PrepGo

AP Calculus BC Practice Quiz: Finding the Area of the Region Bounded by Two Polar Curves

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

Which of the following definite integrals gives the area of the region bounded by the polar curves r = f(θ) and r = g(θ) between θ = α and θ = β, where f(θ) ≥ g(θ) ≥ 0 for the entire interval?

All Questions (7)

Which of the following definite integrals gives the area of the region bounded by the polar curves r = f(θ) and r = g(θ) between θ = α and θ = β, where f(θ) ≥ g(θ) ≥ 0 for the entire interval?

A) ∫[α, β] (f(θ) - g(θ)) dθ

B) 1/2 ∫[α, β] (f(θ)² - g(θ)²) dθ

C) π ∫[α, β] (f(θ)² - g(θ)²) dθ

D) 1/2 ∫[α, β] (f(θ) - g(θ))² dθ

Correct Answer: B

The formula for the area of a region bounded by two polar curves r_outer = f(θ) and r_inner = g(θ) from θ = α to θ = β is given by A = 1/2 ∫[α, β] (r_outer² - r_inner²) dθ. Option B correctly represents this formula. Option A is for Cartesian coordinates. Option C incorrectly includes π, confusing the formula with the disk/washer method for volumes of revolution. Option D incorrectly squares the difference of the radii instead of taking the difference of the squares.

Which integral represents the area of the region inside the cardioid r = 1 + cos(θ) and outside the circle r = 1?

A) 1/2 ∫[0, 2π] ((1 + cos(θ))² - 1²) dθ

B) ∫[0, π/2] ((1 + cos(θ))² - 1²) dθ

C) 1/2 ∫[-π/2, π/2] ((1 + cos(θ))² - 1²) dθ

D) 1/2 ∫[-π/2, π/2] (1² - (1 + cos(θ))²) dθ

Correct Answer: C

First, find the points of intersection by setting the two equations equal: 1 = 1 + cos(θ), which gives cos(θ) = 0. This occurs at θ = -π/2 and θ = π/2. In this interval, the cardioid r = 1 + cos(θ) is the outer curve (r_outer) and the circle r = 1 is the inner curve (r_inner). The area is given by the formula A = 1/2 ∫[α, β] (r_outer² - r_inner²) dθ. Plugging in the curves and bounds gives A = 1/2 ∫[-π/2, π/2] ((1 + cos(θ))² - 1²) dθ.

What is the setup for the definite integral that calculates the area of the region inside the circle r = 4cos(θ) and outside the circle r = 2?

A) 1/2 ∫[0, 2π] ((4cos(θ))² - 2²) dθ

B) ∫[0, π/3] ((4cos(θ))² - 2²) dθ

C) 1/2 ∫[-π/3, π/3] (2² - (4cos(θ))²) dθ

D) 1/2 ∫[-π/3, π/3] ((4cos(θ))² - 2²) dθ

Correct Answer: D

To find the limits of integration, set the radii equal: 4cos(θ) = 2, which simplifies to cos(θ) = 1/2. The principal solutions are θ = -π/3 and θ = π/3. For the region between these angles, r = 4cos(θ) is the outer curve and r = 2 is the inner curve. The area formula is A = 1/2 ∫[α, β] (r_outer² - r_inner²) dθ. Substituting the functions and limits gives A = 1/2 ∫[-π/3, π/3] ((4cos(θ))² - 2²) dθ.

Let R be the region inside the rose curve r = 4sin(2θ) and outside the circle r = 2. Which of the following integrals gives the area of R?

A) 2 ∫[π/12, 5π/12] ((4sin(2θ))² - 2²) dθ

B) 4 ∫[π/12, 5π/12] ((4sin(2θ))² - 2²) dθ

C) 1/2 ∫[0, 2π] ((4sin(2θ))² - 2²) dθ

D) 4 ∫[0, π/12] ((4sin(2θ))² - 2²) dθ

Correct Answer: A

First, find the intersection points: 4sin(2θ) = 2, so sin(2θ) = 1/2. This means 2θ = π/6 or 2θ = 5π/6, giving θ = π/12 and θ = 5π/12. This interval defines the portion of one petal that is outside the circle. The rose curve r = 4sin(2θ) has 4 petals. The area of the region outside the circle in the first quadrant is given by 1/2 ∫[π/12, 5π/12] ((4sin(2θ))² - 2²) dθ. By symmetry, the total area is 4 times this value. So, Total Area = 4 * (1/2) ∫[π/12, 5π/12] ((4sin(2θ))² - 2²) dθ = 2 ∫[π/12, 5π/12] ((4sin(2θ))² - 2²) dθ.

