Source: Paper Homework Assignment 8, Texas A&M University
Tags: polar curves, polar area, area in polar coordinates, r = 8sinθ, cardioid, limaçon, rose curve, polar integration, Math 2414, calculus II
Polar curves are defined by equations of the form r = f(θ). You can convert them to Cartesian form using r² = x² + y² and r sin θ = y, r cos θ = x. The area enclosed by a polar curve between angles α and β is A = (1/2) ∫ r² dθ. For regions between two curves, you subtract the inner curve's area from the outer curve's area inside the same integral.
Polar curve
A curve described by an equation r = f(θ), where the radius varies as a function of angle. Plotted on a polar grid.
Cardioid
A heart-shaped curve of the form r = a ± b cos θ or r = a ± b sin θ, where a = b. Example: r = 2 + 2 cos θ.
Limaçon (with inner loop / without inner loop)
A curve of the form r = a + b cos θ (or sin θ). If a < b, it has an inner loop. If a > b, no inner loop. If a = b, it is a cardioid. Example: r = 3 + 2 cos θ is a limaçon without inner loop.
Rose curve
A curve of the form r = a cos(nθ) or r = a sin(nθ). If n is odd, the rose has n petals. If n is even, it has 2n petals. Example: r = cos 3θ has 3 petals.
Circle in polar form
r = a sin θ gives a circle of diameter a centred at (0, a/2) in Cartesian. r = a cos θ gives a circle of diameter a centred at (a/2, 0).
Polar area formula
A = (1/2) ∫ from α to β of [f(θ)]² dθ. This computes the area swept out by the radius vector between angles α and β.
The standard technique: multiply both sides by r (or use known identities) to introduce x, y, r².
Worked example: r = 8 sin θ
Multiply both sides by r:
r² = 8r sin θ
Since r² = x² + y² and r sin θ = y:
x² + y² = 8y
Rearrange: x² + y² − 8y = 0
Complete the square in y: x² + (y² − 8y + 16) = 16
x² + (y − 4)² = 16
This is a circle with centre (0, 4) and radius 4.
To build the θ vs r table for graphing:
θ | r = 8 sin θ |
|---|---|
0 | 0 |
π/4 | 4√2 ≈ 5.66 |
π/2 | 8 |
3π/4 | 4√2 ≈ 5.66 |
π | 0 |
5π/4 | −4√2 |
3π/2 | −8 |
7π/4 | −4√2 |
2π | 0 |
The curve traces out completely from θ = 0 to θ = π. Between π and 2π, the negative r values retrace the same circle.
The formula:
A = (1/2) ∫ from α to β of r² dθ
The key step is always identifying the correct bounds α and β, which are the angles where the curve begins and ends one full trace of the region you want.
Worked example: area inside r = 8 sin θ
The curve traces from θ = 0 to θ = π (one full circle).
A = (1/2) ∫₀^π (8 sin θ)² dθ = (1/2) ∫₀^π 64 sin²θ dθ = 32 ∫₀^π sin²θ dθ
Apply the half-angle identity: sin²θ = (1 − cos 2θ)/2
= 32 ∫₀^π (1 − cos 2θ)/2 dθ = 16 ∫₀^π (1 − cos 2θ) dθ = 16 [θ − (sin 2θ)/2] from 0 to π = 16 [(π − 0) − (0 − 0)] = 16π units²
Sanity check: this is a circle of radius 4, so πr² = 16π. Confirmed.
This limaçon has a = 2, b = 1, so a > b, meaning no inner loop. The curve is symmetric about the x-axis.
By symmetry, the total area equals twice the area from θ = 0 to θ = π:
A = 2 · (1/2) ∫₀^π (2 + cos θ)² dθ = ∫₀^π (2 + cos θ)² dθ
Expand:
(2 + cos θ)² = 4 + 4 cos θ + cos²θ
Use cos²θ = (1 + cos 2θ)/2:
= 4 + 4 cos θ + (1 + cos 2θ)/2 = 9/4 + 4 cos θ + (cos 2θ)/2
Wait, let's be more careful:
= 4 + 4 cos θ + 1/2 + (cos 2θ)/2 = 9/2 + 4 cos θ + (cos 2θ)/2
Hmm, let's follow the assignment's working exactly. After expanding and simplifying:
∫₀^π (9/4 + 2 cos θ + (1/2) cos²θ) ...
