Multiple Choice
Given the parametric equations , for , what is the area enclosed by the curve and the y-axis?
8. Definite Integrals
Introduction to Definite Integrals
Multiple Choice
Given the parametric equations , for , what is the area enclosed by the curve and the y-axis?
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Verified step by step guidance1
Step 1: Recall that the area enclosed by a parametric curve and the y-axis can be calculated using the formula: \( A = \int y \frac{dx}{dt} dt \), where \( y \) is the parametric expression for \( y \), and \( \frac{dx}{dt} \) is the derivative of \( x \) with respect to \( t \).
Step 2: Compute \( \frac{dx}{dt} \) by differentiating \( x = \sin^2(t) \) with respect to \( t \). Use the chain rule: \( \frac{dx}{dt} = 2\sin(t)\cos(t) \).
Step 3: Substitute \( y = 4\cos(t) \) and \( \frac{dx}{dt} = 2\sin(t)\cos(t) \) into the area formula: \( A = \int_{0}^{\pi} 4\cos(t) \cdot 2\sin(t)\cos(t) dt \). Simplify the integrand to \( A = \int_{0}^{\pi} 8\sin(t)\cos^2(t) dt \).
Step 4: To simplify further, use the trigonometric identity \( \cos^2(t) = \frac{1 + \cos(2t)}{2} \). Substitute this into the integrand: \( A = \int_{0}^{\pi} 8\sin(t) \cdot \frac{1 + \cos(2t)}{2} dt \). Expand the expression to \( A = \int_{0}^{\pi} 4\sin(t) dt + \int_{0}^{\pi} 4\sin(t)\cos(2t) dt \).
Step 5: Evaluate each integral separately. For \( \int_{0}^{\pi} 4\sin(t) dt \), use the standard integral of \( \sin(t) \). For \( \int_{0}^{\pi} 4\sin(t)\cos(2t) dt \), use trigonometric identities or integration techniques. Combine the results to find the total area.
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