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.