Polar Coordinates Polar coordinates ( r , θ ) (r, \theta) ( r , θ ) ( x , y ) (x, y) ( x , y )
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Adjust the parameters in the graph above to explore the relationships between variables.
Board Coverage Board Paper Notes AQA Paper 1 Polar curves, area enclosed, tangents Edexcel FP2 Full coverage: conversion, sketching, area, tangents OCR (A) — Not in OCR (A) specification CIE (9231) P2 Full coverage: curves, area, tangents
1. Converting Between Cartesian and Polar 1.1 Definitions Definition. The polar coordinates ( r , θ ) (r, \theta) ( r , θ ) P P P
r r r O O O P P P radial coordinate )θ \theta θ x x x O P OP O P angular coordinate )The relationship between Cartesian and polar coordinates is:
x = r cos θ , y = r sin θ \boxed{x = r\cos\theta, \qquad y = r\sin\theta} x = r cos θ , y = r sin θ
r 2 = x 2 + y 2 , tan θ = y x \boxed{r^2 = x^2 + y^2, \qquad \tan\theta = \frac{y}{x}} r 2 = x 2 + y 2 , tan θ = x y
1.2 Converting from polar to Cartesian Given ( r , θ ) (r, \theta) ( r , θ ) ( r cos θ , r sin θ ) (r\cos\theta, r\sin\theta) ( r cos θ , r sin θ )
Example. Convert ( 4 , π / 3 ) (4, \pi/3) ( 4 , π /3 )
x = 4 cos ( π / 3 ) = 4 ⋅ 1 2 = 2 x = 4\cos(\pi/3) = 4 \cdot \frac{1}{2} = 2 x = 4 cos ( π /3 ) = 4 ⋅ 2 1 = 2 y = 4 sin ( π / 3 ) = 4 ⋅ 3 2 = 2 3 y = 4\sin(\pi/3) = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} y = 4 sin ( π /3 ) = 4 ⋅ 2 3 = 2 3
The Cartesian coordinates are ( 2 , 2 3 ) (2, 2\sqrt{3}) ( 2 , 2 3 )
1.3 Converting from Cartesian to polar Given ( x , y ) (x, y) ( x , y ) r = x 2 + y 2 r = \sqrt{x^2+y^2} r = x 2 + y 2 θ = arctan ( y / x ) \theta = \arctan(y/x) θ = arctan ( y / x )
Example. Convert ( − 3 , 3 ) (-3, 3) ( − 3 , 3 )
r = 9 + 9 = 3 2 r = \sqrt{9+9} = 3\sqrt{2} r = 9 + 9 = 3 2 θ = π − arctan ( 1 ) = 3 π / 4 \theta = \pi - \arctan(1) = 3\pi/4 θ = π − arctan ( 1 ) = 3 π /4
The polar coordinates are ( 3 2 , 3 π / 4 ) (3\sqrt{2}, 3\pi/4) ( 3 2 , 3 π /4 )
When converting from Cartesian to polar, always check the quadrant of the point. The Calculator value of arctan ( y / x ) \arctan(y/x) arctan ( y / x )
2. Polar Equations of Curves 2.1 Lines and circles Vertical line x = a x = a x = a r cos θ = a r\cos\theta = a r cos θ = a r = a sec θ r = a\sec\theta r = a sec θ
Horizontal line y = b y = b y = b r sin θ = b r\sin\theta = b r sin θ = b r = b cosec θ r = b\cosec\theta r = b cosec θ
Circle centre ( a , 0 ) (a, 0) ( a , 0 ) a a a r = 2 a cos θ r = 2a\cos\theta r = 2 a cos θ
Circle centre ( 0 , a ) (0, a) ( 0 , a ) a a a r = 2 a sin θ r = 2a\sin\theta r = 2 a sin θ
Circle centre origin radius a a a r = a r = a r = a
Proof of the polar equation r = 2 a cos θ r = 2a\cos\theta r = 2 a cos θ A circle with centre ( a , 0 ) (a, 0) ( a , 0 ) a a a ( x − a ) 2 + y 2 = a 2 (x-a)^2 + y^2 = a^2 ( x − a ) 2 + y 2 = a 2
Expanding: x 2 − 2 a x + a 2 + y 2 = a 2 x^2 - 2ax + a^2 + y^2 = a^2 x 2 − 2 a x + a 2 + y 2 = a 2 x 2 + y 2 = 2 a x x^2 + y^2 = 2ax x 2 + y 2 = 2 a x
Substituting x = r cos θ x = r\cos\theta x = r cos θ r 2 = x 2 + y 2 r^2 = x^2 + y^2 r 2 = x 2 + y 2
r 2 = 2 a r cos θ r^2 = 2ar\cos\theta r 2 = 2 a r cos θ
Since r = 0 r = 0 r = 0 r ≠ 0 r \neq 0 r = 0
r = 2 a cos θ ■ \boxed{r = 2a\cos\theta} \quad \blacksquare r = 2 a cos θ ■
2.2 Cardioids A cardioid has equation r = a ( 1 + cos θ ) r = a(1 + \cos\theta) r = a ( 1 + cos θ ) r = a ( 1 + sin θ ) r = a(1 + \sin\theta) r = a ( 1 + sin θ )
Properties of r = a ( 1 + cos θ ) r = a(1 + \cos\theta) r = a ( 1 + cos θ )
Symmetry: symmetric about the initial line (θ = 0 \theta = 0 θ = 0 θ \theta θ − θ -\theta − θ r r r Maximum r r r θ = 0 \theta = 0 θ = 0 r = 2 a r = 2a r = 2 a Minimum r r r θ = π \theta = \pi θ = π r = 0 r = 0 r = 0 Passes through the origin when cos θ = − 1 \cos\theta = -1 cos θ = − 1 θ = π \theta = \pi θ = π 2.3 Rose curves A rose curve (or rhodonea curve) has equation r = a sin n θ r = a\sin n\theta r = a sin n θ r = a cos n θ r = a\cos n\theta r = a cos n θ
Properties:
If n n n odd : the curve has n n n θ \theta θ 0 0 0 π \pi π If n n n even : the curve has 2 n 2n 2 n θ \theta θ 0 0 0 2 π 2\pi 2 π Example. r = a sin 3 θ r = a\sin 3\theta r = a sin 3 θ r = a cos 4 θ r = a\cos 4\theta r = a cos 4 θ
2.4 Spirals An Archimedean spiral has equation r = a θ r = a\theta r = a θ
A logarithmic spiral has equation r = a e b θ r = ae^{b\theta} r = a e b θ
The logarithmic spiral appears frequently in nature (shells, hurricanes) because the angle between The radius and the tangent is constant.
