We are tasked with finding the exact length of the curve given by the parametric equations:

$x = \frac{y^4}{8} + \frac{1}{4}y^2$ with $1 \leq y \leq 3$.

The formula for the length of a curve defined by $x = f(y)$ from $y = a$ to $y = b$ is given by:

$L = \int_a^b \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy$

Step 1: Find $\frac{dx}{dy}$

First, we need to differentiate $x$ with respect to $y$. The function is:

$x = \frac{y^4}{8} + \frac{1}{4}y^2$

We can differentiate each term separately:

$\frac{dx}{dy} = \frac{d}{dy} \left( \frac{y^4}{8} \right) + \frac{d}{dy} \left( \frac{1}{4}y^2 \right)$

$\frac{dx}{dy} = \frac{4y^3}{8} + \frac{2y}{4} = \frac{y^3}{2} + \frac{y}{2}$

Thus, the derivative of $x$ with respect to $y$ is:

$\frac{dx}{dy} = \frac{y^3}{2} + \frac{y}{2}$

Step 2: Set up the integral

We now substitute this expression into the formula for the length of the curve. The formula becomes:

$L = \int_1^3 \sqrt{1 + \left( \frac{y^3}{2} + \frac{y}{2} \right)^2} \, dy$

Step 3: Simplify the expression inside the square root

First, expand the square:

$\left( \frac{y^3}{2} + \frac{y}{2} \right)^2 = \left( \frac{y^3 + y}{2} \right)^2 = \frac{(y^3 + y)^2}{4}$

Now, simplify the expression inside the square root:

$1 + \frac{(y^3 + y)^2}{4} = \frac{4}{4} + \frac{(y^3 + y)^2}{4} = \frac{4 + (y^3 + y)^2}{4}$

Thus, the integral becomes:

$L = \int_1^3 \sqrt{\frac{4 + (y^3 + y)^2}{4}} \, dy$

This simplifies further to:

$L = \int_1^3 \frac{\sqrt{4 + (y^3 + y)^2}}{2} \, dy$

Step 4: Evaluate the integral

This integral does not have an elementary closed-form solution, so it must be solved either numerically or approximated. Using numerical methods (such as Simpson's rule or a calculator), we can approximate the value of this integral.

Would you like me to compute this numerically for you?