Finding the Area of the Region Bounded by Two Polar Curves: Practice Questions (Calculus BC 9.9)

Answer: (B). The limaçon has 1 ≤ r ≤ 5, so it never goes outside the circle (it only touches it at θ = 0). The region is therefore swept for the full turn, 0 ≤ θ ≤ 2π:

Area = ½ ∫ from 0 to 2π of [25 − (3 + 2 cos θ)²] dθ = ½ ∫ from 0 to 2π of (16 − 12 cos θ − 4 cos²θ) dθ = ½(32π − 0 − 4π) = 14π.

Check: the circle’s area is 25π and the limaçon’s is ½ ∫ from 0 to 2π of (3 + 2 cos θ)² dθ = 11π. 25π − 11π = 14π. ✓

• (A) leaves out the ½ from the sector area formula.
• (C) squares the gap: ½ ∫ from 0 to 2π of (5 − 3 − 2 cos θ)² dθ = 6π. Each radius must be squared separately.
• (D) integrates only from 0 to π (the top half) and forgets to double.

Answer: (C). Solving 2 sin θ = 2 − 2 sin θ gives sin θ = ½, so θ = π/6 and θ = 5π/6, both with r = 1. These are two different points. Now check the pole: r = 2 sin θ is 0 at θ = 0 (and π), and r = 2 − 2 sin θ is 0 at θ = π/2. Both curves pass through the pole, at different angles, so the pole is a third shared point.

• (B) counts only the solutions of the equation and misses the pole.
• (A) keeps only one solution of sin θ = ½. Both π/6 and 5π/6 lie in the range of the circle, 0 ≤ θ ≤ π.
• (D) counts the pole twice, because the circle reaches it at both θ = 0 and θ = π. It is one point.

Answer: (A). The curves meet where 4 sin θ = 2√2, so sin θ = √2/2 and θ = π/4 in the first quadrant. They do not share the pole, because r = 2√2 never reaches it. Test θ = π/6: r = 4 sin θ = 2 and r = 2√2 ≈ 2.828, so the circle r = 4 sin θ is nearer the pole on [0, π/4]. At θ = π/2 it has r = 4, so on [π/4, π/2] the circle r = 2√2 is nearer (r² = 8). For a region inside both, use the nearer curve on each piece. The value is 2π − 2 ≈ 4.283.

• (B) uses the farther curve on each piece. Its value, 2π + 2 ≈ 8.283, is larger than a quarter of the disc r ≤ 2√2 (area 2π ≈ 6.283), which is impossible.
• (C) is the area inside r = 4 sin θ and outside r = 2√2 in the first quadrant (it equals 2). That is a different region.
• (D) subtracts over the whole quadrant, but the “outer” curve changes at π/4. The positive and negative parts cancel and the integral equals 0, which cannot be the area.

Answer: (D). With a calculator, 3 sin θ = 2 + cos θ at θ = α ≈ 1.0065 and θ = β ≈ 2.7786 (store both). Test θ = π/2: the circle has r = 3 and the limaçon r = 2, so the circle is outer between α and β. Then

Area = ½ ∫ from α to β of [9 sin²θ − (2 + cos θ)²] dθ ≈ 2.939.

• (A) squares the gap: ½ ∫ from α to β of (3 sin θ − 2 − cos θ)² dθ ≈ 0.626.
• (B) forgets to subtract the inner curve: ½ ∫ from α to β of 9 sin²θ dθ ≈ 5.751.
• (C) leaves out the ½, doubling the answer to ≈ 5.879.

(a) 2 sin 2θ = 1 gives sin 2θ = ½. With 0 < θ < π/2, 2θ = π/6 or 5π/6, so θ = π/12 and θ = 5π/12.

(b) At θ = π/4 the rose has r = 2 and the circle r = 1, so the rose is outer between the two angles: A₁ = ½ ∫ from π/12 to 5π/12 of (4 sin²2θ − 1) dθ.

(c) Use sin²2θ = ½(1 − cos 4θ): 4 sin²2θ − 1 = 1 − 2 cos 4θ. An antiderivative is θ − ½ sin 4θ. [θ − ½ sin 4θ] from π/12 to 5π/12 = (5π/12 − ½ sin(5π/3)) − (π/12 − ½ sin(π/3)) = (5π/12 + √3/4) − (π/12 − √3/4) = π/3 + √3/2. So A₁ = ½(π/3 + √3/2) = π/6 + √3/4 ≈ 0.957.

