A torus is formed when a circle of radius 3 centered at left parenthesis 6 comma 0 right parenthesis is revolved about the​ y-axis.
a. Use the shell method to write an integral for the volume of the torus. 
b. Use the washer method to write an integral for the volume of the torus. 
c. Find the volume of the torus by evaluating one of the two integrals obtained in parts​ (a) and​ (b). (Hint: Both integrals can be evaluated without using the Fundamental Theorem of​ Calculus.)

We are tasked with finding the volume of a torus formed when a circle of radius 3, centered at \( (6, 0) \), is revolved about the \( y \)-axis. We'll use both the shell and washer methods, and finally, we'll compute the volume by evaluating one of these integrals.

### Part (a): Using the Shell Method

The **shell method** is useful when the object is revolved around a vertical axis (here, the \( y \)-axis).

1. **Equation of the circle**:  
   The equation of the circle with radius 3 centered at \( (6, 0) \) is:
   \[
   (x - 6)^2 + y^2 = 9
   \]
   Solving for \( y \), we get:
   \[
   y = \pm \sqrt{9 - (x - 6)^2}
   \]
   
2. **Shell radius and height**:
   - The **radius** of the shell at any \( x \) is the distance from the \( y \)-axis, which is \( x \).
   - The **height** of the shell at any \( x \) is the vertical distance between the top and bottom of the circle, which is \( 2y = 2\sqrt{9 - (x - 6)^2} \).

3. **Integral for volume**:  
   The volume of the torus is given by integrating the volume of the shells. The limits for \( x \) range from \( 6 - 3 = 3 \) to \( 6 + 3 = 9 \), because the circle has a radius of 3 and is centered at \( x = 6 \). Using the shell method formula, the volume is:
   \[
   V = 2\pi \int_{3}^{9} x \cdot 2\sqrt{9 - (x - 6)^2} \, dx
   \]
   Simplifying:
   \[
   V = 4\pi \int_{3}^{9} x \sqrt{9 - (x - 6)^2} \, dx
   \]

### Part (b): Using the Washer Method

The **washer method** works well for solids of revolution when the object is revolved around a vertical axis, like the \( y \)-axis.

1. **Washer radii**:  
   The volume of the torus can be found by revolving the circle around the \( y \)-axis and subtracting the area of the inner hole from the outer area.

   For a slice at height \( y \), the outer radius is the distance from the \( y \)-axis to the right edge of the circle, \( x = 6 + \sqrt{9 - y^2} \), and the inner radius is the distance to the left edge, \( x = 6 - \sqrt{9 - y^2} \).

2. **Integral for volume**:  
   The volume is obtained by integrating the area of the washers from \( y = -3 \) to \( y = 3 \):
   \[
   V = \pi \int_{-3}^{3} \left[ (6 + \sqrt{9 - y^2})^2 - (6 - \sqrt{9 - y^2})^2 \right] \, dy
   \]
   Simplifying the expression inside the integral:
   \[
   V = \pi \int_{-3}^{3} \left( 24\sqrt{9 - y^2} \right) \, dy
   \]
   This further simplifies to:
   \[
   V = 24\pi \int_{-3}^{3} \sqrt{9 - y^2} \, dy
   \]

### Part (c): Finding the Volume

We'll now evaluate one of the integrals. The integral derived from the washer method is easier to solve, as it represents the area of a semicircle.

1. **Integral setup**:
   The integral \( \int_{-3}^{3} \sqrt{9 - y^2} \, dy \) represents the area of a semicircle with radius 3. The formula for the area of a semicircle is:
   \[
   A = \frac{\pi r^2}{2}
   \]
   where \( r = 3 \). Thus, the area is:
   \[
   A = \frac{\pi (3)^2}{2} = \frac{9\pi}{2}
   \]

2. **Final volume**:  
   Using this result in the washer method integral:
   \[
   V = 24\pi \times \frac{9\pi}{2} = 108\pi^2
   \]

Thus, the volume of the torus is \( 108\pi^2 \) cubic units.

---

Would you like further details on any part of this solution?

Here are some related questions:
1. What are the differences between the shell and washer methods in general?
2. How do you derive the area of a semicircle from an integral?
3. What is the geometric significance of the shell method in this context?
4. How would the problem change if the circle were revolved around the \( x \)-axis instead?
5. Can you find the volume of a torus using other methods like Pappus' theorem?

**Tip**: When revolving curves around an axis, visualize the setup to choose the most efficient method (shell or washer).

A torus is formed when a circle of radius 3 centered at (6, 0) is revolved about the y-axis. a. Use the shell method to write an integral for the volume of the torus. b. Use the washer method to write an integral for the volume of the torus. c. Find the volume of the torus by evaluating one of the two integrals obtained in parts (a) and (b).

This detailed guide demonstrates how to calculate the volume of a torus, formed by revolving a circle of radius 3 around the y-axis, using both the shell and washer methods. Includes step-by-step integrals and solutions, suitable for college calculus students.

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