generated by revolving the curve about the x-axis. 

If you need further help solving or simplifying this integral, feel free to ask!

Would you like me to simplify further, or is there anything else you'd like to clarify?

### Related Questions:
1. How would the surface area change if the curve were revolved about the y-axis instead?
2. What is the arc length of the curve over the same interval?
3. Can the integral be solved using numerical methods? If so, how?
4. What is the significance of the function's derivative in determining the surface area?
5. How does changing the interval \([a, b]\) affect the surface area?

**Tip:** When dealing with complex integrals, substitution or numerical methods like Simpson's rule can be highly effective.

Find the area of the surface generated when the given curve is revolved about the x-axis.

y = x^3 + 1/(12x), for 1/2 ≤ x ≤ 1

The area of the surface is ___ square units.
(Type an exact answer, using π as needed.)

Calculate the exact surface area generated when the curve y = x^3 + 1/(12x) is revolved about the x-axis over the interval [1/2, 1]. This problem involves advanced calculus concepts like surface area of revolution and integration techniques, ideal for college-level or advanced high school calculus students.

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