To solve the double integral $\iint_D (4x + 5y) \, dA$ where $D = \{(x, y) \mid x^2 + y^2 \leq 9, x \geq 0\}$, we first identify the region $D$. This region represents the right half of a circle with radius 3 centered at the origin.

Step 1: Change to Polar Coordinates

In polar coordinates, we set:

• $x = r \cos \theta$
• $y = r \sin \theta$
• The Jacobian $dA = r \, dr \, d\theta$

Step 2: Determine the Limits of Integration

Given the constraints:

• $x^2 + y^2 \leq 9$ translates to $r^2 \leq 9$ or $0 \leq r \leq 3$.
• $x \geq 0$ translates to $0 \leq \theta \leq \pi$.

Step 3: Rewrite the Integral

The integral becomes:

$\iint_D (4x + 5y) \, dA = \int_0^{\pi} \int_0^3 (4(r \cos \theta) + 5(r \sin \theta)) r \, dr \, d\theta$

This simplifies to:

$= \int_0^{\pi} \int_0^3 (4r^2 \cos \theta + 5r^2 \sin \theta) \, dr \, d\theta$

Step 4: Evaluate the Integral

First, we can evaluate the inner integral with respect to $r$:

$\int_0^3 (4r^2 \cos \theta + 5r^2 \sin \theta) \, dr$

Calculating this gives:

= \frac{4}{3} (3^3) \cos \theta + \frac{5}{3} (3^3) \sin \theta = \frac{4 \cdot 27}{3} \cos \theta + \frac{5 \cdot 27}{3} \sin \theta = 36 \cos \theta + 45 \sin \theta$$ Now, substituting back into the outer integral: $$\int_0^{\pi} (36 \cos \theta + 45 \sin \theta) \, d\theta$$ ### Step 5: Evaluate the Outer Integral Calculating the two parts separately: 1. For $$36 \cos \theta$$: $$\int_0^{\pi} 36 \cos \theta \, d\theta = 36 [\sin \theta]_0^{\pi} = 36 (0 - 0) = 0$$ 2. For $$45 \sin \theta$$: \[ \int_0^{\pi} 45 \sin \theta \, d\theta = 45 [-\cos \theta]_0^{\pi} = 45 [-(-1 - 1)] = 45 \cdot 2 = 90 \] ### Final Result Combining the results, we have: $$\iint_D (4x + 5y) \, dA = 0 + 90 = 90$$ Thus, the value of the double integral is $$\boxed{90}$$. ### Questions for Further Exploration: 1. How does changing the bounds affect the value of the integral? 2. What happens if we calculate the integral using rectangular coordinates instead? 3. Can you explain why the integral of $$36 \cos \theta$$ resulted in zero? 4. How can we apply this method to different functions over similar regions? 5. What are the geometric interpretations of the integrals calculated? ### Tip: When working with integrals in polar coordinates, always ensure that your limits correctly reflect the circular region you are integrating over.