To find the volume of the solid generated by rotating the region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 9$ about the line $x = 9$, we use the shell method. Here's how:

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Step 1: Understand the geometry

1. The curves $y = \sqrt{x}$, $y = 0$, and $x = 9$ form a region in the $xy$-plane.
2. The axis of rotation is the vertical line $x = 9$.

The shell method formula for volume is: $V = 2\pi \int_{a}^{b} \text{(radius)} \times \text{(height)} \, dx$

• Radius: The distance from the axis of rotation ($x = 9$) to a point on the curve is $9 - x$.
• Height: The height of the shell is the value of $y = \sqrt{x}$.

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Step 2: Set up the integral

The region lies from $x = 0$ to $x = 9$. The volume is: $V = 2\pi \int_{0}^{9} (9 - x)(\sqrt{x}) \, dx$

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Step 3: Simplify the integrand

Expand the integrand: $(9 - x)(\sqrt{x}) = 9\sqrt{x} - x\sqrt{x}$

Express the terms: $9\sqrt{x} = 9x^{1/2}, \quad x\sqrt{x} = x^{3/2}$

Thus: $V = 2\pi \int_{0}^{9} \left( 9x^{1/2} - x^{3/2} \right) dx$

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Step 4: Compute the integral

1. Integrate $9x^{1/2}$: $\int 9x^{1/2} \, dx = 9 \cdot \frac{x^{3/2}}{3/2} = 6x^{3/2}$

2. Integrate $x^{3/2}$: $\int x^{3/2} \, dx = \frac{x^{5/2}}{5/2} = \frac{2}{5}x^{5/2}$

So: $\int_{0}^{9} \left( 9x^{1/2} - x^{3/2} \right) dx = \left[ 6x^{3/2} - \frac{2}{5}x^{5/2} \right]_{0}^{9}$

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Step 5: Evaluate at bounds

1. At $x = 9$:

   • $6x^{3/2} = 6 \cdot (9^{3/2}) = 6 \cdot 27 = 162$
   • $\frac{2}{5}x^{5/2} = \frac{2}{5} \cdot (9^{5/2}) = \frac{2}{5} \cdot 243 = \frac{486}{5}$

   So: $6x^{3/2} - \frac{2}{5}x^{5/2} = 162 - \frac{486}{5} = \frac{810 - 486}{5} = \frac{324}{5}$

2. At $x = 0$:

   • Both $6x^{3/2}$ and $\frac{2}{5}x^{5/2}$ are 0.

Thus: $\int_{0}^{9} \left( 9x^{1/2} - x^{3/2} \right) dx = \frac{324}{5}$

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Step 6: Multiply by $2\pi$

$V = 2\pi \cdot \frac{324}{5} = \frac{648\pi}{5}$

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Final Answer:

$\boxed{\frac{648\pi}{5}}$

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Would you like a detailed step-by-step breakdown of any part of this solution or additional examples?

Here are 5 related questions for practice:

1. How would the volume change if the region were rotated about the $y$-axis?
2. What is the volume of the same region rotated about the line $x = -1$?
3. Can you calculate the volume of the region rotated about $y = 0$?
4. How do you solve a similar problem using the disk/washer method instead?
5. What is the relationship between the shell and washer methods?

Tip: Always sketch the region and axis of rotation to ensure your setup aligns with the geometry of the problem.