3. Sketching Polar Curves 3.1 Systematic method Identify symmetry: Symmetric about the initial line (θ = 0 \theta = 0 θ = 0 θ \theta θ − θ -\theta − θ Symmetric about θ = π / 2 \theta = \pi/2 θ = π /2 θ \theta θ π − θ \pi - \theta π − θ Symmetric about the pole if replacing r r r − r -r − r Find key values: Evaluate r r r θ = 0 , π / 6 , π / 4 , π / 3 , π / 2 , π , 3 π / 2 , 2 π \theta = 0, \pi/6, \pi/4, \pi/3, \pi/2, \pi, 3\pi/2, 2\pi θ = 0 , π /6 , π /4 , π /3 , π /2 , π , 3 π /2 , 2 π
Find where r = 0 r = 0 r = 0 These are points where the curve passes through the pole.
Find maximum ∣ r ∣ |r| ∣ r ∣ Differentiate r r r θ \theta θ d r / d θ = 0 dr/d\theta = 0 d r / d θ = 0
Trace the curve: As θ \theta θ ( r , θ ) (r, \theta) ( r , θ )
Example. Sketch r = 2 + cos θ r = 2 + \cos\theta r = 2 + cos θ 0 ≤ θ ≤ 2 π 0 \leq \theta \leq 2\pi 0 ≤ θ ≤ 2 π
Symmetric about θ = 0 \theta = 0 θ = 0 cos ( − θ ) = cos θ \cos(-\theta) = \cos\theta cos ( − θ ) = cos θ r(0) = 3$$r(\pi/2) = 2$$r(\pi) = 1$$r(3\pi/2) = 2 .r > 0 r > 0 r > 0 θ \theta θ 2 + cos θ ≥ 1 2 + \cos\theta \geq 1 2 + cos θ ≥ 1 The curve is a limacon with no inner loop . A limacon r = a + b cos θ r = a + b\cos\theta r = a + b cos θ b > a b > a b > a a < b ≤ 2 a a < b \leq 2a a < b ≤ 2 a (actually a < 2 b a < 2b a < 2 b a ≥ 2 b a \geq 2b a ≥ 2 b
Inner loop: b > a b > a b > a Dimpled: a < 2 b a < 2b a < 2 b b < a b < a b < a Convex: a ≥ 2 b a \geq 2b a ≥ 2 b Cardioid: a = b a = b a = b 4. Area Enclosed by a Polar Curve Theorem. The area enclosed by the polar curve r = f ( θ ) r = f(\theta) r = f ( θ ) θ = α \theta = \alpha θ = α θ = β \theta = \beta θ = β
A = 1 2 ∫ α β r 2 d θ \boxed{A = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta} A = 2 1 ∫ α β r 2 d θ
Divide the angular range [ α , β ] [\alpha, \beta] [ α , β ] n n n Δ θ = β − α n \Delta\theta = \dfrac{\beta-\alpha}{n} Δ θ = n β − α
Each sector is approximately a circular sector of radius r ( θ i ) r(\theta_i) r ( θ i ) Δ θ \Delta\theta Δ θ
Δ A i ≈ 1 2 r 2 ( θ i ) Δ θ \Delta A_i \approx \frac{1}{2}r^2(\theta_i)\,\Delta\theta Δ A i ≈ 2 1 r 2 ( θ i ) Δ θ
Summing all sectors:
A ≈ ∑ i = 1 n 1 2 r 2 ( θ i ) Δ θ A \approx \sum_{i=1}^{n}\frac{1}{2}r^2(\theta_i)\,\Delta\theta A ≈ ∑ i = 1 n 2 1 r 2 ( θ i ) Δ θ
Taking the limit as n → ∞ n \to \infty n → ∞
A = lim n → ∞ ∑ i = 1 n 1 2 r 2 ( θ i ) Δ θ = 1 2 ∫ α β r 2 d θ ■ A = \lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{2}r^2(\theta_i)\,\Delta\theta = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta \quad \blacksquare A = lim n → ∞ ∑ i = 1 n 2 1 r 2 ( θ i ) Δ θ = 2 1 ∫ α β r 2 d θ ■
Example. Find the area enclosed by one petal of r = cos 3 θ r = \cos 3\theta r = cos 3 θ
One petal is traced from θ = − π / 6 \theta = -\pi/6 θ = − π /6 θ = π / 6 \theta = \pi/6 θ = π /6 r = 0 r = 0 r = 0
