Let's solve the problem step by step: --- ### **a) Sketch the curves** #### Curve 1: \( r = 5 \sin \theta \) - This is a circle centered at \( (0, 2.5) \) with radius \( 2.5 \) in polar coordinates. - It passes through the origin (\( r=0 \) when \( \theta=0 \) or \( \pi \)), and reaches maximum \( r=5 \) at \( \theta = \frac{\pi}{2} \). #### Curve 2: \( r = 2 + \sin \theta \) - This is a limacon with an inner loop if the constant is less than 1, but here, since 2 > 1, it does **not** have an inner loop. - It is always positive and reaches maximum \( r=3 \) at \( \theta = \frac{\pi}{2} \), and minimum \( r=1 \) at \( \theta = \frac{3\pi}{2} \). **Sketch**: Both curves are above the pole. The circle (from \( r = 5 \sin \theta \)) is always farther from the origin than the limacon (from \( r = 2 + \sin \theta \)), except possibly at endpoints. To find where they intersect, set: \[ 5\sin\theta = 2 + \sin\theta \implies 4\sin\theta = 2 \implies \sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}, \frac{5\pi}{6} \] So, they intersect at these values of \(\theta\). --- ### **b) Set up the Integral** We want the area inside \( r = 5 \sin\theta \) and outside \( r = 2 + \sin\theta \), **between their points of intersection**. In polar coordinates, the area between two curves from \( \theta = a \) to \( \theta = b \) is: \[ A = \frac{1}{2} \int_{a}^{b} \left( R_{outer}^2 - R_{inner}^2 \right) d\theta \] Here, - \( R_{outer} = 5\sin\theta \) - \( R_{inner} = 2 + \sin\theta \) - The limits are \( \theta = \frac{\pi}{6} \) to \( \theta = \frac{5\pi}{6} \) (points of intersection). So, \[ A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} \left[ (5\sin\theta)^2 - (2 + \sin\theta)^2 \right] d\theta \] Let's expand: \[ (5\sin\theta)^2 = 25\sin^2\theta \] \[ (2+\sin\theta)^2 = 4 + 4\sin\theta + \sin^2\theta \] So, \[ A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} [25\sin^2\theta - (4 + 4\sin\theta + \sin^2\theta)] d\theta \] \[ = \frac{1}{2} \int_{\pi/6}^{5\pi/6} [24\sin^2\theta - 4 - 4\sin\theta] d\theta \] --- #### **Now, Solve the Integral** \[ A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} [24\sin^2\theta - 4 - 4\sin\theta] d\theta \] Let's integrate term by term: ##### 1. \( \int \sin^2\theta d\theta \) Recall: \[ \sin^2\theta = \frac{1 - \cos 2\theta}{2} \] So, \[ \int \sin^2\theta d\theta = \int \frac{1 - \cos 2\theta}{2} d\theta = \frac{1}{2} \theta - \frac{1}{4}\sin 2\theta \] ##### 2. \( \int d\theta = \theta \) ##### 3. \( \int \sin\theta d\theta = -\cos\theta \) --- So, plugging in: \[ A = \frac{1}{2} \left\{ 24 \int_{\pi/6}^{5\pi/6} \sin^2\theta d\theta - 4 \int_{\pi/6}^{5\pi/6} d\theta - 4 \int_{\pi/6}^{5\pi/6} \sin\theta d\theta \right\} \] \[ = \frac{1}{2} \left\{ 24 \left[ \frac{1}{2}\theta - \frac{1}{4}\sin 2\theta \right]_{\pi/6}^{5\pi/6} - 4 \left[ \theta \right]_{\pi/6}^{5\pi/6} - 4 \left[ -\cos\theta \right]_{\pi/6}^{5\pi/6} \right\} \] \[ = \frac{1}{2} \left\{ 12 \left[ \theta - \frac{1}{2}\sin 2\theta \right]_{\pi/6}^{5\pi/6} - 4 \left[ \theta \right]_{\pi/6}^{5\pi/6} + 4 \left[ \cos\theta \right]_{\pi/6}^{5\pi/6} \right\} \] --- Now, plug in the limits: Let’s compute each part: #### \( \theta \) part: \[ [\theta]_{\pi/6}^{5\pi/6} = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \] #### \( \sin 2\theta \) part: \[ \sin(2 \cdot \frac{5\pi}{6}) = \sin(\frac{5\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2} \] \[ \sin(2 \cdot \frac{\pi}{6}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \] So, \[ [\sin 2\theta]_{\pi/6}^{5\pi/6} = (-\frac{\sqrt{3}}{2}) - (\frac{\sqrt{3}}{2}) = -\sqrt{3} \] #### \( \cos\theta \) part: \[ \cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} \] \[ \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \] So, \[ [\cos\theta]_{\pi/6}^{5\pi/6} = (-\frac{\sqrt{3}}{2}) - (\frac{\sqrt{3}}{2}) = -\sqrt{3} \] --- Now substitute into the expression: \[ A = \frac{1}{2} \left\{ 12 \left[ \frac{2\pi}{3} - \frac{1}{2}(-\sqrt{3}) \right] - 4 \left( \frac{2\pi}{3} \right) + 4 (-\sqrt{3}) \right\} \] \[ = \frac{1}{2} \left\{ 12 \left( \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \right) - \frac{8\pi}{3} - 4\sqrt{3} \right\} \] Evaluate \( 12 \left( \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \right) = 12\cdot\frac{2\pi}{3} + 12\cdot\frac{\sqrt{3}}{2} = 8\pi + 6\sqrt{3} \) So, \[ = \frac{1}{2} \left( 8\pi + 6\sqrt{3} - \frac{8\pi}{3} - 4\sqrt{3} \right) \] \[ = \frac{1}{2} \left( 8\pi - \frac{8\pi}{3} + 6\sqrt{3} - 4\sqrt{3} \right) \] \[ = \frac{1}{2} \left( \frac{24\pi - 8\pi}{3} + 2\sqrt{3} \right) \] \[ = \frac{1}{2} \left( \frac{16\pi}{3} + 2\sqrt{3} \right) \] \[ = \frac{8\pi}{3} + \sqrt{3} \] --- ## **Final Answer** ### **a) Sketch**: Draw the circles for \( r = 5\sin\theta \) and the limacon for \( r = 2 + \sin\theta \), which intersect at \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \). ### **b) Area** \[ \boxed{ A = \frac{8\pi}{3} + \sqrt{3} } \] This is the area inside the first curve and outside the second.

Certainly! Let's go through the detailed steps carefully, including all calculations, explanations, and the final result. --- ## **Part a) Sketch the curves** ### Curve 1: \( r = 5 \sin \theta \) - This is a circle in polar coordinates. - It passes through the pole at \( \theta = 0, \pi \) because \( r=0 \) when \( \sin \theta=0 \). - Maximum \( r=5 \) occurs at \( \theta= \frac{\pi}{2} \) where \( \sin \frac{\pi}{2} = 1 \). - The circle is symmetric about the vertical axis, centered at \( (0, 2.5) \) in Cartesian coordinates. ### Curve 2: \( r = 2 + \sin \theta \) - This is a limacon. - Minimum \( r=1 \) at \( \theta = \frac{3\pi}{2} \). - Maximum \( r=3 \) at \( \theta= \frac{\pi}{2} \). - It is always positive, and it does **not** have an inner loop because the constant term (2) is greater than 1. --- ## **Part b) Find the area inside the first curve \( r=5\sin\theta \) and outside the second \( r=2+\sin\theta \)** ### Step 1: Find the points of intersection Set the two equations equal: \[ 5 \sin \theta = 2 + \sin \theta \] \[ 5 \sin \theta - \sin \theta = 2 \] \[ 4 \sin \theta = 2 \] \[ \sin \theta = \frac{1}{2} \] The solutions in \([0, 2\pi]\): \[ \boxed{ \theta = \frac{\pi}{6}, \quad \theta= \frac{5\pi}{6} } \] These are the points where the curves intersect. The region of interest is between these angles. --- ### Step 2: Set up the integral for the area In polar coordinates, the area between two curves from \( \theta=a \) to \( \theta=b \) is: \[ A = \frac{1}{2} \int_{a}^{b} \left[ R_{outer}^2 - R_{inner}^2 \right] d\theta \] Since the **larger** radius is from \( r=5 \sin \theta \) and the **smaller** from \( r=2 + \sin \theta \), the integral becomes: \[ A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} \left[ (5 \sin \theta)^2 - (2 + \sin \theta)^2 \right] d\theta \] --- ### Step 3: Expand the squares inside the integral \[ (5 \sin \theta)^2 = 25 \sin^2 \theta \] \[ (2 + \sin \theta)^2 = 4 + 4 \sin \theta + \sin^2 \theta \] Substitute back: \[ A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} \left[ 25 \sin^2 \theta - (4 + 4 \sin \theta + \sin^2 \theta) \right] d\theta \] Simplify: \[ A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} \left( 25 \sin^2 \theta - 4 - 4 \sin \theta - \sin^2 \theta \right) d\theta \] \[ A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} \left( 24 \sin^2 \theta - 4 - 4 \sin \theta \right) d\theta \] --- ### Step 4: Break into separate integrals \[ A = \frac{1}{2} \left[ 24 \int_{\pi/6}^{5\pi/6} \sin^2 \theta \, d\theta - 4 \int_{\pi/6}^{5\pi/6} d\theta - 4 \int_{\pi/6}^{5\pi/6} \sin \theta \, d\theta \right] \] --- ### Step 5: Compute each integral #### (i) \( \int \sin^2 \theta \, d\theta \) Recall: \[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \] So, \[ \int \sin^2 \theta \, d\theta = \int \frac{1 - \cos 2\theta}{2} d\theta = \frac{1}{2} \int (1 - \cos 2\theta) d\theta \] \[ = \frac{1}{2} \left[ \theta - \frac{1}{2} \sin 2\theta \right] + C \] --- #### (ii) \( \int d\theta \) \[ = \theta + C \] --- #### (iii) \( \int \sin \theta \, d\theta \) \[ = - \cos \theta + C \] --- ### Step 6: Evaluate the integrals at the limits - \( \theta = \frac{\pi}{6} \) - \( \theta = \frac{5\pi}{6} \) --- ### **Integral 1: \( \int_{\pi/6}^{5\pi/6} \sin^2 \theta \, d\theta \)** \[ = \left[ \frac{1}{2} \left( \theta - \frac{1}{2} \sin 2\theta \right) \right]_{\pi/6}^{5\pi/6} \] Calculate at \( \theta=5\pi/6 \): \[ \frac{1}{2} \left( \frac{5\pi}{6} - \frac{1}{2} \sin \left( 2 \times \frac{5\pi}{6} \right) \right) \] \[ = \frac{1}{2} \left( \frac{5\pi}{6} - \frac{1}{2} \sin \left( \frac{10\pi}{6} \right) \right) \] \[ \sin \left( \frac{10\pi}{6} \right) = \sin \left( \frac{5\pi}{3} \right) = - \frac{\sqrt{3}}{2} \] So, \[ = \frac{1}{2} \left( \frac{5\pi}{6} - \frac{1}{2} \times \left( - \frac{\sqrt{3}}{2} \right) \right) = \frac{1}{2} \left( \frac{5\pi}{6} + \frac{\sqrt{3}}{4} \right) \] Similarly at \( \theta = \pi/6 \): \[ \frac{1}{2} \left( \frac{\pi}{6} - \frac{1}{2} \sin \left( 2 \times \frac{\pi}{6} \right) \right) = \frac{1}{2} \left( \frac{\pi}{6} - \frac{1}{2} \sin \left( \frac{\pi}{3} \right) \right) \] \[ \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \] Thus, \[ = \frac{1}{2} \left( \frac{\pi}{6} - \frac{1}{2} \times \frac{\sqrt{3}}{2} \right) = \frac{1}{2} \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) \] Now, subtract: \[ \left[ \frac{1}{2} \left( \frac{5\pi}{6} + \frac{\sqrt{3}}{4} \right) \right] - \left[ \frac{1}{2} \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) \right] \] \[ = \frac{1}{2} \left( \frac{5\pi}{6} + \frac{\sqrt{3}}{4} - \frac{\pi}{6} + \frac{\sqrt{3}}{4} \right) \] \[ = \frac{1}{2} \left( \frac{4\pi}{6} + \frac{\sqrt{3}}{2} \right) = \frac{1}{2} \left( \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \right) \] \[ = \frac{1}{2} \times \frac{2\pi}{3} + \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\pi}{3} + \frac{\sqrt{3}}{4} \] --- ### **Integral 2: \( \int_{\pi/6}^{5\pi/6} d\theta \)** \[ = \left[ \theta \right]_{\pi/6}^{5\pi/6} = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \] --- ### **Integral 3: \( \int_{\pi/6}^{5\pi/6} \sin \theta \, d\theta \)** \[ = \left[ - \cos \theta \right]_{\pi/6}^{5\pi/6} \] Calculate at \( 5\pi/6 \): \[ - \cos \left( \frac{5\pi}{6} \right) = - \left( - \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{2} \] At \( \pi/6 \): \[ - \cos \left( \frac{\pi}{6} \right) = - \left( \frac{\sqrt{3}}{2} \right) = - \frac{\sqrt{3}}{2} \] Subtract: \[ \frac{\sqrt{3}}{2} - \left( - \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3} \] --- ## **Step 7: Combine all parts** Recall: \[ A = \frac{1}{2} \left[ 24 \times \text{(Integral 1)} - 4 \times \text{(Integral 2)} - 4 \times \text{(Integral 3)} \right] \] Plugging in the values: \[ A = \frac{1}{2} \left[ 24 \times \left( \frac{\pi}{3} + \frac{\sqrt{3}}{4} \right) - 4 \times \frac{2\pi}{3} - 4 \times \sqrt{3} \right] \] Calculate each term: - \( 24 \times \frac{\pi}{3} = 8 \pi \) - \( 24 \times \frac{\sqrt{3}}{4} = 6 \sqrt{3} \) - \( 4 \times \frac{2\pi}{3} = \frac{8\pi}{3} \) - \( 4 \times \sqrt{3} = 4 \sqrt{3} \) Putting it all together: \[ A = \frac{1}{2} \left[ 8 \pi + 6 \sqrt{3} - \frac{8 \pi}{3} - 4 \sqrt{3} \right] \] Combine like terms: \[ A = \frac{1}{2} \left[ \left(8 \pi - \frac{8 \pi}{3}\right) + \left( 6 \sqrt{3} - 4 \sqrt{3} \right) \right] \] Simplify each group: - \( 8 \pi - \frac{8 \pi}{3} = \frac{24 \pi}{3} - \frac{8 \pi}{3} = \frac{16 \pi}{3} \) - \( 6 \sqrt{3} - 4 \sqrt{3} = 2 \sqrt{3} \) Now, \[ A = \frac{1}{2} \left( \frac{16 \pi}{3} + 2 \sqrt{3} \right) \] Distribute \( \frac{1}{2} \): \[ A = \frac{1}{2} \times \frac{16 \pi}{3} + \frac{1}{2} \times 2 \sqrt{3} = \frac{8 \pi}{3} + \sqrt{3} \] --- ## **Final Answer:** \[ \boxed{ \text{Area} = \frac{8 \pi}{3} + \sqrt{3} } \] This is the area inside the curve \( r=5 \sin \theta \) and outside \( r

**Final Answer:** \[ \boxed{ \text{Area} = \frac{8 \pi}{3} + \sqrt{3} } \]

The area between two polar curves from their intersection points is given by \(\frac{1}{2} \int (\text{outer radius}^2 - \text{inner radius}^2) d\theta\).