The problem requires determining the limits of integration for the shaded region $R$ in the polar coordinates, based on the given iterated integral:

$\iint_R f(x, y) \, dA = \int_A^B \int_C^D f(r \cos(\theta), r \sin(\theta)) \, r \, dr \, d\theta$

The shaded region in the figure appears to be a sector of an annulus. Let's analyze the limits based on the region's boundaries:

Observing the Graph:

1. Radius (r): The shaded region extends from an inner radius to an outer radius. The inner radius is 2, and the outer radius is 4.

2. Angle ($\theta$): The shaded sector spans the angles from $\theta = 0$ to $\theta = \frac{\pi}{2}$ (from the positive x-axis to the positive y-axis).

Setting the Limits

From this, we determine:

• $A = 0$: the lower bound of $\theta$.
• $B = \frac{\pi}{2}$: the upper bound of $\theta$.
• $C = 2$: the lower bound of $r$.
• $D = 4$: the upper bound of $r$.

Final Answer

Thus, the limits of integration are:

• $A = 0$
• $B = \frac{\pi}{2}$
• $C = 2$
• $D = 4$

Would you like further details on setting up polar coordinates for integrals, or have any questions?

Here are some related questions for further practice:

1. How would the limits change if the shaded region was a complete circle instead of a quarter-circle?
2. How would you set up the integral if the shaded region spanned from $\theta = \frac{\pi}{4}$ to $\theta = \frac{3\pi}{4}$?
3. What changes in limits are required if the inner radius was instead 1?
4. How would the integral setup differ if the function $f(x, y)$ depended on $r$ alone?
5. Can you express this region in rectangular coordinates and set up the corresponding integral?

Tip: When converting regions to polar coordinates, carefully observe the boundaries of $r$ and $\theta$, as these often define arcs and circular sectors.