A = 1 2 ∫ − π / 6 π / 6 cos 2 3 θ d θ = 1 2 ∫ − π / 6 π / 6 1 + cos 6 θ 2 d θ A = \frac{1}{2}\int_{-\pi/6}^{\pi/6}\cos^2 3\theta\,d\theta = \frac{1}{2}\int_{-\pi/6}^{\pi/6}\frac{1+\cos 6\theta}{2}\,d\theta A = 2 1 ∫ − π /6 π /6 cos 2 3 θ d θ = 2 1 ∫ − π /6 π /6 2 1 + c o s 6 θ d θ
= 1 4 [ θ + sin 6 θ 6 ] − π / 6 π / 6 = 1 4 ( π 6 − ( − π 6 ) ) = π 12 = \frac{1}{4}\left[\theta + \frac{\sin 6\theta}{6}\right]_{-\pi/6}^{\pi/6} = \frac{1}{4}\left(\frac{\pi}{6} - \left(-\frac{\pi}{6}\right)\right) = \frac{\pi}{12} = 4 1 [ θ + 6 s i n 6 θ ] − π /6 π /6 = 4 1 ( 6 π − ( − 6 π ) ) = 12 π
Example. Find the area enclosed by the cardioid r = a ( 1 + cos θ ) r = a(1 + \cos\theta) r = a ( 1 + cos θ )
By symmetry, compute from 0 0 0 π \pi π
A = 2 ⋅ 1 2 ∫ 0 π a 2 ( 1 + cos θ ) 2 d θ = a 2 ∫ 0 π ( 1 + 2 cos θ + cos 2 θ ) d θ A = 2\cdot\frac{1}{2}\int_0^\pi a^2(1+\cos\theta)^2\,d\theta = a^2\int_0^\pi(1+2\cos\theta+\cos^2\theta)\,d\theta A = 2 ⋅ 2 1 ∫ 0 π a 2 ( 1 + cos θ ) 2 d θ = a 2 ∫ 0 π ( 1 + 2 cos θ + cos 2 θ ) d θ
= a 2 ∫ 0 π ( 1 + 2 cos θ + 1 + cos 2 θ 2 ) d θ = a 2 ∫ 0 π ( 3 2 + 2 cos θ + cos 2 θ 2 ) d θ = a^2\int_0^\pi\left(1+2\cos\theta+\frac{1+\cos 2\theta}{2}\right)d\theta = a^2\int_0^\pi\left(\frac{3}{2}+2\cos\theta+\frac{\cos 2\theta}{2}\right)d\theta = a 2 ∫ 0 π ( 1 + 2 cos θ + 2 1 + c o s 2 θ ) d θ = a 2 ∫ 0 π ( 2 3 + 2 cos θ + 2 c o s 2 θ ) d θ
= a 2 [ 3 θ 2 + 2 sin θ + sin 2 θ 4 ] 0 π = a 2 ⋅ 3 π 2 = 3 π a 2 2 = a^2\left[\frac{3\theta}{2} + 2\sin\theta + \frac{\sin 2\theta}{4}\right]_0^\pi = a^2\cdot\frac{3\pi}{2} = \boxed{\frac{3\pi a^2}{2}} = a 2 [ 2 3 θ + 2 sin θ + 4 s i n 2 θ ] 0 π = a 2 ⋅ 2 3 π = 2 3 π a 2
4.2 Area between two polar curves The area between curves r 1 ( θ ) r_1(\theta) r 1 ( θ ) r 2 ( θ ) r_2(\theta) r 2 ( θ ) α \alpha α β \beta β
A = 1 2 ∫ α β [ r 1 2 ( θ ) − r 2 2 ( θ ) ] d θ A = \frac{1}{2}\int_\alpha^\beta \bigl[r_1^2(\theta) - r_2^2(\theta)\bigr]\,d\theta A = 2 1 ∫ α β [ r 1 2 ( θ ) − r 2 2 ( θ ) ] d θ
5. Tangents to Polar Curves Since x = r cos θ x = r\cos\theta x = r cos θ y = r sin θ y = r\sin\theta y = r sin θ θ \theta θ
d x d θ = d r d θ cos θ − r sin θ \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta d θ d x = d θ d r cos θ − r sin θ
d y d θ = d r d θ sin θ + r cos θ \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta d θ d y = d θ d r sin θ + r cos θ
Therefore:
d y d x = d r d θ sin θ + r cos θ d r d θ cos θ − r sin θ \boxed{\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}} d x d y = d θ d r cos θ − r sin θ d θ d r sin θ + r cos θ
This follows directly from the parametric differentiation rule d y d x = d y / d θ d x / d θ \dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta} d x d y = d x / d θ d y / d θ x ( θ ) = r ( θ ) cos θ x(\theta) = r(\theta)\cos\theta x ( θ ) = r ( θ ) cos θ y ( θ ) = r ( θ ) sin θ y(\theta) = r(\theta)\sin\theta y ( θ ) = r ( θ ) sin θ ■ \blacksquare ■
5.2 Tangents at the pole The curve passes through the pole when r = 0 r = 0 r = 0 θ = θ 0 \theta = \theta_0 θ = θ 0 r ( θ 0 ) = 0 r(\theta_0) = 0 r ( θ 0 ) = 0
