(i) The curve cuts the y-axis at \(A = (0, 1)\) and the line \(x = 5\) at \(B = (5, \frac{1}{2})\).

The slope of line \(AB\) is \(\frac{\frac{1}{2} - 1}{5 - 0} = -\frac{1}{10}\).

Using the point-slope form: \(y - 1 = -\frac{1}{10}(x - 0)\).

Simplifying gives: \(y = -\frac{1}{10}x + 1\).

(ii) To find the volume, calculate the volume of revolution for both the curve and the line, then subtract.

For the curve: \(V_{\text{curve}} = \pi \int_{0}^{5} \left( \frac{1}{(3x+1)^{\frac{1}{4}}} \right)^2 \, dx\).

\(= \pi \int_{0}^{5} (3x+1)^{-\frac{1}{2}} \, dx\).

Integrating gives: \(\frac{2\pi}{3} [(3x+1)^{\frac{1}{2}}]_{0}^{5}\).

\(= \frac{2\pi}{3} [4 - 1] = 2\pi\).

For the line: \(V_{\text{line}} = \pi \int_{0}^{5} \left( -\frac{1}{10}x + 1 \right)^2 \, dx\).

\(= \pi \int_{0}^{5} \left( \frac{1}{100}x^2 - \frac{1}{5}x + 1 \right) \, dx\).

Integrating gives: \(\pi \left[ \frac{1}{300}x^3 - \frac{1}{10}x^2 + x \right]_{0}^{5}\).

\(= \pi \left[ \frac{125}{300} - \frac{25}{10} + 5 \right] = \frac{35\pi}{12}\).

The volume of the shaded region is \(V_{\text{line}} - V_{\text{curve}} = \frac{35\pi}{12} - 2\pi = \frac{11\pi}{12}\).