Consider the curve \( f(x)=x^{2} \) over the interval \( 0 \leq x \leq 4 \). Use a calculator to find the surface area generated by revolv the graph of \( f(x) \) about the \( x \)-axis. Round your answer to the nearest hundredth.

It seems that there was an issue with the calculation. Let's try to calculate the integral manually. The integral to be calculated is: \[ S = \int_{0}^{4} 2\pi x^2 \sqrt{1 + (2x)^2} dx \] To solve this integral, we can use the substitution method. Let's substitute \( u = 1 + (2x)^2 \), which means \( du = 4x \cdot 2x \, dx = 8x^2 \, dx \). The integral becomes: \[ S = \frac{1}{4} \int_{1}^{17} 2\pi u^{\frac{1}{2}} \, du \] Now, we can calculate the integral: \[ S = \frac{1}{4} \cdot 2\pi \cdot \left[ \frac{2}{3} u^{\frac{3}{2}} \right]_{1}^{17} \] \[ S = \frac{1}{3} \pi \cdot \left[ 17^{\frac{3}{2}} - 1 \right] \] \[ S = \frac{1}{3} \pi \cdot \left[ 17\sqrt{17} - 1 \right] \] Now, we can calculate the numerical value of \( S \) using a calculator. Calculate the value by following steps: - step0: Calculate: \(\frac{1}{3}\pi \left(17\sqrt{17}-1\right)\) - step1: Multiply: \(\frac{\pi }{3}\left(17\sqrt{17}-1\right)\) - step2: Multiply: \(\frac{\pi \left(17\sqrt{17}-1\right)}{3}\) - step3: Multiply: \(\frac{17\sqrt{17}\times \pi -\pi }{3}\) The surface area generated by revolving the graph of \( f(x) = x^2 \) about the \( x \)-axis over the interval \( 0 \leq x \leq 4 \) is approximately 72.35 square units when rounded to the nearest hundredth.