Example. Find the tangents at the pole for r = sin 3 θ r = \sin 3\theta r = sin 3 θ
r = 0 r = 0 r = 0 sin 3 θ = 0 \sin 3\theta = 0 sin 3 θ = 0 3 θ = 0 , π , 2 π , 3 π 3\theta = 0, \pi, 2\pi, 3\pi 3 θ = 0 , π , 2 π , 3 π θ = 0 , π / 3 , 2 π / 3 , π , 4 π / 3 , 5 π / 3 \theta = 0, \pi/3, 2\pi/3, \pi, 4\pi/3, 5\pi/3 θ = 0 , π /3 , 2 π /3 , π , 4 π /3 , 5 π /3
These give 6 tangent lines at the pole (consistent with the fact that r = sin 3 θ r = \sin 3\theta r = sin 3 θ
Example. Find the equation of the tangent to r = 1 + cos θ r = 1 + \cos\theta r = 1 + cos θ θ = π / 3 \theta = \pi/3 θ = π /3
r = 1 + cos ( π / 3 ) = 3 / 2 r = 1 + \cos(\pi/3) = 3/2 r = 1 + cos ( π /3 ) = 3/2 ( x , y ) = ( r cos θ , r sin θ ) = ( 3 / 4 , 3 3 / 4 ) (x, y) = (r\cos\theta, r\sin\theta) = (3/4, 3\sqrt{3}/4) ( x , y ) = ( r cos θ , r sin θ ) = ( 3/4 , 3 3 /4 )
d r d θ = − sin θ \dfrac{dr}{d\theta} = -\sin\theta d θ d r = − sin θ θ = π / 3 \theta = \pi/3 θ = π /3 d r d θ = − 3 / 2 \dfrac{dr}{d\theta} = -\sqrt{3}/2 d θ d r = − 3 /2
d y d x = ( − 3 / 2 ) ( 3 / 2 ) + ( 3 / 2 ) ( 1 / 2 ) ( − 3 / 2 ) ( 1 / 2 ) − ( 3 / 2 ) ( 3 / 2 ) = − 3 / 4 + 3 / 4 − 3 / 4 − 3 3 / 4 = 0 − 3 = 0 \frac{dy}{dx} = \frac{(-\sqrt{3}/2)(\sqrt{3}/2) + (3/2)(1/2)}{(-\sqrt{3}/2)(1/2) - (3/2)(\sqrt{3}/2)} = \frac{-3/4 + 3/4}{-\sqrt{3}/4 - 3\sqrt{3}/4} = \frac{0}{-\sqrt{3}} = 0 d x d y = ( − 3 /2 ) ( 1/2 ) − ( 3/2 ) ( 3 /2 ) ( − 3 /2 ) ( 3 /2 ) + ( 3/2 ) ( 1/2 ) = − 3 /4 − 3 3 /4 − 3/4 + 3/4 = − 3 0 = 0
The tangent is horizontal: y = 3 3 / 4 y = 3\sqrt{3}/4 y = 3 3 /4
5.3 Horizontal and vertical tangents Horizontal tangents occur when d y d θ = 0 \dfrac{dy}{d\theta} = 0 d θ d y = 0 d x d θ ≠ 0 \dfrac{dx}{d\theta} \neq 0 d θ d x = 0
d r d θ sin θ + r cos θ = 0 \frac{dr}{d\theta}\sin\theta + r\cos\theta = 0 d θ d r sin θ + r cos θ = 0
Vertical tangents occur when d x d θ = 0 \dfrac{dx}{d\theta} = 0 d θ d x = 0 d y d θ ≠ 0 \dfrac{dy}{d\theta} \neq 0 d θ d y = 0
d r d θ cos θ − r sin θ = 0 \frac{dr}{d\theta}\cos\theta - r\sin\theta = 0 d θ d r cos θ − r sin θ = 0
6. Summary of Key Results Result Formula Conversion x = r\cos\theta$$y = r\sin\theta$$r^2 = x^2+y^2 Circle r = 2 a cos θ r = 2a\cos\theta r = 2 a cos θ Centre ( a , 0 ) (a,0) ( a , 0 ) a a a Area A = 1 2 ∫ α β r 2 d θ A = \dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta A = 2 1 ∫ α β r 2 d θ Gradient d y d x = r ′ sin θ + r cos θ r ′ cos θ − r sin θ \dfrac{dy}{dx} = \dfrac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta} d x d y = r ′ cos θ − r sin θ r ′ sin θ + r cos θ
Problems
Problem 1 Convert the Cartesian equation x 2 + y 2 − 4 x = 0 x^2 + y^2 - 4x = 0 x 2 + y 2 − 4 x = 0
Hint 1 Substitute x = r cos θ x = r\cos\theta x = r cos θ r 2 = x 2 + y 2 r^2 = x^2+y^2 r 2 = x 2 + y 2
