Question Define \( Q \) as the region bounded on the right by the graph of \( f(y)=\frac{3}{y} \) and on the left by the \( y \)-axis for \( y \in[1,3] \). Find the volume of the solid of revolution formed by revolving \( Q \) around the \( x \)-axis. - Enter an exact value in terms of \( \pi \).

To find the volume of the solid of revolution formed by revolving the region \( Q \) around the \( x \)-axis, we will use the disk method. The volume \( V \) is given by the integral: \[ V = \pi \int_{y_1}^{y_2} [f(y)]^2 \, dy \] In this case, \( f(y) = \frac{3}{y} \), and the boundaries are \( y_1 = 1 \) and \( y_2 = 3 \). Thus, we set up the integral as follows: \[ V = \pi \int_{1}^{3} \left( \frac{3}{y} \right)^2 \, dy \] Calculating \( \left( \frac{3}{y} \right)^2 \): \[ \left( \frac{3}{y} \right)^2 = \frac{9}{y^2} \] Now, we can write the volume integral: \[ V = \pi \int_{1}^{3} \frac{9}{y^2} \, dy \] Next, we calculate the integral: \[ V = \pi \cdot 9 \int_{1}^{3} y^{-2} \, dy \] The integral of \( y^{-2} \) is: \[ \int y^{-2} \, dy = -\frac{1}{y} \] Evaluating the definite integral from \( 1 \) to \( 3 \): \[ = -\frac{1}{3} - (-\frac{1}{1}) = -\frac{1}{3} + 1 = \frac{2}{3} \] Now substituting back into the volume formula: \[ V = \pi \cdot 9 \cdot \frac{2}{3} = \pi \cdot 6 = 6\pi \] Hence, the volume of the solid of revolution is: \[ \boxed{6\pi} \]