The problem involves finding the volume of a solid of revolution formed by rotating the region $Q$ around the $y$-axis. Let's break down the steps and calculate the volume.

Step 1: Define the region $Q$

• The curve is $f(x) = x$.
• The boundaries are the vertical lines $x = \frac{1}{2}$ and $y = 3$ and $y = 6$.

First, we rewrite the equation $f(x) = x$ in terms of $y$: $x = y.$ This equation will be helpful when rotating around the $y$-axis.

Step 2: Set up the integral using the shell method

Since the region is rotated around the $y$-axis, we will use the shell method. The formula for the volume of a solid of revolution using the shell method is: $V = 2\pi \int_{y_1}^{y_2} ( \text{radius of shell}) \cdot (\text{height of shell}) \, dy.$

In our case:

• The radius of the shell is $x = y$, as the region is rotated around the $y$-axis.
• The height of the shell is the distance between $y = 1/2$ and $x = 0$, which is $\frac{1}{2}$.

So the volume is given by: [ V = 2\pta