Answer 1 r 2 − 4 r cos θ = 0 ⟹ r ( r − 4 cos θ ) = 0 r^2 - 4r\cos\theta = 0 \implies r(r - 4\cos\theta) = 0 r 2 − 4 r cos θ = 0 ⟹ r ( r − 4 cos θ ) = 0 r ≠ 0 r \neq 0 r = 0 r = 4 cos θ r = 4\cos\theta r = 4 cos θ
This is a circle with centre ( 2 , 0 ) (2, 0) ( 2 , 0 ) 2 2 2
Problem 2 Find the area enclosed by one petal of r = sin 2 θ r = \sin 2\theta r = sin 2 θ
Hint 2 One petal of sin 2 θ \sin 2\theta sin 2 θ θ = 0 \theta = 0 θ = 0 θ = π / 2 \theta = \pi/2 θ = π /2
Answer 2 A = 1 2 ∫ 0 π / 2 sin 2 2 θ d θ = 1 2 ∫ 0 π / 2 1 − cos 4 θ 2 d θ = 1 4 [ θ − sin 4 θ 4 ] 0 π / 2 = 1 4 ⋅ π 2 = π 8 A = \dfrac{1}{2}\displaystyle\int_0^{\pi/2}\sin^2 2\theta\,d\theta = \dfrac{1}{2}\int_0^{\pi/2}\dfrac{1-\cos 4\theta}{2}\,d\theta = \dfrac{1}{4}\left[\theta - \dfrac{\sin 4\theta}{4}\right]_0^{\pi/2} = \dfrac{1}{4}\cdot\dfrac{\pi}{2} = \dfrac{\pi}{8} A = 2 1 ∫ 0 π /2 sin 2 2 θ d θ = 2 1 ∫ 0 π /2 2 1 − cos 4 θ d θ = 4 1 [ θ − 4 sin 4 θ ] 0 π /2 = 4 1 ⋅ 2 π = 8 π
Problem 3 Find the area enclosed by the cardioid r = 2 ( 1 − cos θ ) r = 2(1 - \cos\theta) r = 2 ( 1 − cos θ )
Hint 3 Use symmetry about θ = π \theta = \pi θ = π 0 0 0 2 π 2\pi 2 π ( 1 − cos θ ) 2 (1-\cos\theta)^2 ( 1 − cos θ ) 2
Answer 3 A = 1 2 ∫ 0 2 π 4 ( 1 − cos θ ) 2 d θ = 2 ∫ 0 2 π ( 1 − 2 cos θ + cos 2 θ ) d θ A = \dfrac{1}{2}\displaystyle\int_0^{2\pi}4(1-\cos\theta)^2\,d\theta = 2\int_0^{2\pi}(1 - 2\cos\theta + \cos^2\theta)\,d\theta A = 2 1 ∫ 0 2 π 4 ( 1 − cos θ ) 2 d θ = 2 ∫ 0 2 π ( 1 − 2 cos θ + cos 2 θ ) d θ
= 2 ∫ 0 2 π ( 3 2 − 2 cos θ + cos 2 θ 2 ) d θ = 2 [ 3 θ 2 − 2 sin θ + sin 2 θ 4 ] 0 2 π = 2 ⋅ 3 π = 6 π = 2\int_0^{2\pi}\left(\dfrac{3}{2} - 2\cos\theta + \dfrac{\cos 2\theta}{2}\right)d\theta = 2\left[\dfrac{3\theta}{2} - 2\sin\theta + \dfrac{\sin 2\theta}{4}\right]_0^{2\pi} = 2 \cdot 3\pi = 6\pi = 2 ∫ 0 2 π ( 2 3 − 2 cos θ + 2 cos 2 θ ) d θ = 2 [ 2 3 θ − 2 sin θ + 4 sin 2 θ ] 0 2 π = 2 ⋅ 3 π = 6 π
Problem 4 Find d y d x \dfrac{dy}{dx} d x d y r = a ( 1 + sin θ ) r = a(1+\sin\theta) r = a ( 1 + sin θ ) θ = π / 6 \theta = \pi/6 θ = π /6
Hint 4 r = a(1+\sin\theta)$$\dfrac{dr}{d\theta} = a\cos\theta . Substitute into the gradient formula.
Answer 4 At θ = π / 6 \theta = \pi/6 θ = π /6 r = a(1+1/2) = 3a/2$$dr/d\theta = a\sqrt{3}/2 .
d y d x = ( a 3 / 2 ) ( 1 / 2 ) + ( 3 a / 2 ) ( 3 / 2 ) ( a 3 / 2 ) ( 3 / 2 ) − ( 3 a / 2 ) ( 1 / 2 ) = a 3 / 4 + 3 a 3 / 4 3 a / 4 − 3 a / 4 = a 3 0 \dfrac{dy}{dx} = \dfrac{(a\sqrt{3}/2)(1/2) + (3a/2)(\sqrt{3}/2)}{(a\sqrt{3}/2)(\sqrt{3}/2) - (3a/2)(1/2)} = \dfrac{a\sqrt{3}/4 + 3a\sqrt{3}/4}{3a/4 - 3a/4} = \dfrac{a\sqrt{3}}{0} d x d y = ( a 3 /2 ) ( 3 /2 ) − ( 3 a /2 ) ( 1/2 ) ( a 3 /2 ) ( 1/2 ) + ( 3 a /2 ) ( 3 /2 ) = 3 a /4 − 3 a /4 a 3 /4 + 3 a 3 /4 = 0 a 3
The gradient is undefined — the tangent is vertical at this point.
Problem 5 Find the points on r = 4 cos θ r = 4\cos\theta r = 4 cos θ
Hint 5 A tangent parallel to the initial line is horizontal: d y / d θ = 0 dy/d\theta = 0 d y / d θ = 0
Answer 5 r = 4\cos\theta$$dr/d\theta = -4\sin\theta .
