Aerospace Radar Tracking Coverage Sector Area (Aerospace Engineering) Solution

An aerospace radar systems engineer at the UNLV Howard R. Hughes College of Engineering is configuring an automated airspace surveillance phased-array radar unit at Harry Reid International Airport in Las Vegas.

The primary radar is calibrated to filter out near-field ground clutter, establishing an inner blind-zone radius of $r_{\text{inner}} = 10\text{ km}$, while its transmitter power budget limits solid target detection to a maximum instrumental range of $r_{\text{outer}} = 80\text{ km}$. To maintain high tracking data-rate updates along a critical high-density arrival corridor, the antenna electronically steers its beam across an azimuth swath from $\theta_1 = \frac{\pi}{6}\text{ rad}$ ($30^\circ$) to $\theta_2 = \frac{\pi}{3}\text{ rad}$ ($60^\circ$).

Step-by-Step Solution:

1. Express Geographic Corridor as Polar Bounded Domain:
   State the Polar Rectangle Definition (defined in Polar Rectangles and Annular Sectors):

   $$R = \left\{ (r, \theta) \in \mathbb{R}^2 \mid r_{\text{inner}} \le r \le r_{\text{outer}}, \, \theta_1 \le \theta \le \theta_2 \right\}$$
   Substitute the radar system operating parameters:
   $$10 \le r \le 80\text{ km}, \quad \frac{\pi}{6} \le \theta \le \frac{\pi}{3}\text{ rad}$$

2. Formulate the Double Integral for Coverage Area:
   State the Polar Area Double Integral:

   $$\text{Area} = \iint_R 1 \, dA = \int_{\theta_1}^{\theta_2} \int_{r_{\text{inner}}}^{r_{\text{outer}}} r \, dr \, d\theta$$
   Substitute the specific numerical integration boundaries:
   $$\text{Area} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \int_{10}^{80} r \, dr \, d\theta$$

3. Evaluate the Double Integral:
   Apply Fubini's Separability Theorem in Polar Coordinates (defined in Separability in Polar Integrals):

   $$\text{Area} = \left( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d\theta \right) \left( \int_{10}^{80} r \, dr \right)$$

   Evaluate the radial integral:

   $$\int_{10}^{80} r \, dr = \left[ \frac{r^2}{2} \right]_{10}^{80} = \frac{(80)^2}{2} - \frac{(10)^2}{2} = \frac{6400 - 100}{2} = \frac{6300}{2} = 3150\text{ km}^2$$

   Evaluate the angular integral:

   $$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d\theta = [\theta]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi - \pi}{6} = \frac{\pi}{6}\text{ rad}$$

   Multiply the evaluated results:

   $$\text{Area} = \left(\frac{\pi}{6}\right) \times 3150 = 525\pi\text{ km}^2 \approx 1649.336\text{ km}^2$$

Meaning of the Answer & Real-Life Application:

• Quantitative Interpretation: The active geographic surveillance coverage area swept by the phased-array beam is exactly $525\pi\text{ km}^2 \approx 1649.34\text{ km}^2$.
• Engineering Application: In air traffic management and radar receiver architecture, computing the exact ground footprint area allows systems engineers to estimate the expected target population based on peak regional aircraft densities (e.g., aircraft per square kilometer). This directly sizes the real-time Kalman filtering track-processor memory, prevents tracker buffer saturation, and guarantees seamless handoff coordination with adjacent Federal Aviation Administration (FAA) en-route radar sectors across Nevada airspace.