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To solve this problem, we need to determine the area of the region that lies inside \( r = \sqrt{3} \sin \theta \) and outside \( r = \cos \theta \).

### Step-by-Step Solution Outline:

1. **Identify the region of intersection**:
   - The equations \( r = \sqrt{3} \sin \theta \) and \( r = \cos \theta \) intersect where they have the same \( r \)-value.
   - Set \( \sqrt{3} \sin \theta = \cos \theta \).
   - This simplifies to \( \tan \theta = \frac{1}{\sqrt{3}} \), giving \( \theta = \frac{\pi}{6} \).

2. **Determine the limits for \( \theta \)**:
   - Since we are interested in the region inside \( r = \sqrt{3} \sin \theta \), which is a circle, and outside \( r = \cos \theta \), a limaçon, we focus on \( \theta \in \left[ \frac{\pi}{6}, \frac{\pi}{2} \right] \) for the intersection point.
   - Beyond \( \theta = \frac{\pi}{6} \), \( r = \cos \theta \) remains inside \( r = \sqrt{3} \sin \theta \) until \( \theta = \frac{\pi}{2} \).

3. **Set up the integral**:
   - The area in polar coordinates is given by the integral \( \frac{1}{2} \int r^2 \, d\theta \).
   - The region's area inside \( r = \sqrt{3} \sin \theta \) but outside \( r = \cos \theta \) can be found by subtracting the area covered by \( r = \cos \theta \) from the area of \( r = \sqrt{3} \sin \theta \).

   The integral setup for the area is:
   \[
   \frac{1}{2} \int_{\pi/6}^{\pi/2} \left( (\sqrt{3} \sin \theta)^2 - (\cos \theta)^2 \right) \, d\theta
   \]

4. **Matching with options**:
   - Option **(e)** is set up to subtract the integral for \( r = \cos \theta \) from \( r = \sqrt{3} \sin \theta \) in the specified range, matching our integral setup.

### Answer:
The correct answer is **(e)**.

Would you like further details on this solution or an explanation of the polar coordinate area formula?

### Follow-Up Questions

1. How do we derive the area formula in polar coordinates?
2. Why does the intersection point play a crucial role in setting up the limits?
3. How would the solution change if we were looking for the area inside \( r = \cos \theta \) instead?
4. Can you generalize the approach to find areas between any two polar curves?
5. What would the graph of these two curves look like in the \( r\theta \)-plane?

### Tip
Always remember to verify intersections between polar curves by equating them and solving for \( \theta \), as this determines the integral's limits for the region of interest.

Which of the following represents the area of the region that lies inside \( r = \sqrt{3} \sin \theta \) and outside \( r = \cos \theta \)?

Explore the solution for finding the area inside \( r = \sqrt{3} \sin \theta \) and outside \( r = \cos \theta \) using polar coordinates. This problem involves setting up integrals for areas between curves and understanding curve intersections in polar form. Suitable for advanced calculus students.

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