d y d θ = − 4 sin θ sin θ + 4 cos θ cos θ = 4 ( cos 2 θ − sin 2 θ ) = 4 cos 2 θ \dfrac{dy}{d\theta} = -4\sin\theta\sin\theta + 4\cos\theta\cos\theta = 4(\cos^2\theta - \sin^2\theta) = 4\cos 2\theta d θ d y = − 4 sin θ sin θ + 4 cos θ cos θ = 4 ( cos 2 θ − sin 2 θ ) = 4 cos 2 θ
cos 2 θ = 0 ⟹ 2 θ = π / 2 , 3 π / 2 ⟹ θ = π / 4 , 3 π / 4 \cos 2\theta = 0 \implies 2\theta = \pi/2, 3\pi/2 \implies \theta = \pi/4, 3\pi/4 cos 2 θ = 0 ⟹ 2 θ = π /2 , 3 π /2 ⟹ θ = π /4 , 3 π /4
At θ = π / 4 \theta = \pi/4 θ = π /4 r = 2 2 r = 2\sqrt{2} r = 2 2 ( 2 , 2 ) (2, 2) ( 2 , 2 ) θ = 3 π / 4 \theta = 3\pi/4 θ = 3 π /4 r = − 2 2 r = -2\sqrt{2} r = − 2 2 r = 2\sqrt{2}$$\theta = 7\pi/4 Point ( 2 , − 2 ) (2, -2) ( 2 , − 2 )
Problem 6 Find the area of the region inside r = 3 cos θ r = 3\cos\theta r = 3 cos θ r = 1 + cos θ r = 1+\cos\theta r = 1 + cos θ
Hint 6 Find the intersection angles by solving 3 cos θ = 1 + cos θ 3\cos\theta = 1+\cos\theta 3 cos θ = 1 + cos θ r o u t e r 2 − r i n n e r 2 r_{\mathrm{outer}}^2 - r_{\mathrm{inner}}^2 r outer 2 − r inner 2
Answer 6 Intersection: 3 cos θ = 1 + cos θ ⟹ 2 cos θ = 1 ⟹ θ = ± π / 3 3\cos\theta = 1+\cos\theta \implies 2\cos\theta = 1 \implies \theta = \pm\pi/3 3 cos θ = 1 + cos θ ⟹ 2 cos θ = 1 ⟹ θ = ± π /3
By symmetry, compute from 0 0 0 π / 3 \pi/3 π /3
A = 2 ⋅ 1 2 ∫ 0 π / 3 [ 9 cos 2 θ − ( 1 + cos θ ) 2 ] d θ A = 2\cdot\dfrac{1}{2}\displaystyle\int_0^{\pi/3}\bigl[9\cos^2\theta - (1+\cos\theta)^2\bigr]\,d\theta A = 2 ⋅ 2 1 ∫ 0 π /3 [ 9 cos 2 θ − ( 1 + cos θ ) 2 ] d θ
= ∫ 0 π / 3 [ 9 cos 2 θ − 1 − 2 cos θ − cos 2 θ ] d θ = ∫ 0 π / 3 [ 8 cos 2 θ − 1 − 2 cos θ ] d θ = \displaystyle\int_0^{\pi/3}\bigl[9\cos^2\theta - 1 - 2\cos\theta - \cos^2\theta\bigr]\,d\theta = \int_0^{\pi/3}\bigl[8\cos^2\theta - 1 - 2\cos\theta\bigr]\,d\theta = ∫ 0 π /3 [ 9 cos 2 θ − 1 − 2 cos θ − cos 2 θ ] d θ = ∫ 0 π /3 [ 8 cos 2 θ − 1 − 2 cos θ ] d θ
= ∫ 0 π / 3 [ 4 ( 1 + cos 2 θ ) − 1 − 2 cos θ ] d θ = ∫ 0 π / 3 ( 3 + 4 cos 2 θ − 2 cos θ ) d θ = \displaystyle\int_0^{\pi/3}\left[4(1+\cos 2\theta) - 1 - 2\cos\theta\right]d\theta = \int_0^{\pi/3}\left(3 + 4\cos 2\theta - 2\cos\theta\right)d\theta = ∫ 0 π /3 [ 4 ( 1 + cos 2 θ ) − 1 − 2 cos θ ] d θ = ∫ 0 π /3 ( 3 + 4 cos 2 θ − 2 cos θ ) d θ
= [ 3 θ + 2 sin 2 θ − 2 sin θ ] 0 π / 3 = π + 2 ⋅ 3 2 − 2 ⋅ 3 2 = π = \left[3\theta + 2\sin 2\theta - 2\sin\theta\right]_0^{\pi/3} = \pi + 2\cdot\dfrac{\sqrt{3}}{2} - 2\cdot\dfrac{\sqrt{3}}{2} = \pi = [ 3 θ + 2 sin 2 θ − 2 sin θ ] 0 π /3 = π + 2 ⋅ 2 3 − 2 ⋅ 2 3 = π
Problem 7 Convert ( − 2 2 , 2 2 ) (-2\sqrt{2}, 2\sqrt{2}) ( − 2 2 , 2 2 )
Hint 7 r = x 2 + y 2 r = \sqrt{x^2+y^2} r = x 2 + y 2 θ \theta θ
Answer 7 r = 8 + 8 = 4 r = \sqrt{8+8} = 4 r = 8 + 8 = 4
tan θ = 2 2 − 2 2 = − 1 \tan\theta = \dfrac{2\sqrt{2}}{-2\sqrt{2}} = -1 tan θ = − 2 2 2 2 = − 1 θ = 3 π / 4 \theta = 3\pi/4 θ = 3 π /4
Polar coordinates: ( 4 , 3 π / 4 ) (4, 3\pi/4) ( 4 , 3 π /4 )
Problem 8 Sketch the curve r = θ r = \theta r = θ 0 ≤ θ ≤ 4 π 0 \leq \theta \leq 4\pi 0 ≤ θ ≤ 4 π
Hint 8 This is an Archimedean spiral. As θ \theta θ r r r
Answer 8 This is an Archimedean spiral . Key points:
At θ = 0 \theta = 0 θ = 0 r = 0 r = 0 r = 0 At θ = π / 2 \theta = \pi/2 θ = π /2 r = π / 2 r = \pi/2 r = π /2 θ = π / 2 \theta = \pi/2 θ = π /2 At θ = π \theta = \pi θ = π r = π r = \pi r = π x x x At θ = 2 π \theta = 2\pi θ = 2 π r = 2 π r = 2\pi r = 2 π x x x At θ = 4 π \theta = 4\pi θ = 4 π r = 4 π r = 4\pi r = 4 π The spiral winds outward with equal spacing between successive turns.
