Answer
If the curve $y = f(x)$, $a \leq x \leq b$, is rotated about the horizontal line $y = c$, where $f(x) < c$, the surface area of the resulting solid of revolution can be calculated using the formula: $S = 2\pi \int_{a}^{b} (c - f(x)) \sqrt{1 + \left(f'(x)\right)^2} dx$.
Work Step by Step
\begin{align*}
S &= 2\pi \int_a^b (c - f(x)) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx \
&= 2\pi \int_a^b (c - f(x)) \sqrt{1 + \left(\frac{df}{dx}\right)^2} , dx
\end{align*}
where $\frac{df}{dx}$ represents the derivative of $f(x)$.
Note that the function $c - f(x)$ represents the distance between the curve $y=f(x)$ and the line $y=c$ at any given point. We then multiply this distance by the arc length element $\sqrt{1 + \left(\frac{df}{dx}\right)^2}$ to get the surface area of the small element that is rotated about the line $y=c$.
We can then evaluate this integral using standard techniques of integration to find the area of the resulting surface.
