To find the exact surface area generated by rotating the curve $y = \sqrt{x^2 + 1}$ from $x = 0$ to $x = 6$ about the x-axis, we can use the formula for the surface area of revolution:

$S = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Step 1: Differentiate $y = \sqrt{x^2 + 1}$

We first need to find the derivative of $y = \sqrt{x^2 + 1}$.

$\frac{dy}{dx} = \frac{d}{dx} \left( \sqrt{x^2 + 1} \right) = \frac{2x}{2\sqrt{x^2 + 1}} = \frac{x}{\sqrt{x^2 + 1}}$

Step 2: Set up the integral

Now, substitute $y = \sqrt{x^2 + 1}$ and $\frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}}$ into the formula for the surface area.

$S = 2\pi \int_{0}^{6} \sqrt{x^2 + 1} \sqrt{1 + \left( \frac{x}{\sqrt{x^2 + 1}} \right)^2} \, dx$

Simplifying the expression under the square root:

$1 + \left( \frac{x}{\sqrt{x^2 + 1}} \right)^2 = 1 + \frac{x^2}{x^2 + 1} = \frac{x^2 + 1 + x^2}{x^2 + 1} = \frac{2x^2 + 1}{x^2 + 1}$

Thus, the integral becomes:

$S = 2\pi \int_{0}^{6} \sqrt{x^2 + 1} \cdot \sqrt{\frac{2x^2 + 1}{x^2 + 1}} \, dx = 2\pi \int_{0}^{6} \sqrt{2x^2 + 1} \, dx$

Step 3: Solve the integral

Now we need to evaluate the integral $\int \sqrt{2x^2 + 1} \, dx$. This integral can be computed using a substitution method or a table of integrals. Let’s solve this step:

Let $u = 2x^2 + 1$, then $du = 4x dx$, leading to:

$S = 2\pi \cdot \left[ \frac{1}{6} (2x^2 + 1)^{3/2} \right]_{0}^{6}$

Substituting the limits:

$S = 2\pi \left[ \frac{1}{6} \left( 2(6^2) + 1 \right)^{3/2} - \frac{1}{6} (2(0^2) + 1)^{3/2} \right]$

Simplifying:

$S = 2\pi \left[ \frac{1}{6} \left( 73^{3/2} - 1^{3/2} \right) \right]$

Final answer:

The exact surface area is:

$S = \frac{\pi}{3} \left( 73^{3/2} - 1 \right)$

Would you like more details on any of the steps? Here are some related questions:

1. How do you find the surface area when rotating around the y-axis?
2. Can you use numerical methods to approximate this integral?
3. How does this method change if the function is rotated about a different axis?
4. What other applications are there for surface area of revolution in physics or engineering?
5. How do different functions affect the complexity of surface area integrals?

Tip: Always double-check the function's derivative carefully before applying it in surface area calculations.