To find the surface area of revolution of the curve $y = 7\sqrt{x}$ 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-by-step solution:

1. Given function: $y = 7\sqrt{x}$ We need to compute $\frac{dy}{dx}$ (the derivative of $y$ with respect to $x$).

2. Derivative $\frac{dy}{dx}$: $y = 7x^{1/2}$ Using the power rule for differentiation: $\frac{dy}{dx} = 7 \times \frac{1}{2}x^{-1/2} = \frac{7}{2\sqrt{x}}$

3. Surface Area formula: Now, substitute $y = 7\sqrt{x}$ and $\frac{dy}{dx} = \frac{7}{2\sqrt{x}}$ into the surface area formula: $S = 2\pi \int_{1.2}^{3.2} 7\sqrt{x} \sqrt{1 + \left( \frac{7}{2\sqrt{x}} \right)^2} \, dx$

4. Simplifying inside the square root: $1 + \left( \frac{7}{2\sqrt{x}} \right)^2 = 1 + \frac{49}{4x}$ So, the integral becomes: $S = 2\pi \int_{1.2}^{3.2} 7\sqrt{x} \sqrt{1 + \frac{49}{4x}} \, dx$

This integral is a bit complex and is best evaluated using numerical methods.

I can compute this definite integral to find the surface area numerically. Let's proceed with that.The surface area of revolution of the curve $y = 7\sqrt{x}$ about the x-axis, for $1.2 < x < 3.2$, is approximately 334.31 square units.

Would you like further details on any step or have any questions?

Here are some related questions for practice:

1. What is the surface area of revolution for the curve $y = x^2$ about the x-axis between $x = 1$ and $x = 4$?
2. How would the surface area change if the curve is rotated about the y-axis instead?
3. Can you compute the surface area of revolution for $y = \frac{1}{x}$ between $x = 2$ and $x = 5$?
4. What is the arc length of $y = 7\sqrt{x}$ between $x = 1.2$ and $x = 3.2$?
5. What happens to the surface area if the curve is $y = 7\sqrt{x+1}$?

Tip: When integrating functions with square roots, double-check the algebraic manipulations under the square root for accuracy.