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 $\frac{\pi }{5}\le x\le \frac{\pi }{4}$; about the $x$-axis
a. Choose the correct answer below.
$2\pi {\int }_{\pi /5}^{\pi /4}\tan x\sqrt{1+{\sec }^{4}x}dx$
${\int }_{\pi /5}^{\pi /4}\tan x\sqrt{1+{\sec }^{4}x}dx$
$2\pi {\int }_{\pi /5}^{\pi /4}x\sqrt{1+{\sec }^{4}x}dx$
$2\pi {\int }_{\pi /5}^{\pi /4}\tan x\sqrt{1+{\sec }^{2}x}dx$
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 $\frac{\pi }{5}\le x\le \frac{\pi }{4}$ is revolved about the x-axis.

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

Step 3 ：In this case, $f(x)=\tan x$ and $f^{\prime }(x)={\sec }^{2}x$. 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.