Actually, the assignment computes (1/2)A first, then doubles. The final result is:
A = (9/4)θ + 2 sin θ − (1/8) sin 2θ | from 0 to π
= 9π/4 + 0 − 0 − (0 + 0 − 0) = 9π/4
Since this was (1/2)A:
A = 9π/2 units²
This is a 3-petalled rose (odd n = 3).
One petal spans from where cos 3θ = 0, which gives 3θ = π/2, so θ = π/6. The petal from θ = −π/6 to θ = π/6, or equivalently 0 to π/6 for half a petal.
Area of one petal:
(1/6)A_total ... but let's compute one petal directly.
One-sixth of the total area (one petal = 1/3 of total, but the assignment computes 1/6 of total first):
(1/6)A = ∫₀^(π/6) (1/2) cos²3θ dθ
Use cos²3θ = (1 + cos 6θ)/2:
= ∫₀^(π/6) (1/4)(1 + cos 6θ) dθ = (1/4)[θ + (sin 6θ)/6] from 0 to π/6 = (1/4)[π/6 + 0 − 0] = π/24
So A = π/24 for one sixth-petal, giving A = π/4 units² for the full rose (all three petals).
The assignment specifically requires finding the area of the shaded leaf directly, without using symmetry shortcuts.
The leaf spans from θ = π/2 to θ = 5π/6 (one of the three petals).
A = ∫(π/2)^(5π/6) (1/2)(2 cos 3θ)² dθ = ∫(π/2)^(5π/6) (1/2) · 4 cos²3θ dθ = 2 ∫_(π/2)^(5π/6) cos²3θ dθ
Apply the identity cos²3θ = (1 + cos 6θ)/2:
= 2 ∫(π/2)^(5π/6) (1 + cos 6θ)/2 dθ = ∫(π/2)^(5π/6) (1 + cos 6θ) dθ = [θ + (sin 6θ)/6] from π/2 to 5π/6
Evaluate at 5π/6: 5π/6 + (1/6) sin(5π) = 5π/6 + 0 = 5π/6
Evaluate at π/2: π/2 + (1/6) sin(3π) = π/2 + 0 = π/2
= 5π/6 − π/2 = (5π − 3π)/6 = 2π/6
A = π/3 units²
When finding the area of a region that lies outside one curve and inside another, the formula is:
A = (1/2) ∫_α^β [r_outer² − r_inner²] dθ
Step 1: Find the intersection points by setting the two r-expressions equal. Step 2: Determine which curve is outer and which is inner on the interval. Step 3: Integrate.
Worked example: area outside r = 4 and inside r = 3 + 2 cos θ
Find intersections:
3 + 2 cos θ = 4 → 2 cos θ = 1 → cos θ = 1/2 → θ = π/3 and θ = 5π/3 (which is −π/3)
On the interval [−π/3, π/3], the limaçon r = 3 + 2 cos θ is outer (r > 4), and the circle r = 4 is inner.
By symmetry about the x-axis:
A = 2 · (1/2) ∫₀^(π/3) [(3 + 2 cos θ)² − 4²] dθ
Expand (3 + 2 cos θ)² = 9 + 12 cos θ + 4 cos²θ
So the integrand becomes:
9 + 12 cos θ + 4 cos²θ − 16 = 4 cos²θ + 12 cos θ − 7
Use cos²θ = (1 + cos 2θ)/2:
= 4 · (1 + cos 2θ)/2 + 12 cos θ − 7 = 2 + 2 cos 2θ + 12 cos θ − 7 = 2 cos 2θ + 12 cos θ − 5
Integrate from 0 to π/3:
= [sin 2θ + 12 sin θ − 5θ] from 0 to π/3
At π/3: sin(2π/3) + 12 sin(π/3) − 5π/3 = √3/2 + 12 · √3/2 − 5π/3 = √3/2 + 6√3 − 5π/3
At 0: 0 + 0 − 0 = 0
= √3/2 + 6√3 − 5π/3 = (√3 + 12√3)/2 − 5π/3 = 13√3/2 − 5π/3
Wait, let me recheck. 12 sin(π/3) = 12 · (√3/2) = 6√3. And sin(2π/3) = √3/2.
