MATH U101 · Module 1 · Clinic

Polar Problem Clinic

From formulas to finished problems — the two checklists that turn §11.5 area and arc-length questions into routine, plus six worked examples in the exact shapes the tutorial sheet uses.

If the pieces work but the problems don't — read this first

There is a very specific stage of learning where you can differentiate sin θ, integrate cos 2θ, recite the identities — and still get zero full problems right. That stage is normal, it is where most of the class quietly is, and it is not a knowledge gap. A §11.5 problem chains five or six small steps, and one slip anywhere sinks the whole answer. Nobody's school syllabus drilled that chain — it's new to the BITSAT kids too.

The fix is not more theory. It is (1) running every problem through the same written checklist until the order is automatic, and (2) noticing that these problems are engineered to collapse — if your algebra is turning ugly, you skipped a step, and the checklist tells you which one.

1 · The arc-length recipe §11.5

The game

Every polar arc-length problem in this course is built so that r² + (r′)² collapses to a constant (or something one identity away from one). The whole question is really testing whether you simplify before differentiating and keep track of where you are in the formula. Run these six steps in this order, every time:

Write the formula first, with the limits already in place: L = ∫_a^b √(r² + (r′)²) dθ.
Simplify r before touching it. If r has a square root, hunt for a perfect square inside (table below). Differentiating an unsimplified root is the single most common way these problems go wrong.
Differentiate the simplified r to get r′ = dr/dθ.
Square both and add. Expand r² and (r′)²; expect sin²θ + cos²θ = 1 to make the cross-terms cancel.
Take the square root of the whole sum — once. If the sum is 8, the integrand is √8 = 2√2, and that's what gets integrated.
Integrate and evaluate the limits: antiderivative first, then top limit minus bottom limit. Remember ∫ c dθ = cθ — a constant integrates to constant × θ.

Spot the perfect square

Four double-angle rearrangements do all the simplifying in step 2. These are the same identities as the power-reduce pair on the flashcards, read in the other direction:

You see (under a √)	It equals	So the √ becomes
1 + cos 2θ	2 cos²θ	√2 \|cos θ\|
1 − cos 2θ	2 sin²θ	√2 \|sin θ\|
1 + sin 2θ	(sin θ + cos θ)²	\|sin θ + cos θ\|
1 − sin 2θ	(cos θ − sin θ)²	\|cos θ − sin θ\|

Unpack this step

Why the sin 2θ ones are perfect squares: expand (sin θ + cos θ)² = sin²θ + 2 sin θ cos θ + cos²θ = 1 + sin 2θ, using sin²+cos² = 1 and 2 sin θ cos θ = sin 2θ.

The |…| matters: √(x²) = |x|, not x. Check the sign of the thing inside over your θ-interval; on every problem in this course the interval is chosen so it's non-negative and the bars drop — but you must say so, and that check is worth marks.

Classic trap — the outer square root

You compute r² + (r′)² = 2 and then wonder where √2 came from. It came from the formula itself: the √ was sitting in ∫√(r² + (r′)²) dθ all along, waiting for you to substitute. If the sum were 9 the integrand would be 3; if 16, then 4; the sum being 2 gives √2. Nobody claimed 2 = √2.

Worked example ALength of r = 3 sin θ + 3 cos θ, 0 ≤ θ ≤ π/2

Formula with limits: L = ∫₀^π/2 √(r² + (r′)²) dθ.
No square root in r — nothing to simplify. On to step 3.
r′ = 3 cos θ − 3 sin θ (differentiate term by term).
Square and add: r² = 9 + 18 sin θ cos θ and (r′)² = 9 − 18 sin θ cos θ, so the cross-terms cancel: r² + (r′)² = 18.
Unpack this step

(3 sin θ + 3 cos θ)² = 9 sin²θ + 18 sin θ cos θ + 9 cos²θ, and 9 sin²θ + 9 cos²θ = 9. Same expansion with a minus for (r′)².
Outer root, once: integrand is √18 = 3√2.
L = ∫₀^π/2 3√2 dθ = [3√2 θ]₀^π/2 = 3π√2⁄2 ≈ 6.66.
Sanity check: the endpoints are (3, 0) and (0, 3) in Cartesian; the straight line between them is ≈ 4.24, and an arc bulging between them being ≈ 6.66 is plausible. A result smaller than the straight line would mean an error.

