Question: Let $g(y) = \frac{y^3}{6} + \frac{1}{2y}$. Calculate the arc length of the graph of $g(y)$ over the interval $[1,4]$.
1. **State the problem:** We want to find the arc length of the curve defined by $$g(y) = \frac{y^3}{6} + \frac{1}{2y}$$ over the interval $$y \in [1,4].$$
2. **Formula for arc length:** The arc length $L$ of a function $g(y)$ from $y=a$ to $y=b$ is given by:
$$
L = \int_a^b \sqrt{1 + \left(\frac{dg}{dy}\right)^2} \, dy
$$
3. **Find the derivative $g'(y)$:**
$$
g(y) = \frac{y^3}{6} + \frac{1}{2y} = \frac{y^3}{6} + \frac{1}{2} y^{-1}
$$
Using power rule:
$$
g'(y) = \frac{3y^2}{6} - \frac{1}{2} y^{-2} = \frac{y^2}{2} - \frac{1}{2y^2}
$$
4. **Square the derivative:**
$$
\left(g'(y)\right)^2 = \left(\frac{y^2}{2} - \frac{1}{2y^2}\right)^2 = \left(\frac{y^2}{2}\right)^2 - 2 \cdot \frac{y^2}{2} \cdot \frac{1}{2y^2} + \left(\frac{1}{2y^2}\right)^2
$$
Calculate each term:
$$
= \frac{y^4}{4} - 2 \cdot \frac{y^2}{2} \cdot \frac{1}{2y^2} + \frac{1}{4y^4} = \frac{y^4}{4} - \frac{1}{2} + \frac{1}{4y^4}
$$
5. **Add 1 inside the square root:**
$$
1 + \left(g'(y)\right)^2 = 1 + \frac{y^4}{4} - \frac{1}{2} + \frac{1}{4y^4} = \frac{1}{2} + \frac{y^4}{4} + \frac{1}{4y^4}
$$
6. **Rewrite the expression under the root:**
$$
\frac{1}{2} + \frac{y^4}{4} + \frac{1}{4y^4} = \frac{2}{4} + \frac{y^4}{4} + \frac{1}{4y^4} = \frac{y^4 + 2 + \frac{1}{y^4}}{4}
$$
7. **Recognize a perfect square:**
$$
y^4 + 2 + \frac{1}{y^4} = \left(y^2 + \frac{1}{y^2}\right)^2
$$
8. **So the integrand simplifies to:**
$$
\sqrt{1 + \left(g'(y)\right)^2} = \sqrt{\frac{\left(y^2 + \frac{1}{y^2}\right)^2}{4}} = \frac{y^2 + \frac{1}{y^2}}{2}
$$
9. **Set up the integral for arc length:**
$$
L = \int_1^4 \frac{y^2 + \frac{1}{y^2}}{2} \, dy = \frac{1}{2} \int_1^4 \left(y^2 + y^{-2}\right) dy
$$
10. **Integrate term-by-term:**
$$
\int y^2 dy = \frac{y^3}{3}, \quad \int y^{-2} dy = -\frac{1}{y}
$$
11. **Evaluate the definite integral:**
$$
L = \frac{1}{2} \left[ \frac{y^3}{3} - \frac{1}{y} \right]_1^4 = \frac{1}{2} \left( \left(\frac{4^3}{3} - \frac{1}{4}\right) - \left(\frac{1^3}{3} - 1\right) \right)
$$
Calculate inside the brackets:
$$
= \frac{1}{2} \left( \frac{64}{3} - \frac{1}{4} - \frac{1}{3} + 1 \right) = \frac{1}{2} \left( \frac{64}{3} - \frac{1}{3} + 1 - \frac{1}{4} \right)
$$
$$
= \frac{1}{2} \left( \frac{63}{3} + \frac{4}{4} - \frac{1}{4} \right) = \frac{1}{2} \left( 21 + \frac{3}{4} \right) = \frac{1}{2} \cdot \frac{87}{4} = \frac{87}{8}
$$
12. **Final answer:**
$$
\boxed{\frac{87}{8}}
$$
This is the exact arc length of the graph of $g(y)$ over $[1,4]$.