Answer is: option3

Solution:

Step 1: Identify the Bounds

• The region is in the first quadrant, so \( x \) ranges from 0 to \( \frac{\pi}{3} \).
• The lower bound for \( y \) is 0 (the x-axis), and the upper bound is \( y = \sec x \).

Step 2: Set Up the Integral

The volume \( V \) of the solid of revolution is given by:

\[ V = \pi \int_a^b \left[ f(x) \right]^2 dx \]

Here, \( f(x) = \sec x \), \( a = 0 \), and \( b = \frac{\pi}{3} \). So:

\[ V = \pi \int_0^{\frac{\pi}{3}} \sec^2 x \, dx \]

Step 3: Compute the Integral

The integral of \( \sec^2 x \) is \( \tan x \):

\[ \int \sec^2 x \, dx = \tan x + C \]

Evaluate from 0 to \( \frac{\pi}{3} \):

\[ V = \pi \left[ \tan \left( \frac{\pi}{3} \right) - \tan(0) \right] \]

\[ V = \pi \left[ \sqrt{3} - 0 \right] = \sqrt{3}\pi \]

Step 4: Match the Answer

The volume is \( \sqrt{3}\pi \), which corresponds to option C.