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 $A=2\pi {\int }_a^{b}f(x)\sqrt{1+(f^{\prime }(x){)}^2}dx$, where $f(x)$ is the function being revolved, $f^{\prime }(x)$ is its derivative, and [a, b] is the interval over which the function is being revolved.

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

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

Step 5 ：We can simplify the integrand a bit by noting that $-\frac{x^3}{2}\ge 0$ 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 $\boxed{{\displaystyle \frac{2\pi }{27}-\frac{13\pi \sqrt{13}}{216}}}$.