Write an integral for the area of the surface generated by revolving the curve \( y=\cos (3 x) \) about the \( x \)-axis on \( -\frac{\pi}{9} \leq x \leq \frac{\pi}{9} \). An integral is \( \square \). (Type an exact answer, using \( \pi \) as needed.)

To find the area of the surface generated by revolving the curve \( y = \cos(3x) \) about the \( x \)-axis, we can use the formula for the surface area of revolution: \[ S = 2\pi \int_{a}^{b} y \sqrt{(y')^2 + 1} \, dx \] where \( y' \) is the derivative of \( y \) with respect to \( x \). First, we find the derivative: \[ y' = \frac{d}{dx} (\cos(3x)) = -3\sin(3x) \] Next, we can substitute into the formula. The limits of integration are \( -\frac{\pi}{9} \) and \( \frac{\pi}{9} \), and we have: \[ S = 2\pi \int_{-\frac{\pi}{9}}^{\frac{\pi}{9}} \cos(3x) \sqrt{(-3\sin(3x))^2 + 1} \, dx \] This simplifies to: \[ S = 2\pi \int_{-\frac{\pi}{9}}^{\frac{\pi}{9}} \cos(3x) \sqrt{9\sin^2(3x) + 1} \, dx \] So, the integral is: \[ 2\pi \int_{-\frac{\pi}{9}}^{\frac{\pi}{9}} \cos(3x) \sqrt{9\sin^2(3x) + 1} \, dx. \]