Consider the curve $f(x)=10 x^{7}$ over the interval $0 \leq x \leq 1$. Use a calculator to find the surface area generated by revolving the graph of $f(x)$ about the $x$-axis. Round your answer to the nearest hundredth.

Step 1 ：We are given the function \(f(x) = 10x^7\) and we are asked to find the surface area generated by revolving the graph of this function about the x-axis over the interval \(0 \leq x \leq 1\).

Step 2 ：The formula for the surface area \(A\) of a solid of revolution generated by revolving a curve \(y=f(x)\), \(a \leq x \leq b\) about the x-axis is given by: \[A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + [f'(x)]^2} dx\]

Step 3 ：First, we need to find the derivative of \(f(x)\), denoted as \(f'(x)\). For \(f(x) = 10x^7\), the derivative \(f'(x) = 70x^6\).

Step 4 ：Substitute \(f(x)\) and \(f'(x)\) into the formula and evaluate the integral, we get: \[A = 5\pi\gamma(2/3)\hyper((-1/2, 2/3), (5/3,), 4900\exp_polar(I\pi))/(3\gamma(5/3))\]

Step 5 ：Evaluating this expression, we find that \(A \approx 314.343419899690\).

Step 6 ：Rounding to the nearest hundredth, we get \(A \approx 314.34\).

Step 7 ：Final Answer: The surface area generated by revolving the graph of \(f(x)=10 x^{7}\) about the x-axis over the interval \(0 \leq x \leq 1\) is approximately \(\boxed{314.34}\).