Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the line x = 6. y=sqrt(x) x=4 y=0

We are tasked with finding the volume of the solid generated by revolving the region bounded by \( y = \sqrt{x} \), \( x = 4 \), and \( y = 0 \) about the line \( x = 6 \). The most appropriate method here is the **shell method** because we are revolving around a vertical line that is not the y-axis.

### Step-by-Step Solution:

#### 1. **General Formula for Shell Method**
For revolving around a vertical line, the formula for the shell method is:

\[
V = 2\pi \int_{a}^{b} \text{(radius)} \times \text{(height)} \, dx
\]

#### 2. **Radius of a Typical Shell**
The radius of a shell is the horizontal distance from the line of rotation \( x = 6 \) to a point on the curve \( x = x \). Thus, the radius is:

\[
\text{radius} = 6 - x
\]

#### 3. **Height of a Typical Shell**
The height of the shell is given by the function \( y = \sqrt{x} \). So the height is:

\[
\text{height} = \sqrt{x}
\]

#### 4. **Limits of Integration**
We are integrating along the x-axis from \( x = 0 \) to \( x = 4 \), as these are the bounds of the region.

Thus, the volume integral is:

\[
V = 2\pi \int_{0}^{4} (6 - x) \cdot \sqrt{x} \, dx
\]

#### 5. **Simplifying the Integral**
We need to expand the integrand before integrating:

\[
V = 2\pi \int_{0}^{4} (6\sqrt{x} - x\sqrt{x}) \, dx
\]

This can be broken down into two terms:

\[
V = 2\pi \left( \int_{0}^{4} 6\sqrt{x} \, dx - \int_{0}^{4} x\sqrt{x} \, dx \right)
\]

Now, let's solve these two integrals separately.

#### 6. **First Integral**
The first term involves \( 6\sqrt{x} = 6x^{1/2} \):

\[
\int 6x^{1/2} \, dx = 6 \cdot \frac{x^{3/2}}{3/2} = 4x^{3/2}
\]

Evaluating from \( 0 \) to \( 4 \):

\[
4(4)^{3/2} - 4(0)^{3/2} = 4 \cdot 8 = 32
\]

#### 7. **Second Integral**
The second term involves \( x\sqrt{x} = x^{3/2} \):

\[
\int x^{3/2} \, dx = \frac{x^{5/2}}{5/2} = \frac{2x^{5/2}}{5}
\]

Evaluating from \( 0 \) to \( 4 \):

\[
\frac{2(4)^{5/2}}{5} - \frac{2(0)^{5/2}}{5} = \frac{2 \cdot 32}{5} = \frac{64}{5}
\]

#### 8. **Final Volume Calculation**
Now substitute these values back into the volume expression:

\[
V = 2\pi \left( 32 - \frac{64}{5} \right)
\]

\[
V = 2\pi \left( \frac{160}{5} - \frac{64}{5} \right) = 2\pi \cdot \frac{96}{5}
\]

\[
V = \frac{192\pi}{5}
\]

Thus, the volume of the solid is:

\[
V = \frac{192\pi}{5} \, \text{cubic units}
\]

### Would you like any more details or explanations? 

Here are some related questions you might find helpful:

1. How does the shell method differ from the disk/washer method?
2. What is the advantage of using the shell method in this problem?
3. How would the setup change if we revolved the region about the y-axis instead of \( x = 6 \)?
4. Can you set up the same volume using the washer method?
5. How would the limits of integration change if the region were bounded by different lines?

**Tip:** When choosing between the shell method and the washer method, consider the axis of rotation. If the axis is parallel to the y-axis, the shell method is often simpler.

Explore how to calculate the volume of the solid generated by revolving the region bounded by y=sqrt(x), x=4, and y=0 around the line x=6 using the shell method. This detailed solution includes step-by-step integration with a breakdown of key calculus concepts, such as the shell method, and appropriate formulas.

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