To find the exact area of the surface obtained by rotating the curve about the $x$-axis, we use the formula for the surface area of a curve rotated around the $x$-axis:

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

where $y$ is the function, and $\frac{dy}{dx}$ is its derivative. In this case, we are given:

$y = \frac{x^3}{3} + \frac{1}{4x}$

over the interval $\frac{1}{2} \leq x \leq 1$.

Step 1: Compute the derivative $\frac{dy}{dx}$

We differentiate $y = \frac{x^3}{3} + \frac{1}{4x}$ with respect to $x$:

$\frac{dy}{dx} = x^2 - \frac{1}{4x^2}$

Step 2: Set up the surface area integral

Now, substitute $y = \frac{x^3}{3} + \frac{1}{4x}$ and $\frac{dy}{dx} = x^2 - \frac{1}{4x^2}$ into the surface area formula. The integral becomes:

$A = 2\pi \int_{\frac{1}{2}}^1 \left(\frac{x^3}{3} + \frac{1}{4x}\right) \sqrt{1 + \left( x^2 - \frac{1}{4x^2} \right)^2} \, dx$

Step 3: Simplify the integrand

First, simplify the expression inside the square root:

$\left( x^2 - \frac{1}{4x^2} \right)^2 = x^4 - 2 \cdot \frac{1}{4} + \frac{1}{16x^4} = x^4 - \frac{1}{2} + \frac{1}{16x^4}$

Now add 1 to this result:

$1 + \left( x^2 - \frac{1}{4x^2} \right)^2 = 1 + x^4 - \frac{1}{2} + \frac{1}{16x^4} = x^4 + \frac{1}{2} + \frac{1}{16x^4}$

Thus, the surface area integral becomes:

$A = 2\pi \int_{\frac{1}{2}}^1 \left(\frac{x^3}{3} + \frac{1}{4x}\right) \sqrt{x^4 + \frac{1}{2} + \frac{1}{16x^4}} \, dx$

Step 4: Solve the integral

At this stage, the integral is quite complex and may require numerical methods or further simplification for an exact answer. However, we can set it up for computation:

$A = 2\pi \int_{\frac{1}{2}}^1 \left(\frac{x^3}{3} + \frac{1}{4x}\right) \sqrt{x^4 + \frac{1}{2} + \frac{1}{16x^4}} \, dx$

The final answer can be expressed as this integral, or computed numerically.

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Would you like a numerical evaluation of the integral or more details on a specific part?

Here are some related questions you might consider:

1. How does the surface area formula differ when rotating around the y-axis?
2. What are alternative methods to approximate this integral?
3. Can this formula be applied to other types of functions?
4. What happens if the curve is rotated around another axis?
5. How do changes in the interval $\left[\frac{1}{2}, 1\right]$ affect the surface area?

Tip: In many cases, complex integrals like this one can be tackled using numerical approximation techniques, such as Simpson's rule or using a computer algebra system.