We are tasked with finding the volume of the solid generated by revolving the region $R$ about the $x$-axis using the shell method.

Step 1: Understanding the Problem

The region $R$ is bounded by:

• $y = \sqrt{x}$ (the curve),
• $y = 0$ (the $x$-axis),
• $x = 9$ (a vertical line at $x = 9$).

We are rotating this region around the $x$-axis.

Step 2: The Shell Method Formula

To use the shell method, we consider vertical shells, and the volume of each shell is given by the formula: $V = 2 \pi \int_{a}^{b} \text{(radius)} \times \text{(height)} \, dx$ Where:

• The radius is the distance from the axis of rotation (in this case, the $x$-axis), which is simply $y$.
• The height of the shell is the value of the function, which in this case is $\sqrt{x}$.

Step 3: Setting up the Integral

The shell method requires an integral in terms of $x$. For this setup:

• The limits of integration will range from $x = 0$ to $x = 9$ (the bounds of the region $R$).
• The radius is $y$, which is $\sqrt{x}$.
• The height of each shell is simply the function value, which is $\sqrt{x}$.

Thus, the integral for the volume is: $V = 2\pi \int_{0}^{9} x \cdot \sqrt{x} \, dx$

Step 4: Answer Choices

• Part 1: The integral is set up as $\int_{0}^{9} x \cdot \sqrt{x} \, dx$, so the correct answer is in the form:

  A. Integral from 0 to 9 of $x \cdot \sqrt{x} \, dx$

• Part 2: To compute the volume, we first simplify the integrand: $x \cdot \sqrt{x} = x^{3/2}$ Thus, the integral becomes: $V = 2 \pi \int_{0}^{9} x^{3/2} \, dx$

  Now, integrate $x^{3/2}$: $\int x^{3/2} \, dx = \frac{2}{5} x^{5/2}$ Applying the limits from $0$ to $9$: $V = 2 \pi \left[ \frac{2}{5} x^{5/2} \right]_{0}^{9}$ Evaluating the definite integral: $V = 2 \pi \left( \frac{2}{5} \left(9^{5/2} - 0^{5/2}\right) \right)$ Since $9^{5/2} = (3^2)^{5/2} = 3^5 = 243$, we get: $V = 2 \pi \times \frac{2}{5} \times 243 = \frac{4 \pi \times 243}{5} = \frac{972 \pi}{5}$

Final Answer:

• The integral for the volume is $\int_{0}^{9} x \cdot \sqrt{x} \, dx$.
• The volume is $\frac{972\pi}{5}$.

Let me know if you need any more details or have any other questions!

Related Questions:

1. How do we compute the volume using the disk method instead of the shell method?
2. What would happen if we revolved the region around the $y$-axis instead?
3. How can we apply the shell method to a region bounded by more complex curves?
4. What is the geometric interpretation of the radius in the shell method?
5. Can we use the shell method for regions that are rotated about other lines, such as $y = 3$?

Tip: Always check if the function is easier to handle with shells (vertical slices) or disks (horizontal slices) depending on the axis of rotation.