Worked example BLength of r = √(4 + 4 cos 2θ), −π/2 ≤ θ ≤ π/2

Formula with limits: L = ∫_−π/2^π/2 √(r² + (r′)²) dθ.
Simplify first — this is the step the problem is testing. 4 + 4 cos 2θ = 4(1 + cos 2θ) = 8 cos²θ, so r = √(8 cos²θ) = 2√2 |cos θ|. On [−π/2, π/2], cos θ ≥ 0, so the bars drop: r = 2√2 cos θ.
r′ = −2√2 sin θ. (Compare: differentiating the original square root needs the chain rule twice and produces a fraction that never simplifies cleanly. If that's the road you're on, go back to step 2.)
r² + (r′)² = 8 cos²θ + 8 sin²θ = 8.
Integrand: √8 = 2√2.
L = [2√2 θ]_−π/2^π/2 = 2√2(π/2) − 2√2(−π/2) = 2π√2. Watch the double negative at the lower limit — subtracting a negative adds.
Sanity check: r = 2√2 cos θ is a circle of radius √2 traced exactly once on this interval, and its circumference is 2π√2. The answer is the circumference — everything agrees.

2 · Area problems: the limits routine §11.5

The game

The area formulas are short — A = ½ ∫ r² dθ inside one curve, A = ½ ∫ (R² − r²) dθ between an outer R and an inner r. All the difficulty lives in the limits. Find them with the same routine every time:

Sketch first. A rough sketch, always. You cannot pick limits for a region you haven't seen.
Set r = 0 — solving it gives the angles where a loop begins and ends (that's what "the curve passes through the pole" means).
Set r₁ = r₂ for two curves — solving it gives the crossing angles. A repeated or extra root can be a touch (tangency), not a crossing — example G.
Check the pole separately. Two curves can both pass through the pole at different θ — an intersection step 3 never finds.
Ask: is the region swept exactly once as θ runs through your limits? If the boundary switches from one curve to another partway, split the integral at the switch angle.

Worked example CArea enclosed by the limaçon r = 2 − sin θ

Sketch/limits: here r = 2 − sin θ is never 0 (it stays between 1 and 3), so there's no loop to worry about and the whole curve is traced once as θ runs 0 → 2π. Limits: 0 to 2π.
A = ½ ∫₀^2π (2 − sin θ)² dθ = ½ ∫₀^2π (4 − 4 sin θ + sin²θ) dθ.
Integrate the three pieces over a full period: ∫ 4 = 8π; ∫ 4 sin θ = 0 (full period of sin); ∫ sin²θ = π by power-reduce sin²θ = (1 − cos 2θ)/2.
A = ½(8π − 0 + π) = 9π⁄2.
Compare with the tutorial's r = 2 − cos θ: swapping cos for sin only rotates the curve a quarter turn, and rotating a region never changes its area — so both give 9π/2. If your two answers differ, the slip is in the integration, not the setup.

Example D: the shared lens. Below the dashed ray θ = π/4 its edge is the sin-circle, above it the cos-circle.

Example E: inside r = 2 sin θ, below r = csc θ (the line y = 1).

Example F: r = 1 + 2 cos θ. Shaded: inside the outer loop, outside the inner one.

Worked example DArea shared by the circles r = 4 cos θ and r = 4 sin θ

Sketch: two circles of radius 2 — one sitting on the x-axis, one on the y-axis — overlapping in a lens (first figure above).
Crossings: 4 cos θ = 4 sin θ ⇒ tan θ = 1 ⇒ θ = π/4. Pole check: the cos-circle hits the pole at θ = π/2, the sin-circle at θ = 0 — different angles, but both pass through it, so the pole is the lens's second corner.
Which curve bounds the lens? For 0 ≤ θ ≤ π/4 the nearer boundary is 4 sin θ (it's the smaller of the two there); past π/4 they swap. So split at π/4: A = ½ ∫₀^π/4 (4 sin θ)² dθ + ½ ∫_π/4^π/2 (4 cos θ)² dθ.
The lens is symmetric about the ray θ = π/4, so the two halves are equal: A = 2 · ½ ∫₀^π/4 16 sin²θ dθ = 16 ∫₀^π/4 sin²θ dθ.
Power-reduce and evaluate: 16 [θ/2 − (sin 2θ)/4]₀^π/4 = 16(π/8 − 1/4) = 2π − 4 ≈ 2.28.
Unpack this step

sin²θ = (1 − cos 2θ)/2 integrates to θ/2 − (sin 2θ)/4. At π/4: π/8 − (sin(π/2))/4 = π/8 − 1/4; at 0: 0.
Sanity check: each disc has area 4π ≈ 12.6, and the lens is a small sliver of one — 2.28 fits. A lens bigger than a whole disc means the limits doubled something.

