A student computes the area of the cardioid r = 1 + cos(θ) using A = ∫₀²π r dθ. What is wrong with this setup?
AThe bounds should be 0 to π because the cardioid is symmetric about the x-axis
BThe integrand should be (1/2)r², not r — the area of each infinitesimal sector is (1/2)r² dθ
CThe formula should use dr instead of dθ
DNothing is wrong; the cardioid area formula integrates r directly
Polar area sums infinitesimal circular sectors, each with area (1/2)r² dθ — not rectangles of height r and width dθ. The 1/2 factor always appears in polar area and comes from the sector area formula. The bounds 0 to 2π are correct for the cardioid (it traces once over a full revolution), but the integrand is wrong. The correct formula is A = (1/2) ∫₀²π (1 + cos θ)² dθ.
Question 2 Multiple Choice
Which expression correctly gives the area of the region that lies outside r = 1 + cos(θ) and inside r = 3 cos(θ), over the angular range where r_outer > r_inner?
A(1/2) ∫ (3cosθ − 1 − cosθ)² dθ
B∫ (3cosθ − 1 − cosθ) dθ
C(1/2) ∫ (9cos²θ − (1 + cosθ)²) dθ
D(1/2) ∫ (3cosθ − 1 − cosθ) dθ
The area between two polar curves is (1/2) ∫ (r_outer² − r_inner²) dθ — not (1/2)(r_outer − r_inner)². This is the polar analogue of the Cartesian formula ∫(f − g) dx, but with r² in place of r (because sectors, not rectangles, are the infinitesimal pieces). Option A squares the difference, which is a very common error. Option B drops the 1/2 and also omits the squaring. Option D is wrong for the same reason as A without the squaring of the whole expression.
Question 3 True / False
Two polar curves can intersect at the pole (origin) even when they reach r = 0 at completely different values of θ.
TTrue
FFalse
Answer: True
The pole is a single geometric point, but different curves can pass through it at different angles — i.e., at different values of θ where r = 0. For example, r = sin(θ) reaches 0 at θ = 0 and θ = π, while r = cos(θ) reaches 0 at θ = π/2 and θ = 3π/2. Both curves pass through the origin, but setting sin(θ) = cos(θ) would miss this intersection. When computing area between curves near the pole, you must check for this separately.
Question 4 True / False
The area between two polar curves r_outer and r_inner is computed as (1/2) ∫ (r_outer − r_inner)² dθ.
TTrue
FFalse
Answer: False
The correct formula is (1/2) ∫ (r_outer² − r_inner²) dθ — the difference of squares, not the square of the difference. This follows from subtracting the inner sector from the outer sector: (1/2)r_outer²dθ − (1/2)r_inner²dθ = (1/2)(r_outer² − r_inner²)dθ. The error of squaring the whole difference is common but changes the result significantly.
Question 5 Short Answer
Why does the polar area formula A = (1/2) ∫_α^β r² dθ include a factor of 1/2, and where does it come from?
Think about your answer, then reveal below.
Model answer: The 1/2 comes from the formula for the area of a circular sector. A sector with radius r and central angle dθ covers the fraction dθ/(2π) of the full circle, giving area πr² · dθ/(2π) = (1/2)r² dθ. Polar area is built by summing infinitely many such thin sectors as dθ → 0, so every term in the Riemann sum carries the 1/2 factor. Unlike Cartesian integration, which uses rectangular slices of area f(x)dx, polar integration uses pie-slice sectors whose area formula inherently includes 1/2.
The 1/2 is not an artifact of averaging or a special case — it is always present in polar area because the infinitesimal geometric piece is a sector, not a rectangle. Students who forget it consistently undercount polar areas by a factor of 2.