A small surface of area \(A=1 \mathrm{~cm}^{2}\) is subjected to incident radiation of constant intensity \(I_{i}=2.2 \times 10^{4} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{sr}\) over the entire hemisphere. Determine the rate at which radiation energy is incident on the surface through \((a) 0 \leq \theta\) \(\leq 45^{\circ}\) and (b) \(45 \leq \theta \leq 90^{\circ}\), where \(\theta\) is the angle a radiation beam makes with the normal of the surface.

The given area is 1 cm², but we need it to be in the same units as the intensity (W/m²), so we must convert from cm² to m²: 1 cm² = 1 x 10⁻⁴ m² The intensity is given as \(2.2 × 10^{4} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{sr}\).

The equation to calculate the radiation energy incident on the surface is: \(\frac{dE}{dt} = I_{i} \int_{\theta_{1}}^{\theta_{2}} A \cos \theta d\Omega\) Where: - \(\frac{dE}{dt}\) is the rate of radiation energy incident on the surface - \(I_{i}\) is the radiation intensity (\(2.2 × 10^{4} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{sr}\)) - A is the area of the surface (1 x 10⁻⁴ m²) - \(\cos \theta\) is the cosine of the angle between the incident radiation and the normal of the surface - \(d\Omega\) is the differential solid angle - \(\theta_1\) and \(\theta_2\) are the limits of integration for the angle

For the first range of angles, \(\theta_1 = 0\) and \(\theta_2 = 45^\circ\). We'll integrate using polar coordinates, with \(d\Omega = sin\theta d\theta d\phi\). We'll integrate with respect to \(\theta\) from 0 to 45 degrees and with respect to \(\phi\) from 0 to 2\(\pi\) since the radiation is distributed over the entire hemisphere. Convert the limits of integration to radians: \(\theta_1 = 0\), \(\theta_2 = \frac{\pi}{4}\) \(\phi_1 = 0\), \(\phi_2 = 2 \pi\) Now, we can calculate the rate of radiation energy incident on the surface for \(0 \leq \theta \leq 45^\circ\): \(\frac{dE}{dt} = I_{i} A \int_{\theta_1}^{\theta_2} \int_{\phi_1}^{\phi_2} \cos\theta \sin\theta d\theta d\phi\) Substitute the values of \(I_{i}\), A, \(\theta_1\), \(\theta_2\), \(\phi_1\), and \(\phi_2\): \(\frac{dE}{dt} = (2.2 × 10^{4} \mathrm{~W/m^2 \cdot sr}) (1 × 10^{-4} \mathrm{m^2}) \int_{0}^{\pi/4} \int_{0}^{2\pi} \cos\theta \sin\theta d\theta d\phi\) Solve the integral: \(\frac{dE}{dt} \approx 2.2 \times 10^{4} \mathrm{~W/m^2 \cdot sr} \times 1 \times 10^{-4} \mathrm{m^2} \times 999.73 \mathrm{sr}\) \(\frac{dE}{dt} \approx 21931.3 \) W

For the second range of angles, \(\theta_1 = 45^\circ\) and \(\theta_2 = 90^\circ\). Convert the limits of integration to radians: \(\theta_1 = \frac{\pi}{4}\), \(\theta_2 = \frac{\pi}{2}\) \(\phi_1 = 0\), \(\phi_2 = 2 \pi\) Now, we can calculate the rate of radiation energy incident on the surface for \(45 \leq \theta \leq 90^\circ\): \(\frac{dE}{dt} = I_{i} A \int_{\theta_1}^{\theta_2} \int_{\phi_1}^{\phi_2} \cos\theta \sin\theta d\theta d\phi\) Substitute the values of \(I_{i}\), A, \(\theta_1\), \(\theta_2\), \(\phi_1\), and \(\phi_2\): \(\frac{dE}{dt} = (2.2 × 10^{4} \mathrm{~W/m^2 \cdot sr}) (1 × 10^{-4} \mathrm{m^2}) \int_{\pi/4}^{\pi/2} \int_{0}^{2\pi} \cos\theta \sin\theta d\theta d\phi\) Solve the integral: \(\frac{dE}{dt} \approx 2.2 \times 10^{4} \mathrm{~W/m^2 \cdot sr} \times 1 \times 10^{-4} \mathrm{m^2} \times 467.24 \mathrm{sr}\) \(\frac{dE}{dt} \approx 10279.3\) W So, the rate at which radiation energy is incident on the surface is approximately 21931.3 W for the range \(0 \leq \theta \leq 45^\circ\), and 10279.3 W for the range \(45 \leq \theta \leq 90^\circ\).