Let's break down the problem and solve both parts (a) and (b).

Part (a) - Expression for the Shaded Area

The problem describes two intersecting circles of radius $r$, with an angle $2\theta$ subtended by the common chord at the centers of the circles.

The shaded area represents the intersection of the two circles, which can be found by subtracting the area of the triangular region from the area of the sector for each circle. Since both circles have identical areas of intersection, we'll find the shaded region for one side and then double it.

1. Area of Sector:
   The area of the sector of the circle subtended by angle $2\theta$ is given by: $\text{Area of sector} = \frac{1}{2} r^2 (2\theta) = r^2 \theta$

2. Area of Triangle:
   The area of the isosceles triangle formed by the radii and the chord can be found using basic trigonometry: $\text{Area of triangle} = \frac{1}{2} r^2 \sin(2\theta)$

3. Area of Segment (for one circle):
   The area of the segment (the part of the circle minus the triangular area) is:

   = r^2 \theta - \frac{1}{2} r^2 \sin(2\theta)$$

4. Total Shaded Area (for both circles):
   Since the shaded area is symmetric for both circles, the total shaded area is:

   = 2r^2 \theta - r^2 \sin(2\theta)$$

Thus, the expression for the shaded area is: $\boxed{2r^2 \theta - r^2 \sin(2\theta)}$

Part (b) - Finding $\theta$

Now, we are told that the shaded area is equal to $\frac{1}{4}$ of the area of one circle. The area of one circle is $\pi r^2$, so: $2r^2 \theta - r^2 \sin(2\theta) = \frac{1}{4} \pi r^2$

We can cancel out $r^2$ from both sides: $2 \theta - \sin(2\theta) = \frac{\pi}{4}$

Multiplying through by 4: $8 \theta - 4 \sin(2\theta) = \pi$

Thus, we