Write an integral for the area of the surface generated by revolving the curve $$ y=\cos (3 x) $$ about the $$ x $$-axis on $$ -\frac{\pi}{7} \leq x \leq \frac{\pi}{7} $$. An integral is $$ \square $$. (Type an exact answer, using $$ \pi $$ as needed.)

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To find the area of the surface generated by revolving the curve $$ y = \cos(3x) $$ about the $$ x $$-axis from $$ x = -\frac{\pi}{7} $$ to $$ x = \frac{\pi}{7} $$, we can use the formula for the surface area $$ S $$ of a solid of revolution:

$$
S = 2\pi \int_a^b y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
$$

where $$ y = \cos(3x) $$, $$ a = -\frac{\pi}{7} $$, and $$ b = \frac{\pi}{7} $$.

1. First, we need to find $$ \frac{dy}{dx} $$:
   $$
   y = \cos(3x) \implies \frac{dy}{dx} = -3\sin(3x)
   $$

2. Next, we calculate $$ \left( \frac{dy}{dx} \right)^2 $$:
   $$
   \left( \frac{dy}{dx} \right)^2 = (-3\sin(3x))^2 = 9\sin^2(3x)
   $$

3. Now, we can substitute $$ y $$ and $$ \left( \frac{dy}{dx} \right)^2 $$ into the surface area formula:
   $$
   S = 2\pi \int_{-\frac{\pi}{7}}^{\frac{\pi}{7}} \cos(3x) \sqrt{1 + 9\sin^2(3x)} \, dx
   $$

Thus, the integral for the area of the surface generated by revolving the curve about the $$ x $$-axis is:

$$
\boxed{2\pi \int_{-\frac{\pi}{7}}^{\frac{\pi}{7}} \cos(3x) \sqrt{1 + 9\sin^2(3x)} \, dx}
$$
The integral for the area is $$ 2\pi \int_{-\frac{\pi}{7}}^{\frac{\pi}{7}} \cos(3x) \sqrt{1 + 9\sin^2(3x)} \, dx $$.

Write an integral for the area of the surface generated by revolving the curve \( y=\cos (3 x) \) about the \( x \)-axis on \( -\frac{\pi}{7} \leq x \leq \frac{\pi}{7} \). An integral is \( \square \). (Type an exact answer, using \( \pi \) as needed.)

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