AP Calculus BC Practice Quiz: Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

Correct Answer: D

The formula for the area of a region defined by a polar curve r = f(θ) from θ = α to θ = β is A = (1/2) ∫[from α to β] r^2 dθ. The curve r = 3 sin(θ) is a circle traced completely as θ goes from 0 to π. Therefore, the correct integral is (1/2) ∫[from 0 to π] (3 sin(θ))^2 dθ.

Correct Answer: B

The area of a polar region is given by A = (1/2) ∫ r^2 dθ. For the rose curve r = 4 cos(2θ), a single petal is traced out as r goes from 0 to its maximum and back to 0. This occurs when cos(2θ) goes from 0 to 1 and back to 0. For the petal symmetric about the polar axis, this happens when 2θ goes from -π/2 to π/2, which means θ goes from -π/4 to π/4. Thus, the integral for the area of one petal is (1/2) ∫[from -π/4 to π/4] (4 cos(2θ))^2 dθ.

Correct Answer: C

The concept of finding area in polar coordinates extends the idea from rectangular coordinates. Instead of summing infinitesimal rectangles, the polar area formula is derived by partitioning the region into an infinite number of small sectors of a circle. The area of a single sector with radius r and infinitesimal angle dθ is dA = (1/2)r^2 dθ. Integrating this expression from θ = α to θ = β gives the total area.

Correct Answer: B

The area is A = (1/2) ∫[from 0 to 2π] (2 + 2 cos(θ))^2 dθ. The cardioid is traced once from 0 to 2π. A = (1/2) ∫[from 0 to 2π] (4 + 8 cos(θ) + 4 cos^2(θ)) dθ. Using the identity cos^2(θ) = (1 + cos(2θ))/2, we get A = ∫[from 0 to 2π] (2 + 4 cos(θ) + 2 cos^2(θ)) dθ = ∫[from 0 to 2π] (2 + 4 cos(θ) + 1 + cos(2θ)) dθ = ∫[from 0 to 2π] (3 + 4 cos(θ) + cos(2θ)) dθ. Evaluating the integral gives [3θ + 4 sin(θ) + (1/2)sin(2θ)] from 0 to 2π, which is (3(2π) + 0 + 0) - (0) = 6π.

Correct Answer: C

The inner loop of a limaçon is traced when r passes through 0. We find the bounds by setting r = 0: 2 - 4 sin(θ) = 0, which gives sin(θ) = 1/2. In the interval [0, 2π], this occurs at θ = π/6 and θ = 5π/6. The inner loop is traced as θ varies between these two angles. Therefore, the area of the inner loop is given by the definite integral A = (1/2) ∫[from π/6 to 5π/6] (2 - 4 sin(θ))^2 dθ.

Correct Answer: C

The area formula is A = (1/2) ∫ r^2 dθ. Here, r^2 is given directly as 16 cos(2θ). The curve is defined only when cos(2θ) ≥ 0, which occurs for θ in [-π/4, π/4] (right loop) and [3π/4, 5π/4] (left loop). The total area is the sum of the areas of the two loops. By symmetry, we can find the area of the right loop and double it. Area = 2 * (1/2) ∫[from -π/4 to π/4] 16 cos(2θ) dθ = ∫[from -π/4 to π/4] 16 cos(2θ) dθ = 16 * [(1/2)sin(2θ)] from -π/4 to π/4 = 8 * [sin(π/2) - sin(-π/2)] = 8 * [1 - (-1)] = 16.

Correct Answer: B

For a rose curve r = a sin(nθ) where n is odd, the entire curve (n petals) is traced once as θ varies from 0 to π. Thus, the total area could be (1/2) ∫[from 0 to π] (2 sin(3θ))^2 dθ. Alternatively, we can find the area of one petal and multiply by the number of petals (3). One petal is traced as 3θ goes from 0 to π, which means θ goes from 0 to π/3. Therefore, the area of one petal is (1/2) ∫[from 0 to π/3] (2 sin(3θ))^2 dθ. The total area is 3 times this value, which corresponds to option B.