(d) Each petal reaches r = 2, beyond the circle, in the same way, so R has four congruent pieces: Area = 4A₁ = 2π/3 + √3 ≈ 3.826. Two petals are traced with r < 0 (for example, π/2 < θ < π draws the fourth-quadrant petal), but r² is the same, so each piece still has area A₁. Each petal is swept once for 0 ≤ θ ≤ 2π, so do not double count.

Suggested mark points (4): 1 for both angles π/12 and 5π/12; 1 for the integral with ½, the squares and correct limits; 1 for the exact value π/6 + √3/4; 1 for multiplying by 4 with a reason.

Common errors: ½ ∫ (2 sin 2θ − 1)² dθ (squaring the gap); using 4 sin²2θ = 2 + 2 cos 4θ (wrong identity sign); integrating the rose alone, which gives the whole rose area 2π.

(a) 2 sin θ = 2 − 2 sin θ gives sin θ = ½: the points are (1, π/6) and (1, 5π/6).

(b) At θ = π/12: circle r ≈ 0.518, cardioid r ≈ 1.482, so the circle is nearer on [0, π/6]. At θ = π/3: circle r ≈ 1.732, cardioid r ≈ 0.268, so the cardioid is nearer on [π/6, π/2].

(c) Both curves use only sin θ, so S is symmetric about the y-axis. Double the right-hand half: Area = 2[ ½ ∫ from 0 to π/6 of (2 sin θ)² dθ + ½ ∫ from π/6 to π/2 of (2 − 2 sin θ)² dθ ] = ∫ from 0 to π/6 of 4 sin²θ dθ + ∫ from π/6 to π/2 of (2 − 2 sin θ)² dθ.

(d) First integral: 4 sin²θ = 2 − 2 cos 2θ, so [2θ − sin 2θ] from 0 to π/6 = π/3 − √3/2. Second integral: (2 − 2 sin θ)² = 4 − 8 sin θ + 4 sin²θ = 6 − 8 sin θ − 2 cos 2θ. An antiderivative is 6θ + 8 cos θ − sin 2θ. At π/2: 3π. At π/6: π + 4√3 − √3/2 = π + 7√3/2. Difference: 2π − 7√3/2. Area = π/3 − √3/2 + 2π − 7√3/2 = 7π/3 − 4√3 ≈ 0.402 square units.

Total: 5 points. Acceptable alternative for (c): without symmetry, ½ ∫ from 0 to π/6 of (2 sin θ)² dθ + ½ ∫ from π/6 to 5π/6 of (2 − 2 sin θ)² dθ + ½ ∫ from 5π/6 to π of (2 sin θ)² dθ, which also gives 7π/3 − 4√3. Using the farther curve on each piece earns neither the (c) points nor the (d) point; it gives 5π/3 + 4√3 − 8 ≈ 4.164, more than the circle’s whole area π.

(a) 3 + 2.5 sin θ = 4.5 gives sin θ = 0.6, so θ = α ≈ 0.6435 and θ = β = π − α ≈ 2.4981.

(b) At θ = π/2 the bed edge has r = 5.5 and the circle r = 4.5, so the bed edge is outer for α < θ < β. Elsewhere the bed edge is inside the circle (its r is between 0.5 and 4.5). The dry part is

½ ∫ from α to β of [ (3 + 2.5 sin θ)² − 4.5² ] dθ ≈ 5.966 m².

(c) Area of the whole bed = ½ ∫ from 0 to 2π of (3 + 2.5 sin θ)² dθ = 97π/8 ≈ 38.092 m². Watered part ≈ 38.092 − 5.966 = 32.126 m².

(d) 32.126 ÷ 38.092 ≈ 0.843, so about 84.3% of the bed is watered. That is more than 80%, so the claim is correct.

Total: 5 points. Acceptable alternative for (c): ½ ∫ from 0 to α of (3 + 2.5 sin θ)² dθ + ½ ∫ from α to β of 4.5² dθ + ½ ∫ from β to 2π of (3 + 2.5 sin θ)² dθ ≈ 32.126 m². A correct (c) or (d) from a wrong (b) value carried forward can still earn those points. Squaring the gap in (b) gives about 0.484 m² and loses the (b) points.