Worked example EArea inside r = 2 sin θ and below the line r = csc θ

Translate the strange curve first: r = csc θ means r sin θ = 1, i.e. y = 1 — a horizontal line. (Same move in general: r = a csc θ is the line y = a; r = a sec θ is the vertical line x = a.)
Sketch: r = 2 sin θ is the circle of radius 1 centred at (0, 1); the line y = 1 cuts it (second figure above). We want the part of the disc below the line.
Crossings: 2 sin θ = csc θ ⇒ 2 sin²θ = 1 ⇒ sin θ = 1/√2 ⇒ θ = π/4, 3π/4 — the Cartesian points (1, 1) and (−1, 1).
Walk θ from 0 to π and watch which curve is the region's outer edge: from 0 to π/4 it's the circle, from π/4 to 3π/4 the line is nearer (the region stops at the line), from 3π/4 to π the circle again. Split accordingly: A = ½ ∫₀^π/4 4 sin²θ dθ + ½ ∫_π/4^3π/4 csc²θ dθ + ½ ∫_3π/4^π 4 sin²θ dθ.
New tool, needed exactly here: ∫ csc²θ dθ = −cot θ + C.
Unpack this step

Check it the safe way — differentiate back: (−cot θ)′ = csc²θ, the mirror of (tan θ)′ = sec²θ that's already on your flashcards.
Circle pieces (equal by symmetry): 2 · ½ ∫₀^π/4 4 sin²θ dθ = 4(π/8 − 1/4) = π/2 − 1. Line piece: ½[−cot θ]_π/4^3π/4 = ½((1) − (−1)) = 1.
A = (π/2 − 1) + 1 = π⁄2.
Sanity check — and this one is beautiful: the line y = 1 passes through the circle's centre, so the region is exactly half of a disc of radius 1: ½ · π · 1² = π/2. The calculus and the geometry agree perfectly.

Worked example FThe region between the loops of r = 1 + 2 cos θ

Sketch: with b > a in a + b cos θ, the limaçon has an inner loop (third figure above). Shaded region: inside the big loop, outside the small one — so the plan is outer loop area − inner loop area.
Loop limits — set r = 0: 1 + 2 cos θ = 0 ⇒ cos θ = −1/2 ⇒ θ = 2π/3, 4π/3. The inner loop is traced for 2π/3 ≤ θ ≤ 4π/3 (where r is negative); the outer loop for −2π/3 ≤ θ ≤ 2π/3.
Expand once, use it twice: (1 + 2 cos θ)² = 1 + 4 cos θ + 4 cos²θ = 3 + 4 cos θ + 2 cos 2θ, with antiderivative F(θ) = 3θ + 4 sin θ + sin 2θ.
Unpack this step

4 cos²θ = 4 · (1 + cos 2θ)/2 = 2 + 2 cos 2θ by power-reduce, and ∫ 2 cos 2θ dθ = sin 2θ (divide by the inner 2).
Inner loop: A_in = ½[F]_2π/3^4π/3 = ½(2π − 3√3) = π − 3√3⁄2 ≈ 0.54.
Outer loop: A_out = ½[F]_−2π/3^2π/3 = ½(4π + 3√3) = 2π + 3√3⁄2 ≈ 8.88.
Between the loops: A = A_out − A_in = (2π + 3√3⁄2) − (π − 3√3⁄2) = π + 3√3 ≈ 8.34.
Sanity check: the whole curve fits inside a circle of radius 3 (area ≈ 28), the outer loop is a bit less than a third of that, and the hole removed is small — the sizes all fit.

Worked example GA line cutting a limaçon — with a touch point Quiz 1, 08 Sep

The area inside r = 3 + 2 cos θ and to the right of the line x = −1. (Quiz 1 gave the two integrals and asked only for the four limits.)

