Let's solve **Question 1** step-by-step: --- ### **1. Find the area of the region bounded by circles of radius 1 and 2 (both centered at the origin) and bounded between the polar curves \( r = 4\sin\theta \) and \( r = 2 \).** #### **Step 1: Sketch and Understand the Problem** - Circles: \( r = 1 \) and \( r = 2 \) - Curves: \( r = 4\sin\theta \) (a limacon, passes through origin at \( \theta = 0, \pi \)), and \( r = 2 \) (a circle) - We want the area between these two curves, but only in the region where \( 1 \leq r \leq 2 \). #### **Step 2: Find Points of Intersection** Set \( 4\sin\theta = 2 \): \[ 4\sin\theta = 2 \implies \sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6},\, \frac{5\pi}{6} \] So, the two curves intersect at \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \). #### **Step 3: Set up the Area Integral** The area between two polar curves \( r_1(\theta) \) and \( r_2(\theta) \) from \( \theta = a \) to \( \theta = b \) is: \[ A = \frac{1}{2} \int_a^b \left( r_1^2 - r_2^2 \right) d\theta \] Here: - Outer curve: \( r = 4\sin\theta \) - Inner curve: \( r = 2 \) - Limits: \( \theta = \frac{\pi}{6} \) to \( \theta = \frac{5\pi}{6} \) \[ A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} \left( (4\sin\theta)^2 - (2)^2 \right) d\theta \] #### **Step 4: Simplify the Integrand** \[ (4\sin\theta)^2 = 16\sin^2\theta \] \[ (2)^2 = 4 \] So, \[ A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (16\sin^2\theta - 4) d\theta \] \[ A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} 16\sin^2\theta d\theta - \frac{1}{2} \int_{\pi/6}^{5\pi/6} 4 d\theta \] \[ A = 8 \int_{\pi/6}^{5\pi/6} \sin^2\theta d\theta - 2 \int_{\pi/6}^{5\pi/6} d\theta \] #### **Step 5: Integrate \(\sin^2\theta\)** Recall: \[ \sin^2\theta = \frac{1 - \cos 2\theta}{2} \] So, \[ 8 \int_{\pi/6}^{5\pi/6} \sin^2\theta d\theta = 8 \int_{\pi/6}^{5\pi/6} \frac{1 - \cos 2\theta}{2} d\theta = 4 \int_{\pi/6}^{5\pi/6} (1 - \cos 2\theta) d\theta \] \[ = 4 \left[ \int_{\pi/6}^{5\pi/6} 1 d\theta - \int_{\pi/6}^{5\pi/6} \cos 2\theta d\theta \right] \] \[ = 4 \left[ \theta \bigg|_{\pi/6}^{5\pi/6} - \frac{1}{2} \sin 2\theta \bigg|_{\pi/6}^{5\pi/6} \right] \] #### **Step 6: Compute the Integrals** First, \( \theta \) from \( \frac{\pi}{6} \) to \( \frac{5\pi}{6} \): \[ \Delta\theta = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \] Now, \( \sin 2\theta \): - At \( \theta = \frac{\pi}{6} \): \( 2\theta = \frac{\pi}{3} \), \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \) - At \( \theta = \frac{5\pi}{6} \): \( 2\theta = \frac{5\pi}{3} \), \( \sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2} \) So, \[ \sin 2\theta \bigg|_{\pi/6}^{5\pi/6} = \sin \frac{5\pi}{3} - \sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = -\sqrt{3} \] So, \[ -\frac{1}{2} \sin 2\theta \bigg|_{\pi/6}^{5\pi/6} = -\frac{1}{2}(-\sqrt{3}) = \frac{\sqrt{3}}{2} \] Plug back in: \[ 4 \left[ \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \right] = \frac{8\pi}{3} + 2\sqrt{3} \] Now, subtract the other term: \[ -2 \int_{\pi/6}^{5\pi/6} d\theta = -2 \left( \frac{5\pi}{6} - \frac{\pi}{6} \right) = -2 \left( \frac{4\pi}{6} \right) = -2 \left( \frac{2\pi}{3} \right) = -\frac{4\pi}{3} \] #### **Step 7: Final Area** Combine: \[ A = \left( \frac{8\pi}{3} + 2\sqrt{3} \right) - \frac{4\pi}{3} \] \[ A = \left( \frac{8\pi}{3} - \frac{4\pi}{3} \right) + 2\sqrt{3} \] \[ A = \frac{4\pi}{3} + 2\sqrt{3} \] --- ### **Final Answer:** \[ \boxed{A = \frac{4\pi}{3} + 2\sqrt{3}} \] This is the area of the region bounded by the circles of radii 1 and 2, and between the polar curves \( r = 4\sin\theta \) and \( r = 2 \).