Which of the following represents the area of the region common to the interiors of the circles r = 2sin(θ) and r = 2cos(θ)?

A) 1/2 ∫[0, π/2] ((2sin(θ))² + (2cos(θ))²) dθ

B) 1/2 ∫[0, π/4] (2cos(θ))² dθ + 1/2 ∫[π/4, π/2] (2sin(θ))² dθ

C) 1/2 ∫[0, π/4] (2sin(θ))² dθ + 1/2 ∫[π/4, π/2] (2cos(θ))² dθ

D) ∫[0, π/4] (2sin(θ) - 2cos(θ))² dθ

Correct Answer: C

The two circles intersect when 2sin(θ) = 2cos(θ), which means tan(θ) = 1, so θ = π/4. The region common to both circles is bounded by r = 2sin(θ) for 0 ≤ θ ≤ π/4, and by r = 2cos(θ) for π/4 ≤ θ ≤ π/2. Because the bounding curve changes, the integral must be split into two parts. The area is the sum of the areas of these two sectors. Area = (Area from 0 to π/4) + (Area from π/4 to π/2) = 1/2 ∫[0, π/4] (2sin(θ))² dθ + 1/2 ∫[π/4, π/2] (2cos(θ))² dθ.

Let R be the region that lies inside the circle r = 3 and also inside the circle r = 6sin(θ). The area of R is given by which expression?

A) ∫[0, π/6] (6sin(θ))² dθ + ∫[π/6, π/2] 9 dθ

B) 1/2 ∫[π/6, 5π/6] (3² - (6sin(θ))²) dθ

C) 1/2 ∫[0, π] (3)² dθ + 1/2 ∫[0, π] (6sin(θ))² dθ

D) 1/2 ∫[0, π/6] (6sin(θ))² dθ + 1/2 ∫[π/6, 5π/6] (3)² dθ + 1/2 ∫[5π/6, π] (6sin(θ))² dθ

Correct Answer: A

The intersection occurs when 3 = 6sin(θ), so sin(θ) = 1/2, which gives θ = π/6 and θ = 5π/6. The region's boundary is defined by r = 6sin(θ) from θ=0 to θ=π/6, then by r=3 from θ=π/6 to θ=5π/6, and again by r=6sin(θ) from θ=5π/6 to θ=π. This suggests a sum of three integrals as in option D. However, we can use symmetry about the y-axis (θ=π/2). The total area is twice the area of the right half: Area = 2 * [1/2 ∫[0, π/6] (6sin(θ))² dθ + 1/2 ∫[π/6, π/2] (3)² dθ]. This simplifies to ∫[0, π/6] (6sin(θ))² dθ + ∫[π/6, π/2] 9 dθ, which matches option A.

A region R is bounded by the polar curves r = θ and r = 2θ for 0 ≤ θ ≤ π. Which definite integral gives the area of R?

A) 1/2 ∫[0, π] (θ²) dθ

B) 1/2 ∫[0, π] (2θ - θ)² dθ

C) 1/2 ∫[0, π] (3θ²) dθ

D) 1/2 ∫[0, π] (4θ² + θ²) dθ

Correct Answer: C

The region is bounded by two spirals. For any θ > 0 in the given interval, 2θ > θ, so r_outer = 2θ and r_inner = θ. The area is calculated using the formula A = 1/2 ∫[α, β] (r_outer² - r_inner²) dθ. Substituting the given curves and bounds: A = 1/2 ∫[0, π] ((2θ)² - (θ)²) dθ = 1/2 ∫[0, π] (4θ² - θ²) dθ = 1/2 ∫[0, π] (3θ²) dθ.