To find the volume of the solid formed by rotating the region $Q$ around the $y$-axis, we will use the method of cylindrical shells.

Step 1: Define the region $Q$

The region $Q$ is bounded by:

• The function $f(x) = x$, which is $x = y$ (or $y = x$ rearranged).
• The vertical line $x = \frac{1}{2}$.
• The horizontal lines $y = 3$ and $y = 6$.

The region lies in the first quadrant.

Step 2: Set up the volume integral using the method of cylindrical shells

We rotate the region around the $y$-axis. The method of cylindrical shells is given by the formula:

$V = 2 \pi \int_{a}^{b} x \cdot f(x) \, dx$

In our case:

• The rotation is around the $y$-axis.
• We use vertical strips parallel to the $x$-axis.

We will integrate with respect to $y$, so let's express $x$ in terms of $y$. From $f(x) = x$, we have $x = y$. Thus, our limits are from $y = 3$ to $y = 6$.

The distance from the $y$-axis to the region is $x$, which is equal to $y$, and the height of each shell is given by the difference ( \left( 1 - x \right) = 1 - \frac{1}{2} = height

The cylindrical shell volume will be, upon integrating this over the function over that height interval so our total is given, based on now computing bounds formating with pi and its given so.