https://file.aiquickdraw.com/mathbot/file/1729715180035a8k4eir9.jpg

The problem in the image asks for two things:

1. **Finding the interval of \(\theta\)-values** that correspond to the shaded region.
2. **Finding the area** of the region described by the polar curve \(r = 3 - \theta\).

### Step 1: Interval of \(\theta\)-values
In polar coordinates, the curve is given as \(r = 3 - \theta\). For the region to be shaded, \(r\) must remain non-negative. Hence, we need to solve for when \(r = 3 - \theta \geq 0\).

\[
3 - \theta \geq 0
\]
\[
\theta \leq 3
\]

Since \(r\) must be positive for all points, the range of \(\theta\) starts from \(0\) and ends at \(3\). Therefore, the interval of \(\theta\)-values is:

\[
0 \leq \theta \leq 3
\]

### Step 2: Area of the Region
The area \(A\) in polar coordinates is given by the formula:

\[
A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta
\]

Here, the limits of integration are from \(0\) to \(3\), and \(r = 3 - \theta\). So the area becomes:

\[
A = \frac{1}{2} \int_{0}^{3} (3 - \theta)^2 \, d\theta
\]

Expanding the integrand:

\[
(3 - \theta)^2 = 9 - 6\theta + \theta^2
\]

Now, integrate term by term:

\[
A = \frac{1}{2} \left[ \int_0^3 9 \, d\theta - \int_0^3 6\theta \, d\theta + \int_0^3 \theta^2 \, d\theta \right]
\]

\[
A = \frac{1}{2} \left[ 9\theta \bigg|_0^3 - 6 \cdot \frac{\theta^2}{2} \bigg|_0^3 + \frac{\theta^3}{3} \bigg|_0^3 \right]
\]

Evaluating each term:

1. \(9\theta \bigg|_0^3 = 9(3) - 9(0) = 27\)
2. \(\frac{6\theta^2}{2} \bigg|_0^3 = 3(3^2) - 3(0^2) = 27\)
3. \(\frac{\theta^3}{3} \bigg|_0^3 = \frac{3^3}{3} - \frac{0^3}{3} = 9\)

Now substitute these values:

\[
A = \frac{1}{2} \left[ 27 - 27 + 9 \right] = \frac{1}{2} \times 9 = 4.5
\]

### Final Answer:
- The interval of \(\theta\)-values is \(0 \leq \theta \leq 3\).
- The area of the region is **4.5 square units**.

Would you like further clarification or details on any of the steps? 

Here are 5 related questions to deepen understanding:
1. How would the area calculation change if the function was \(r = 4 - \theta\)?
2. What happens to the shaded region if \(\theta\) extends beyond 3?
3. Can you calculate the area if the interval for \(\theta\) was given from 1 to 3?
4. How would this process differ if the equation \(r = 3 - \theta\) was altered to \(r = \theta - 3\)?
5. What would the graph look like if \(r = 3 + \theta\)?

**Tip:** In polar coordinates, always verify the sign of \(r\) before determining the interval of \(\theta\), as negative \(r\)-values reflect points across the origin.

Which interval of θ-values corresponds to the shaded region in Figure 13? Find the area of the region.

Determine the interval of θ-values for the polar curve r = 3 - θ and calculate the area of the shaded region. Explore concepts in polar coordinates and the use of the polar area formula in calculus.

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