Answer
It is true. The surface area and volume of the solid obtained by rotating this curve about the x-axis are equal for any interval $a \leq x \leq b$.
Work Step by Step
Let's consider the curve $y = \frac{e^x}{2} + \frac{e^{-x}}{2}$ and an interval $a \leq x \leq b$. The surface area of the solid obtained by rotating this curve about the x-axis can be found using the formula for the surface area of a surface of revolution. The formula is $A = 2\pi \int_{a}^{b} y \sqrt{1 + (\frac{dy}{dx})^2} dx$, where $y$ is a function of $x$ and the surface is obtained by rotating the curve $y = f(x)$ about the x-axis from $x = a$ to $x = b$. In this case, we can express the derivative of $y$ with respect to $x$ as $\frac{dy}{dx} = \frac{e^x}{2} - \frac{e^{-x}}{2}$. Substituting these values into the formula for the surface area of a surface of revolution, we get:
$$A = 2\pi \int_{a}^{b} (\frac{e^x}{2} + \frac{e^{-x}}{2}) \sqrt{1 + (\frac{e^x}{2} - \frac{e^{-x}}{2})^2} dx$$
Now let's consider the volume of the solid obtained by rotating this curve about the x-axis. The volume can be found using the disk method. The formula for the volume of a solid of revolution using the disk method is $V = \pi \int_{a}^{b} [f(x)]^2 dx$, where $f(x)$ is the function being rotated about the x-axis from $x = a$ to $x = b$. In this case, we have:
$$V = \pi \int_{a}^{b} (\frac{e^x}{2} + \frac{e^{-x}}{2})^2 dx$$
$$= \pi \int_{a}^{b} (\frac{e^{2x}}{4} + e^x e^{-x} + \frac{e^{-2x}}{4}) dx$$
$$= \pi \int_{a}^{b} (\frac{e^{2x}}{4} + 1 + \frac{e^{-2x}}{4}) dx$$
$$= \pi [\frac{e^{2x}}{8} + x -\frac{e^{-2x}}{8}]_{a}^{b}$$
$$= 2\pi [\frac{(e^{b}-e^{-b})}{4}] - 2\pi [\frac{(e^{a}-e^{-a})}{4}] +\pi(b-a) $$
$$= 2\pi [\int_{a}^{b}\frac{(e^{x}-e^{-x})}{4}]dx+\pi(b-a) $$
$$= 2\pi [\int_{a}^{b}\sinh(x)]dx+\pi(b-a) $$
Now we can see that both expressions for the surface area and volume contain an integral with $\sinh(x)$ as its integrand. Therefore, if we rotate the curve $y =\frac{e^x}{2} + \frac{e^{-x}}{2}$ about the x-axis, the area of the resulting surface is indeed equal to the enclosed volume for any interval $a \leq x \leq b$
