What is the area of the region bounded by the lines , , and the curves and ?
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9. Graphical Applications of Integrals
Area Between Curves
Multiple Choice
Find the area enclosed by one loop of the polar curve .
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Verified step by step guidance1
Understand the problem: We are tasked with finding the area enclosed by one loop of the polar curve r = 4 sin(3θ). In polar coordinates, the area enclosed by a curve is given by the formula A = (1/2) ∫ r² dθ, where the limits of integration correspond to one complete loop of the curve.
Determine the limits of integration: For the curve r = 4 sin(3θ), one loop occurs when r completes a full cycle from 0 back to 0. This happens when 3θ varies from 0 to π (since sin(3θ) = 0 at these points). Therefore, θ varies from 0 to π/3 for one loop.
Set up the integral: Substitute r = 4 sin(3θ) into the area formula. The integral becomes A = (1/2) ∫[0 to π/3] (4 sin(3θ))² dθ.
Simplify the integrand: Expand (4 sin(3θ))² to get 16 sin²(3θ). The integral now becomes A = (1/2) ∫[0 to π/3] 16 sin²(3θ) dθ. Factor out the constant 16/2 = 8, so the integral simplifies to A = 8 ∫[0 to π/3] sin²(3θ) dθ.
Use the power-reduction identity: To integrate sin²(3θ), use the identity sin²(x) = (1 - cos(2x))/2. Substituting this, the integral becomes A = 8 ∫[0 to π/3] (1 - cos(6θ))/2 dθ. Simplify further to A = 4 ∫[0 to π/3] (1 - cos(6θ)) dθ. Now, split the integral into two parts: A = 4 [∫[0 to π/3] 1 dθ - ∫[0 to π/3] cos(6θ) dθ].
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