Area of a Polar Region or the Area Bounded by a Single Polar Curve: Practice Questions (Calculus BC 9.8)

Answer: (C). Area = ½ ∫ from 0 to π/3 of 3² dθ = ½ × 9 × π/3 = 3π/2. This is a sector of a circle of radius 3, one sixth of the full disc (9π ÷ 6 = 3π/2).

• (A) uses r instead of r²: ½ × 3 × π/3 = π/2.
• (B) uses ∫ r dθ with no ½ and no square: 3 × π/3 = π.
• (D) squares r but forgets the ½: 9 × π/3 = 3π.

Answer: (B). r = 4 sin θ is a circle of radius 2 through the pole. It is traced exactly once as θ goes from 0 to π: r = 0 at both ends and r > 0 in between. The integral equals 8 ∫ from 0 to π of sin²θ dθ = 8 × π/2 = 4π, which is π(2)², as expected.

• (A) traces the circle twice (for π < θ < 2π, r < 0 and the same points are plotted again), so it gives 8π, double the area.
• (C) integrates r, not ½ r². Its value is 8, which is not the area.
• (D) covers only the half of the circle to the right of the y-axis, giving 2π.

Answer: (B). Consecutive zeros of r: sin 2θ = 0 at θ = 0 and θ = π/2, with r > 0 between, so one petal lies in the first quadrant. Area = ½ ∫ from 0 to π/2 of 9 sin²2θ dθ = (9/2) ∫ from 0 to π/2 of ½(1 − cos 4θ) dθ = (9/4)[θ − ¼ sin 4θ] from 0 to π/2 = (9/4)(π/2) = 9π/8.

• (A) integrates ½ r instead of ½ r²: ½ ∫ from 0 to π/2 of 3 sin 2θ dθ = 3/2.
• (C) forgets the ½ in front of the integral.
• (D) is the area of all four petals, ½ ∫ from 0 to 2π of 9 sin²2θ dθ.

Answer: (B). For 0 ≤ θ ≤ π, r = θ ≥ 0, so the spiral stays in the upper half-plane and the ray from the pole sweeps the region exactly once. Area = ½ ∫ from 0 to π of θ² dθ = ½ [θ³/3] from 0 to π = π³/6.

• (A) integrates ½ r instead of ½ r²: ½ ∫ θ dθ = π²/4.
• (C) forgets the ½: ∫ θ² dθ = π³/3.
• (D) treats the region as a sector of radius π (the final r) and angle π: ½ π² × π. But r is not constant; it grows from 0 to π.

r = 3 + 2 sin θ ≥ 1 > 0, so the curve is traced once for 0 ≤ θ ≤ 2π. Expand: r² = 9 + 12 sin θ + 4 sin²θ.

(a) ∫ from 0 to 2π: 9 → 18π; 12 sin θ → 0; 4 sin²θ = 2(1 − cos 2θ) → 4π. Area = ½(18π + 0 + 4π) = 11π (about 34.558).

(b) Below the x-axis means π ≤ θ ≤ 2π (there sin θ ≤ 0 and r > 0). ∫ from π to 2π: 9 → 9π; 12 sin θ → 12[−cos θ] from π to 2π = 12(−1 − 1) = −24; 4 sin²θ → 2π. Area = ½(9π − 24 + 2π) = 11π/2 − 12 (about 5.279).

Check: the lower half is much smaller than the upper half, 11π/2 + 12 ≈ 29.279. That fits the shape: r is as small as 1 at θ = 3π/2 but as large as 5 at θ = π/2.

Suggested mark points (3): 1 for the correct expansion of r² and the ½ ∫ setup; 1 for 11π; 1 for the limits π to 2π and 11π/2 − 12.

Common error: forgetting the 12 sin θ term (it vanishes in (a) but not in (b)), which gives 11π/2 for (b), exactly half of (a). The region is not symmetric about the x-axis.

(a) For π/4 < θ < 3π/4, 2θ is between π/2 and 3π/2, so cos 2θ < 0. Then r² = 9 cos 2θ would be negative, and no real r has a negative square. So there are no points on those rays.

(b) Here r² is given directly, so there is nothing to square. Area = ½ ∫ from −π/4 to π/4 of 9 cos 2θ dθ = (9/2)[½ sin 2θ] from −π/4 to π/4 = (9/2)(½ − (−½)) = 9/2.

(c) The curve is symmetric about the y-axis (replacing θ by π − θ leaves cos 2θ unchanged), so the left loop also has area 9/2. Total = 9.

(d) On [0, 2π] the integrand 9 cos 2θ is negative on intervals where the curve has no points at all (from part (a)). Those negative values cancel the positive ones. The polar area formula only applies over intervals where the curve exists and is traced once, so ½ r² ≥ 0 throughout. Splitting the curve into the two loops, as in (b) and (c), gives the correct total 9.

Total: 5 points. Acceptable alternative for (c): 4 × ½ ∫ from 0 to π/4 of 9 cos 2θ dθ = 4 × 9/4 = 9, using symmetry in both axes.

(a) r = 0 when cos θ = −1/3: θ = cos⁻¹(−1/3) ≈ 1.911 and θ = 2π − cos⁻¹(−1/3) ≈ 4.373.

(b) Between these angles cos θ < −1/3, so r < 0: the points are plotted on the opposite side of the pole and trace the inner loop once. Inner loop area = ½ ∫ from 1.911 to 4.373 of (1 + 3 cos θ)² dθ ≈ 2.528.

(c) ½ ∫ from 0 to 2π of (1 + 3 cos θ)² dθ = ½(2π + 0 + 9π) = 11π/2 ≈ 17.279. On [0, 2π] the curve traces the outer loop once and the inner loop once. The inner loop lies inside the outer loop, so its area has been swept twice: once as part of the outer region and once on its own. Correct area enclosed by the outer loop = 17.279 − 2.528 ≈ 14.751.

(d) Inside the outer loop, outside the inner loop: 14.751 − 2.528 ≈ 12.223 (equivalently 17.279 − 2 × 2.528).

Total: 6 points. Answers within 0.001 are accepted; small differences may come from rounding the limits, so keep them stored in the calculator. An integral over [0, 1.911] ∪ [4.373, 2π] for the outer loop (giving 14.751 directly) is an acceptable alternative for (c).