Let’s solve **Question 1** step by step: --- ### **Question 1** **Find the area of the region bounded by:** - Circles of radius \( a \) and \( b \) (\( a < b \), both centered at origin) - Bounded between the polar curves \( r = a \) and \( r = b \) - Bounded between the lines \( \theta = 0 \) and \( \theta = \frac{\pi}{3} \) --- #### **Step 1: Visualize the Region** - The circles \( r = a \) and \( r = b \) are concentric. - The region is a "sector ring" or "annular sector" between angles \( 0 \) and \( \frac{\pi}{3} \). --- #### **Step 2: Area in Polar Coordinates** The area in polar coordinates between two curves \( r_1(\theta) \) and \( r_2(\theta) \) from \( \theta = \alpha \) to \( \theta = \beta \) is: \[ A = \frac{1}{2} \int_{\alpha}^{\beta} \left( r_2^2 - r_1^2 \right) d\theta \] Here, - \( r_2 = b \), \( r_1 = a \) - \( \alpha = 0 \), \( \beta = \frac{\pi}{3} \) --- #### **Step 3: Plug in the Values** \[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} (b^2 - a^2) d\theta \] \[ = \frac{1}{2} (b^2 - a^2) \int_{0}^{\frac{\pi}{3}} d\theta \] \[ = \frac{1}{2} (b^2 - a^2) \left[ \theta \right]_{0}^{\frac{\pi}{3}} \] \[ = \frac{1}{2} (b^2 - a^2) \left( \frac{\pi}{3} - 0 \right) \] \[ = \frac{1}{2} (b^2 - a^2) \cdot \frac{\pi}{3} \] \[ = \frac{\pi}{6} (b^2 - a^2) \] --- ### **Final Answer** \[ \boxed{\frac{\pi}{6} (b^2 - a^2)} \] --- **Explanation:** - We used the formula for area in polar coordinates. - Subtracted the area enclosed by the smaller circle from that of the larger, only over the angle \( \frac{\pi}{3} \). - The answer is the area of the "ring sector" between \( r = a \) and \( r = b \) from \( \theta = 0 \) to \( \theta = \frac{\pi}{3} \).