To find the volume of the solid generated by revolving the region bounded by the curves $y = 4x - 3$, $y = \sqrt{x}$, and $x = 0$ about the $y$-axis, we will use the shell method. The formula for the volume using the shell method is:

$V = 2\pi \int_{a}^{b} \left( \text{radius} \times \text{height} \right) \, dx$

Step 1: Set up the integral

• The radius is the distance from the $y$-axis, which is just $x$ (since we're rotating around the $y$-axis).
• The height of the shell at each $x$ is the difference between the two functions, $y = 4x - 3$ (upper curve) and $y = \sqrt{x}$ (lower curve).

Thus, the height of the shell is:

$\text{height} = (4x - 3) - \sqrt{x}$

Step 2: Find the limits of integration

We need to find the limits of integration, which correspond to the values of $x$ where the curves $y = 4x - 3$ and $y = \sqrt{x}$ intersect. Set them equal to each other to solve for $x$:

$4x - 3 = \sqrt{x}$

Square both sides to eliminate the square root:

$(4x - 3)^2 = x$

Expanding the left-hand side:

$16x^2 - 24x + 9 = x$

Rearrange the equation:

$16x^2 - 25x + 9 = 0$

Now solve this quadratic equation using the quadratic formula:

$x = \frac{-(-25) \pm \sqrt{(-25)^2 - 4(16)(9)}}{2(16)}$

$x = \frac{25 \pm \sqrt{625 - 576}}{32} = \frac{25 \pm \sqrt{49}}{32} = \frac{25 \pm 7}{32}$

So, $x = \frac{32}{32} = 1$ or $x = \frac{18}{32} = \frac{9}{16}$.

Thus, the limits of integration are from $x = \frac{9}{16}$ to $x = 1$.

Step 3: Set up the integral

The volume is given by:

$V = 2\pi \int_{\frac{9}{16}}^1 x \left( (4x - 3) - \sqrt{x} \right) \, dx$

Simplify the integrand:

$V = 2\pi \int_{\frac{9}{16}}^1 x \left( 4x - 3 - \sqrt{x} \right) \, dx$

Step 4: Integrate

Now, we break this integral into three parts and integrate each term:

$V = 2\pi \left( \int_{\frac{9}{16}}^1 4x^2 \, dx - \int_{\frac{9}{16}}^1 3x \, dx - \int_{\frac{9}{16}}^1 x^{3/2} \, dx \right)$

1. First term:

$\int_{\frac{9}{16}}^1 4x^2 \, dx = \left[ \frac{4x^3}{3} \right]_{\frac{9}{16}}^1 = \frac{4}{3}(1^3) - \frac{4}{3}\left( \frac{9}{16} \right)^3$

2. Second term:

$\int_{\frac{9}{16}}^1 3x \, dx = \left[ \frac{3x^2}{2} \right]_{\frac{9}{16}}^1$