Problem 9 Find the equation of the tangent to r = 2 + sin θ r = 2 + \sin\theta r = 2 + sin θ θ = π / 2 \theta = \pi/2 θ = π /2
Hint 9 Find the Cartesian coordinates of the point, then compute d y / d x dy/dx d y / d x
Answer 9 At θ = π / 2 \theta = \pi/2 θ = π /2 r = 3 r = 3 r = 3 ( x , y ) = ( 3 cos ( π / 2 ) , 3 sin ( π / 2 ) ) = ( 0 , 3 ) (x, y) = (3\cos(\pi/2), 3\sin(\pi/2)) = (0, 3) ( x , y ) = ( 3 cos ( π /2 ) , 3 sin ( π /2 )) = ( 0 , 3 )
d r / d θ = cos θ dr/d\theta = \cos\theta d r / d θ = cos θ θ = π / 2 \theta = \pi/2 θ = π /2 d r / d θ = 0 dr/d\theta = 0 d r / d θ = 0
d y d x = 0 ⋅ 1 + 3 ⋅ 0 0 ⋅ 0 − 3 ⋅ 1 = 0 − 3 = 0 \dfrac{dy}{dx} = \dfrac{0\cdot 1 + 3\cdot 0}{0\cdot 0 - 3\cdot 1} = \dfrac{0}{-3} = 0 d x d y = 0 ⋅ 0 − 3 ⋅ 1 0 ⋅ 1 + 3 ⋅ 0 = − 3 0 = 0
The tangent is horizontal: y = 3 y = 3 y = 3
Problem 10 Find the area enclosed by the limacon r = 1 + 2 cos θ r = 1 + 2\cos\theta r = 1 + 2 cos θ
Hint 10 The inner loop occurs where r < 0 r < 0 r < 0 1 + 2 cos θ < 0 1 + 2\cos\theta < 0 1 + 2 cos θ < 0 θ \theta θ 1 2 r 2 d θ \frac{1}{2}r^2\,d\theta 2 1 r 2 d θ
Answer 10 r = 0 r = 0 r = 0 1 + 2 cos θ = 0 ⟹ cos θ = − 1 / 2 ⟹ θ = 2 π / 3 , 4 π / 3 1 + 2\cos\theta = 0 \implies \cos\theta = -1/2 \implies \theta = 2\pi/3, 4\pi/3 1 + 2 cos θ = 0 ⟹ cos θ = − 1/2 ⟹ θ = 2 π /3 , 4 π /3
The inner loop is traced from θ = 2 π / 3 \theta = 2\pi/3 θ = 2 π /3 θ = 4 π / 3 \theta = 4\pi/3 θ = 4 π /3
A = 1 2 ∫ 2 π / 3 4 π / 3 ( 1 + 2 cos θ ) 2 d θ A = \dfrac{1}{2}\displaystyle\int_{2\pi/3}^{4\pi/3}(1+2\cos\theta)^2\,d\theta A = 2 1 ∫ 2 π /3 4 π /3 ( 1 + 2 cos θ ) 2 d θ
= 1 2 ∫ 2 π / 3 4 π / 3 ( 1 + 4 cos θ + 4 cos 2 θ ) d θ = 1 2 ∫ 2 π / 3 4 π / 3 ( 3 + 4 cos θ + 2 cos 2 θ ) d θ = \dfrac{1}{2}\displaystyle\int_{2\pi/3}^{4\pi/3}(1+4\cos\theta+4\cos^2\theta)\,d\theta = \dfrac{1}{2}\int_{2\pi/3}^{4\pi/3}\left(3+4\cos\theta+2\cos 2\theta\right)d\theta = 2 1 ∫ 2 π /3 4 π /3 ( 1 + 4 cos θ + 4 cos 2 θ ) d θ = 2 1 ∫ 2 π /3 4 π /3 ( 3 + 4 cos θ + 2 cos 2 θ ) d θ
= 1 2 [ 3 θ + 4 sin θ + sin 2 θ ] 2 π / 3 4 π / 3 = \dfrac{1}{2}\left[3\theta + 4\sin\theta + \sin 2\theta\right]_{2\pi/3}^{4\pi/3} = 2 1 [ 3 θ + 4 sin θ + sin 2 θ ] 2 π /3 4 π /3
= 1 2 [ ( 4 π − 2 3 + 3 / 2 ) − ( 2 π + 2 3 − 3 / 2 ) ] = \dfrac{1}{2}\left[\left(4\pi - 2\sqrt{3} + \sqrt{3}/2\right) - \left(2\pi + 2\sqrt{3} - \sqrt{3}/2\right)\right] = 2 1 [ ( 4 π − 2 3 + 3 /2 ) − ( 2 π + 2 3 − 3 /2 ) ]
= 1 2 [ 2 π − 3 3 ] = π − 3 3 2 = \dfrac{1}{2}\left[2\pi - 3\sqrt{3}\right] = \pi - \dfrac{3\sqrt{3}}{2} = 2 1 [ 2 π − 3 3 ] = π − 2 3 3
7. Advanced Worked Examples Example 7.1: Area between two curves with careful intersection analysis Problem. Find the area of the region that lies inside both r = 1 + cos θ r = 1 + \cos\theta r = 1 + cos θ r = 3 cos θ r = 3\cos\theta r = 3 cos θ
Solution. Setting 1 + cos θ = 3 cos θ 1 + \cos\theta = 3\cos\theta 1 + cos θ = 3 cos θ
1 = 2 cos θ ⟹ θ = ± π 3 1 = 2\cos\theta \implies \theta = \pm\frac{\pi}{3} 1 = 2 cos θ ⟹ θ = ± 3 π
