Use the shell method to find the volume of the solid generated by revolving the region bounded by y equals 4 x minus 3​, y equals StartRoot x EndRoot​, and x equals 0 about the​ y-axis.

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:**

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




Use the shell method to find the volume of the solid generated by revolving the region bounded by y equals 4x minus 3, y equals sqrt(x), and x equals 0 about the y-axis.

Find the volume of the solid generated by revolving the region bounded by y = 4x - 3, y = sqrt(x), and x = 0 about the y-axis using the shell method. This solution includes key calculus concepts, detailed steps for integration, and application of the quadratic formula. Ideal for students in Grades 11-12 and college calculus courses.

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