Compute the area enclosed by one loop of r = 1 + cos θ.

Unlock all questions

This demo includes only 20 questions. Upgrade to access hundreds of questions, flashcards, exam simulations, and disable ads.

Full question bankExam simulationsFlashcards
From $25.99Unlock all

Get ready for your Honors Mathematics 3 exam with our engaging quizzes. Use flashcards and multiple choice questions with explanations to enhance your study. Prepare effectively for the test!

Multiple Choice

Compute the area enclosed by one loop of r = 1 + cos θ.

Explanation:
In polar coordinates, the area swept by a curve r = f(θ) as θ runs over an interval is A = (1/2) ∫ r^2 dθ over that interval. For r = 1 + cos θ, the loop is completed as θ goes from 0 to 2π (r is nonnegative and the cardioid closes at θ = π). So compute the area as A = (1/2) ∫_0^{2π} (1 + cos θ)^2 dθ. Expand and integrate: (1 + cos θ)^2 = 1 + 2 cos θ + cos^2 θ. Over 0 to 2π, ∫ 1 dθ = 2π, ∫ 2 cos θ dθ = 0, and ∫ cos^2 θ dθ = ∫ (1 + cos 2θ)/2 dθ = π. Sum gives 2π + π = 3π. Multiply by 1/2 to obtain A = (1/2)·3π = 3π/2. Therefore, the area enclosed by one loop is 3π/2.

In polar coordinates, the area swept by a curve r = f(θ) as θ runs over an interval is A = (1/2) ∫ r^2 dθ over that interval. For r = 1 + cos θ, the loop is completed as θ goes from 0 to 2π (r is nonnegative and the cardioid closes at θ = π). So compute the area as A = (1/2) ∫_0^{2π} (1 + cos θ)^2 dθ.

Expand and integrate: (1 + cos θ)^2 = 1 + 2 cos θ + cos^2 θ. Over 0 to 2π, ∫ 1 dθ = 2π, ∫ 2 cos θ dθ = 0, and ∫ cos^2 θ dθ = ∫ (1 + cos 2θ)/2 dθ = π. Sum gives 2π + π = 3π. Multiply by 1/2 to obtain A = (1/2)·3π = 3π/2.

Therefore, the area enclosed by one loop is 3π/2.

More Multiple Choice Questions
Subscribe

Get the latest from Passetra

You can unsubscribe at any time. Read our privacy policy