Question
Let $f(x)=-\frac{x^{3}}{2}$ over the interval $[-1,0]$. Find the surface area of the surface generated by revolving the graph of $f(x)$ around the $x$-axis.
Enter an exact answer.

Step 1 ：We are given the function \(f(x) = -\frac{x^{3}}{2}\) and we are asked to find the surface area of the surface generated by revolving the graph of this function around the x-axis over the interval [-1, 0].

Step 2 ：The formula for the surface area of a solid of revolution is \(A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2} dx\), where \(f(x)\) is the function being revolved, \(f'(x)\) is its derivative, and [a, b] is the interval over which the function is being revolved.

Step 3 ：In this case, \(f(x) = -\frac{x^{3}}{2}\), \(f'(x) = -\frac{3x^{2}}{2}\), and the interval is [-1, 0].

Step 4 ：So, we need to compute the integral \(A = 2\pi \int_{-1}^{0} -\frac{x^{3}}{2} \sqrt{1 + \left(-\frac{3x^{2}}{2}\right)^2} dx\).

Step 5 ：We can simplify the integrand a bit by noting that \(-\frac{x^{3}}{2} \geq 0\) for \(x \in [-1, 0]\), so we can take the absolute value inside the square root: \(A = 2\pi \int_{-1}^{0} -\frac{x^{3}}{2} \sqrt{1 + \left(\frac{3x^{2}}{2}\right)^2} dx\).

Step 6 ：Now we can compute this integral to find the surface area.

Step 7 ：The final answer is \(\boxed{\frac{2\pi}{27} - \frac{13\pi\sqrt{13}}{216}}\).