So the result is √3/2 + 6√3 − 5π/3.
Combine the √3 terms: √3/2 + 6√3 = √3/2 + 12√3/2 = 13√3/2.
So: 13√3/2 − 5π/3.
But we need to double (the factor of 2 from symmetry was already included in the setup). Actually, looking again, the 2 · (1/2) cancelled, so the integral as stated gives the full area:
A = (13√3/2 − 5π/3) units²
Or equivalently: A = (6√3 + (−10π + 3√3)/6), which the assignment writes as (6√3 + (−10π + 3√3)/6) units².
Polar area (single curve): A = (1/2) ∫_α^β r² dθ
Polar area (between curves): A = (1/2) ∫_α^β [r_outer² − r_inner²] dθ
Half-angle identity (essential for integration): sin²θ = (1 − cos 2θ)/2 cos²θ = (1 + cos 2θ)/2
Common polar curve to Cartesian conversions: r = a sin θ → x² + (y − a/2)² = (a/2)² (circle) r = a cos θ → (x − a/2)² + y² = (a/2)² (circle)
Rose curve petal count: r = cos(nθ) or r = sin(nθ): n petals if n is odd, 2n petals if n is even
⚠️ The half-angle identities sin²θ = (1 − cos 2θ)/2 and cos²θ = (1 + cos 2θ)/2 appear in nearly every polar area computation. Have them memorised cold.
⚠️ Choosing the correct integration bounds is the most common source of error. Always check: where does the curve (or the region) start and end? Plot a quick table of θ vs r values if unsure.
⚠️ For rose curves, one petal of r = cos(nθ) spans from −π/(2n) to π/(2n). The total area of all petals can be found by multiplying one petal's area by the number of petals.
⚠️ When finding area between two curves, always verify which curve is outer and which is inner on your interval. A sign error here flips the answer.
⚠️ "Do not use symmetry" is a common exam instruction. In that case, you must integrate over the exact bounds of the specific region, not halve/double.
⚠️ r = 8 sin θ traces its full circle from 0 to π (not 0 to 2π). Using 0 to 2π would double-count.
Q: What is the Cartesian equation of the polar curve r = 6 cos θ?
A: Multiply both sides by r: r² = 6r cos θ → x² + y² = 6x → (x − 3)² + y² = 9. This is a circle centred at (3, 0) with radius 3.
Q: How many petals does the rose r = sin 4θ have?
A: Since n = 4 is even, the rose has 2n = 8 petals.
Q: Set up (but don't evaluate) the integral for the area of one petal of r = 3 sin 2θ.
A: One petal of sin 2θ spans from θ = 0 to θ = π/2 (where sin 2θ ≥ 0). A = (1/2) ∫₀^(π/2) (3 sin 2θ)² dθ = (9/2) ∫₀^(π/2) sin²2θ dθ.
Q: What trig identity do you need to integrate sin²3θ?
A: sin²3θ = (1 − cos 6θ)/2. The argument of cosine doubles: 3θ becomes 6θ.
Q: For the region between r = 4 and r = 3 + 2 cos θ, why do the intersection points matter?
A: The intersection angles (θ = ±π/3) define the bounds of integration. Outside those bounds, the circle r = 4 is larger than the limaçon, so no enclosed region exists there.
polar area, area in polar coordinates, polar area formula, half-angle identity, double angle identity, rose curve area, cardioid area, limaçon area, area between polar curves, polar intersection points, r = a sin theta circle, r = a cos theta circle, completing the square polar, petal area, MATH 2414, Texas A&M, calculus II polar integration