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Assume that  𝑓(𝑥)≥0
  and that  𝑓′
  exists and is continuous on the interval  [𝑎,𝑏].
  The surface area  𝑆
  of the surface obtained by rotating the graph of  𝑓
  about the  𝑥-
 axis for  𝑎≤𝑥≤𝑏
  is equal to
𝑆=2𝜋∫𝑏𝑎𝑓(𝑥)1+𝑓′(𝑥)2‾‾‾‾‾‾‾‾‾‾√𝑑𝑥
 
Find  𝑓′(𝑥)
  using the Power Rule for derivatives.

Substitute  𝑎,
   𝑏,
   𝑓,
  and  𝑓′
  into the surface area formula and simplify.

Evaluate the integral with the Substitution Method.

Let's carefully go through the steps again to correctly compute the surface area of the surface of revolution for the function \( y = 3x^3 \) about the x-axis over the interval \([0, 7]\).

1. **Find \( f(x) \) and \( f'(x) \):**
   \[ f(x) = 3x^3 \]
   \[ f'(x) = 9x^2 \]

2. **Substitute \( f(x) \) and \( f'(x) \) into the surface area formula:**
   \[ S = 2\pi \int_{0}^{7} 3x^3 \sqrt{1 + (9x^2)^2} \, dx \]
   \[ S = 2\pi \int_{0}^{7} 3x^3 \sqrt{1 + 81x^4} \, dx \]

3. **Simplify the integrand:**
   \[ S = 6\pi \int_{0}^{7} x^3 \sqrt{1 + 81x^4} \, dx \]

4. **Use substitution to evaluate the integral:**
   Let \( u = 1 + 81x^4 \). Then,
   \[ du = 324x^3 \, dx \]
   \[ x^3 \, dx = \frac{du}{324} \]

   Change the limits of integration:
   When \( x = 0 \), \( u = 1 \).
   When \( x = 7 \), \( u = 1 + 81(7)^4 \).

   Calculate \( 81(7)^4 \):
   \[ 81 \cdot 7^4 = 81 \cdot 2401 = 194481 \]
   So, \( u = 1 + 194481 = 194482 \).

5. **Rewrite the integral in terms of \( u \):**
   \[ S = 6\pi \int_{1}^{194482} \sqrt{u} \cdot \frac{du}{324} \]
   \[ S = \frac{6\pi}{324} \int_{1}^{194482} u^{1/2} \, du \]
   \[ S = \frac{\pi}{54} \int_{1}^{194482} u^{1/2} \, du \]

6. **Evaluate the integral:**
   \[ \int u^{1/2} \, du = \frac{2}{3} u^{3/2} \]

   Apply the limits:
   \[ S = \frac{\pi}{54} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{194482} \]
   \[ S = \frac{\pi}{54} \cdot \frac{2}{3} \left( 194482^{3/2} - 1^{3/2} \right) \]
   \[ S = \frac{\pi}{81} \left( 194482^{3/2} - 1 \right) \]

7. **Simplify the result:**
   \[ 194482^{3/2} = (194482^{1/2})^3 = (441.016)^3 \approx 85766161.2 \]
   So,
   \[ S = \frac{\pi}{81} \left( 85766161.2 - 1 \right) \]
   \[ S = \frac{\pi}{81} \cdot 85766160.2 \]
   \[ S \approx 1058551.6\pi \]

Thus, the correct surface area \( S \) is approximately \( 1058551.6\pi \).

Do you have any questions or need further details?

Here are some questions you might find interesting:

1. What are the steps to find the derivative of a polynomial function?
2. How do you determine the limits of integration after a substitution?
3. Can you explain the process of simplifying the integrand before integration?
4. What are common mistakes to avoid when applying the substitution method?
5. How can we check if our solution is correct?

Tip: When using the substitution method, carefully transform both the integrand and the limits of integration to avoid errors.

Learn how to calculate the surface area of revolution for a function rotated about the x-axis. This example involves finding f(x) = 3x^3, f'(x), integrating using substitution, and evaluating the integral to find the surface area. Suitable for college-level students.

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