A torus is formed when a circle of radius 5 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). (Hint: Both integrals can be evaluated without using the Fundamental Theorem of Calculus.)

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### Problem 3c: Find the Volume of the Torus

To find the volume of the torus formed by revolving the circle of radius 5 centered at $$(6, 0)$$ about the $$y$$-axis, we can use the **shell method** integral derived in part (a):

$$
V = 4\pi \int_{1}^{11} x \sqrt{25 - (x - 6)^2} \, dx
$$

However, evaluating this integral directly can be complex. Instead, we'll use a geometric approach based on the known formula for the volume of a torus.

### Geometric Approach

A torus generated by revolving a circle with radius $$r$$ around an axis at a distance $$R$$ from the center of the circle has a volume given by:

$$
V = 2\pi^2 R r^2
$$

In our case:
- The radius of the circle being revolved ($$r$$) is 5.
- The distance from the center of the circle to the $$y$$-axis ($$R$$) is 6.

Plugging these values into the formula:

$$
V = 2\pi^2 \times 6 \times 5^2 = 2\pi^2 \times 6 \times 25 = 300\pi^2
$$

### Final Answer

**c.** The volume of the torus is $$300\pi^2$$ cubic units.

Answer: Problem c Answer

Using the shell method, the volume is 4 π times the integral from 1 to 11 of x √(25 – (x – 6)²) dx. Thus,

V = 4π ∫₁¹¹ x √[25 − (x−6)^2] dx

which evaluates to 300 π².
**c.** The volume of the torus is $$300\pi^2$$ cubic units.

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