a. Write the integral that gives the area of the surface generated when the curve is revolved about the given axis.
b. Use a calculator or software to approximate the surface area.
$y=\tan x$, for $\frac{\pi}{5} \leq x \leq \frac{\pi}{4}$; about the $x$-axis
a. Choose the correct answer below.
$2 \pi \int_{\pi / 5}^{\pi / 4} \tan x \sqrt{1+\sec ^{4} x} d x$
$\int_{\pi / 5}^{\pi / 4} \tan x \sqrt{1+\sec ^{4} x} d x$
$2 \pi \int_{\pi / 5}^{\pi / 4} x \sqrt{1+\sec ^{4} x} d x$
$2 \pi \int_{\pi / 5}^{\pi / 4} \tan x \sqrt{1+\sec ^{2} x} d x$
b. The area of the surface is square units.
(Do not round until the final answer. Then round to the nearest hundredth as needed.)

Step 1 ：First, we need to understand the problem. We are asked to find the surface area generated when the curve $y=\tan x$ for $\frac{\pi}{5} \leq x \leq \frac{\pi}{4}$ is revolved about the x-axis.

Step 2 ：The formula for the surface area of a solid of revolution is $2\pi \int_{a}^{b} f(x) \sqrt{1+[f'(x)]^2} dx$, where $f(x)$ is the function being revolved, and $f'(x)$ is its derivative.

Step 3 ：In this case, $f(x) = \tan x$ and $f'(x) = \sec^2 x$. So, the integral becomes $2\pi \int_{\pi / 5}^{\pi / 4} \tan x \sqrt{1+\sec ^{4} x} dx$.

Step 4 ：So, the correct answer for part a is $2\pi \int_{\pi / 5}^{\pi / 4} \tan x \sqrt{1+\sec ^{4} x} dx$.

Step 5 ：For part b, we need to evaluate the integral. This can be done using a calculator or software, and the result should be rounded to the nearest hundredth.