Translate the line: x = −1 ⇒ r cos θ = −1 ⇒ r = −sec θ, a positive radius only where cos θ < 0, i.e. on π/2 < θ < 3π/2. (Vertical line → ±sec; horizontal → ±cosec; the sign follows the side of the origin.)
Sketch: a dimpled limaçon (3 > 2, no inner loop), leftmost point at x = −9/8, so the line x = −1 clips two thin slivers off its left side.
Crossings: 3 + 2 cos θ = −sec θ; multiply by cos θ: 2 cos²θ + 3 cos θ + 1 = 0 ⇒ (2 cos θ + 1)(cos θ + 1) = 0. Roots cos θ = −½ (θ = 2π/3, 4π/3 — true crossings, at (−1, ±√3)) and cos θ = −1 (θ = π — the curve touches the line at (−1, 0) and returns). A tangency root is normal here: keep it, note it, and let the two crossings set the limits.
Walk θ once round from −2π/3: the ray leaves the region through the curve for −2π/3 ≤ θ ≤ 2π/3, and through the line for 2π/3 ≤ θ ≤ 4π/3 (the touch at π does not switch the boundary). So A = ½ ∫_−2π/3^2π/3 (3 + 2 cos θ)² dθ + ½ ∫_2π/3^4π/3 sec²θ dθ; by symmetry about the x-axis, A = ∫₀^2π/3 (3 + 2 cos θ)² dθ + ∫_2π/3^π sec²θ dθ — the quiz's boxes. (Mirror halves, e.g. −2π/3 → 0 and π → 4π/3, are equally correct.)
Value, for the check: (3 + 2 cos θ)² = 11 + 12 cos θ + 2 cos 2θ integrates to 22π/3 + 11√3/2 on [0, 2π/3]; ∫_2π/3^π sec² = [tan θ] = √3. A = 22π/3 + 13√3/2 ≈ 34.30, versus the whole limaçon's 11π ≈ 34.56: the slivers are ≈ 0.26. Sensible.

Example G: shaded = inside r = 3 + 2 cos θ, right of x = −1. Crossings at 2π/3 and 4π/3; the touch at π. Full step-by-step and a zoom on the slivers: Quiz 1 post-mortem.

Classic trap — the pole hides from the algebra

In example D, solving r₁ = r₂ found only θ = π/4 — yet the lens visibly has two corners. The second is the pole, which each circle visits at a different θ, so the equation never sees it. Every intersection problem needs the pole checked separately, by asking "does each curve pass through r = 0 at some angle?"

3 · Now do the tutorial sheet

Each worked example above is deliberately the same shape as a tutorial problem with different numbers — so the tutorial stays yours to solve. Attempt each with the matching example open beside you and the recipe steps written at the top of your page:

Tutorial problem	Shape	Twin above
4 — area of limaçon 2 − cos θ	single-curve area, power-reduce	C
5 — length of 2 sin θ + 2 cos θ	arc length, cross-terms cancel	A
6 — length of √(1 + cos 2θ)	arc length, simplify-r-first	B
7 — area shared by 2 cos θ, 2 sin θ	lens, split at the crossing	D
8 — inside 4 sin θ, below 3 csc θ	circle + horizontal line	E
9 — shaded region of 1 + 2 sin θ	outer loop − inner loop	F

Two ground rules while you work: write the checklist step numbers in the margin as you go (it makes the skipped step visible), and when an answer comes out wrong, find which step broke before redoing anything — that's the difference between practising and flailing. For raw speed on the individual moves, the polar pipeline drill serves a fresh randomized set every visit.

Ask an AI well

If you use ChatGPT/Claude on these, make it coach rather than solve — copying a worked solution feels like understanding but doesn't survive contact with a blank page. Three prompts that keep you the one doing the work:

“I'm working on a polar area problem from Thomas §11.5. Don't solve it. Ask me for my first step, tell me if it's right, and only then let me take the next one.”

“Here is my full working for finding the length of r = … . Find the first line where I went wrong, tell me why it's wrong, and stop there — don't show me the rest.”

“Give me two new problems exactly like Thomas §11.5 arc length ones, with final answers only, so I can practise the simplify-first pattern. Nothing beyond §11.5.”