The problem involves finding the volume of a solid formed by rotating a region $Q$ around the $y$-axis.

Step-by-step solution:

The region $Q$ is bounded by:

• The function $f(x) = x$,
• The line $x = 1$,
• The lines $y = 1$ and $y = 3$.

1. Express $x$ in terms of $y$:

Since $f(x) = x$, we can solve for $x$ in terms of $y$. From $y = x$, we have: $x = y$

Thus, the region is bounded between $y = 1$ and $y = 3$, and the upper limit on $x$ is $x = 1$.

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

The formula for the volume of a solid of revolution using cylindrical shells is: $V = 2\pi \int_{y_{\text{min}}}^{y_{\text{max}}} \text{radius} \times \text{height} \, dy$

Here, the radius is $y$ (since we're rotating around the $y$-axis), and the height is $1 - x = 1 - y$. The limits of integration are from $y = 1$ to $y = 3$.

Thus, the volume integral becomes: $V = 2\pi \int_{1}^{3} y(1 - y) \, dy$

3. Evaluate the integral:

First, expand the integrand: $y(1 - y) = y - y^2$

Now, integrate term by term: $\int_1^3 (y - y^2) \, dy = \int_1^3 y \, dy - \int_1^3 y^2 \, dy$

Each integral can be computed as follows: $\int_1^3 y \, dy = \left[ \frac{y^2}{2} \right]_1^3 = \frac{3^2}{2} - \frac{1^2}{2} = \frac{9}{2} - \frac{1}{2} = \frac{8}{2} = 4$ $\int_1^3 y^2 \, dy = \left[ \frac{y^3}{3} \right]_1^3 = \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$

Thus, the integral becomes: $V = 2\pi \left( 4 - \frac{26}{3} \right) = 2\pi \left( \frac{12}{3} - \frac{26}{3} \right) = 2\pi \left( \frac{-14}{3} \right) = \frac{-28\pi}{3}$

Since the volume is negative, we take the absolute value. Hence, the exact volume is: $V = \frac{28\pi}{3}$

Final Answer:

$V = \frac{28\pi}{3}$

Do you need any further details or clarification?

Related Questions:

1. What is the volume if the region was rotated around the $x$-axis instead?
2. How would the volume change if the upper bound for $y$ was increased to 4?
3. What is the surface area of the resulting solid of revolution?
4. How can we solve this using the disc method?
5. What happens if we rotate the region around the line $x = 2$?

Tip:

Always consider which method—cylindrical shells or discs—is simpler based on the axis of rotation when finding volumes of revolution.