To find the volume $V$ of the solid obtained by rotating the region bounded by the curves $y = x^2$ and $y = 3x$ about the $y$-axis, we will use the method of cylindrical shells. This method is suitable for rotating a region around a vertical axis such as the $y$-axis.

Step-by-Step Solution

1. Identify the region of interest:

   The curves given are: $y = x^2 \quad \text{(parabola)}$ $y = 3x \quad \text{(line)}$

   First, we find the points of intersection between these two curves. Set $x^2 = 3x$ to find the $x$-values where the curves intersect.

   $x^2 = 3x \implies x^2 - 3x = 0 \implies x(x - 3) = 0$

   The solutions to this equation are $x = 0$ and $x = 3$. So, the curves intersect at $(0, 0)$ and $(3, 9)$.

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

   When rotating around the $y$-axis, the radius of a shell is given by the $x$-coordinate, and the height of a shell is the difference in $y$-values between the line and the parabola: $\text{Height} = 3x - x^2$

   The volume $V$ of the solid is: $V = \int_{a}^{b} 2\pi x (\text{Height}) \, dx$ Here, $a = 0$ and $b = 3$.

   Substituting the height: $V = \int_{0}^{3} 2\pi x (3x - x^2) \, dx$

3. Simplify and compute the integral:

   Expand the integrand: $V = 2\pi \int_{0}^{3} (3x^2 - x^3) \, dx$

   Integrate term by term: $V = 2\pi \left[ \int_{0}^{3} 3x^2 \, dx - \int_{0}^{3} x^3 \, dx \right]$

   Compute each integral: $\int 3x^2 \, dx = x^3 \quad \text{and} \quad \int x^3 \, dx = \frac{x^4}{4}$

   Evaluate these from $0$ to $3$: $V = 2\pi \left[ \left(x^3 \right) \Big|_{0}^{3} - \left(\frac{x^4}{4}\right) \Big|_{0}^{3} \right]$

   Plugging in the limits: $V = 2\pi \left[(3^3 - 0^3) - \left(\frac{3^4}{4} - \frac{0^4}{4}\right)\right]$

   Simplify: $V = 2\pi \left[(27) - \left(\frac{81}{4}\right)\right]$ $V = 2\pi \left[27 - 20.25\right]$ $V = 2\pi (6.75)$

   Finally, compute the volume: $V = 13.5\pi$

Final Answer

The volume of the solid obtained by rotating the region bounded by $y = x^2$ and $y = 3x$ about the $y$-axis is $\boxed{13.5\pi}$.

---

Would you like more details or have any questions about this problem?

Here are some related questions for further exploration:

1. How would the volume change if the region was rotated about the line $x = 4$ instead of the $y$-axis?
2. What is the volume if the region between $y = x^2$ and $y = 3x$ is rotated about the $x$-axis?
3. Can we use the washer method to solve this problem? If so, how?
4. What would be the volume if the region is rotated about the line $y = -1$?
5. How do the methods of cylindrical shells and washers differ, and when is each method more appropriate?

Tip: Always sketch the region and axis of rotation to visualize the problem better and understand the method to apply.