To find the surface area of the solid generated by revolving the region bounded by the curves $y = x^2$, $y = 0$, $x = 0$, and $x = \sqrt{7}$ about the x-axis, we use the formula for the surface area of a solid of revolution about the x-axis:

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

Here, $y = x^2$, so we have the following steps:

1. Derivative of $y$:
   $\frac{dy}{dx} = 2x$

2. Substitute into the formula:
   The surface area formula becomes: $S = 2\pi \int_{0}^{\sqrt{7}} x^2 \sqrt{1 + (2x)^2} \, dx$ Simplify: $S = 2\pi \int_{0}^{\sqrt{7}} x^2 \sqrt{1 + 4x^2} \, dx$

Next, I will compute this integral and round the answer to four decimal places.

Let me calculate this for you.The surface area of the solid generated by revolving the region bounded by the curves $y = x^2$, $y = 0$, $x = 0$, and $x = \sqrt{7}$ about the x-axis is approximately 159.2 square units.

Would you like more details on the calculation or have any questions?

Here are 5 related questions to further explore:

1. How would the surface area change if the region were rotated about the y-axis?
2. What is the volume of the solid generated by the same region when revolved around the x-axis?
3. How does the surface area formula differ when revolving around a line other than the x-axis?
4. Can the same method be applied to find the surface area when the function is more complex?
5. How does the surface area change if the limits of integration are modified?

Tip: When finding the surface area of a solid of revolution, always double-check that the function and its derivative are correctly squared and simplified in the formula.