To find the volume of the solid formed when the shaded region is rotated about the y-axis, we use the formula for the volume of revolution:

\(V = \pi \int_{a}^{b} x^2 \, dy\)

Given \(x = \frac{12}{y^2} - 2\), we have:

\(x^2 = \left( \frac{12}{y^2} - 2 \right)^2\)

Expanding, \(x^2 = \frac{144}{y^4} - \frac{48}{y^2} + 4\).

The volume is:

\(V = \pi \int_{1}^{2} \left( \frac{144}{y^4} - \frac{48}{y^2} + 4 \right) \, dy\)

Integrating term by term:

\(\int \frac{144}{y^4} \, dy = -\frac{144}{3y^3} = -\frac{48}{y^3}\)

\(\int \frac{48}{y^2} \, dy = -\frac{48}{y}\)

\(\int 4 \, dy = 4y\)

Thus,

\(V = \pi \left[ -\frac{48}{y^3} - \frac{48}{y} + 4y \right]_{1}^{2}\)

Evaluating from 1 to 2:

\(V = \pi \left( \left( -\frac{48}{8} - \frac{48}{2} + 8 \right) - \left( -48 - 48 + 4 \right) \right)\)

\(V = \pi \left( -6 - 24 + 8 + 92 \right)\)

\(V = \pi \times 22\)

Therefore, the volume is \(22\pi\).