Both curves are symmetric about the initial line, so we compute from 0 0 0 π / 3 \pi/3 π /3
For 0 ≤ θ ≤ π / 3 0 \leq \theta \leq \pi/3 0 ≤ θ ≤ π /3 3 cos θ ≥ 1 + cos θ 3\cos\theta \geq 1 + \cos\theta 3 cos θ ≥ 1 + cos θ 2 cos θ ≥ 1 2\cos\theta \geq 1 2 cos θ ≥ 1 r outer = 3 cos θ r_{\text{outer}} = 3\cos\theta r outer = 3 cos θ r inner = 1 + cos θ r_{\text{inner}} = 1 + \cos\theta r inner = 1 + cos θ
A = 2 ⋅ 1 2 ∫ 0 π / 3 [ 9 cos 2 θ − ( 1 + cos θ ) 2 ] d θ = ∫ 0 π / 3 [ 9 cos 2 θ − 1 − 2 cos θ − cos 2 θ ] d θ A = 2\cdot\frac{1}{2}\int_0^{\pi/3}\bigl[9\cos^2\theta - (1+\cos\theta)^2\bigr]\,d\theta = \int_0^{\pi/3}\bigl[9\cos^2\theta - 1 - 2\cos\theta - \cos^2\theta\bigr]\,d\theta A = 2 ⋅ 2 1 ∫ 0 π /3 [ 9 cos 2 θ − ( 1 + cos θ ) 2 ] d θ = ∫ 0 π /3 [ 9 cos 2 θ − 1 − 2 cos θ − cos 2 θ ] d θ
= ∫ 0 π / 3 [ 8 cos 2 θ − 2 cos θ − 1 ] d θ = ∫ 0 π / 3 [ 4 ( 1 + cos 2 θ ) − 2 cos θ − 1 ] d θ = \int_0^{\pi/3}\bigl[8\cos^2\theta - 2\cos\theta - 1\bigr]\,d\theta = \int_0^{\pi/3}\bigl[4(1+\cos 2\theta) - 2\cos\theta - 1\bigr]\,d\theta = ∫ 0 π /3 [ 8 cos 2 θ − 2 cos θ − 1 ] d θ = ∫ 0 π /3 [ 4 ( 1 + cos 2 θ ) − 2 cos θ − 1 ] d θ
= ∫ 0 π / 3 ( 3 + 4 cos 2 θ − 2 cos θ ) d θ = [ 3 θ + 2 sin 2 θ − 2 sin θ ] 0 π / 3 = \int_0^{\pi/3}(3 + 4\cos 2\theta - 2\cos\theta)\,d\theta = \left[3\theta + 2\sin 2\theta - 2\sin\theta\right]_0^{\pi/3} = ∫ 0 π /3 ( 3 + 4 cos 2 θ − 2 cos θ ) d θ = [ 3 θ + 2 sin 2 θ − 2 sin θ ] 0 π /3
= π + 2 ⋅ 3 2 − 2 ⋅ 3 2 = π = \pi + 2\cdot\frac{\sqrt{3}}{2} - 2\cdot\frac{\sqrt{3}}{2} = \pi = π + 2 ⋅ 2 3 − 2 ⋅ 2 3 = π
Example 7.2: Converting Cartesian to polar and sketching Problem. Convert x 2 + y 2 = 2 y x^2 + y^2 = 2y x 2 + y 2 = 2 y
Solution. Substituting x = r\cos\theta$$y = r\sin\theta$$r^2 = x^2 + y^2 :
r 2 = 2 r sin θ ⟹ r = 2 sin θ ( r ≠ 0 ) r^2 = 2r\sin\theta \implies r = 2\sin\theta \quad (r \neq 0) r 2 = 2 r sin θ ⟹ r = 2 sin θ ( r = 0 )
This is a circle with centre ( 0 , 1 ) (0, 1) ( 0 , 1 ) 1 1 1 r = 2 a sin θ r = 2a\sin\theta r = 2 a sin θ a = 1 a = 1 a = 1
The curve passes through the pole at θ = 0 \theta = 0 θ = 0 θ = π \theta = \pi θ = π r = 2 r = 2 r = 2 θ = π / 2 \theta = \pi/2 θ = π /2
Example 7.3: Finding where tangents are vertical or horizontal Problem. For the cardioid r = 2 ( 1 − cos θ ) r = 2(1 - \cos\theta) r = 2 ( 1 − cos θ )
Solution. r = 2(1 - \cos\theta)$$\dfrac{dr}{d\theta} = 2\sin\theta .
Horizontal tangents occur when d y d θ = 0 \dfrac{dy}{d\theta} = 0 d θ d y = 0
d r d θ sin θ + r cos θ = 0 ⟹ 2 sin 2 θ + 2 ( 1 − cos θ ) cos θ = 0 \frac{dr}{d\theta}\sin\theta + r\cos\theta = 0 \implies 2\sin^2\theta + 2(1 - \cos\theta)\cos\theta = 0 d θ d r sin θ + r cos θ = 0 ⟹ 2 sin 2 θ + 2 ( 1 − cos θ ) cos θ = 0
2 sin 2 θ + 2 cos θ − 2 cos 2 θ = 0 ⟹ 2 ( 1 − cos 2 θ ) + 2 cos θ − 2 cos 2 θ = 0 2\sin^2\theta + 2\cos\theta - 2\cos^2\theta = 0 \implies 2(1 - \cos^2\theta) + 2\cos\theta - 2\cos^2\theta = 0 2 sin 2 θ + 2 cos θ − 2 cos 2 θ = 0 ⟹ 2 ( 1 − cos 2 θ ) + 2 cos θ − 2 cos 2 θ = 0
2 − 2 cos 2 θ + 2 cos θ − 2 cos 2 θ = 0 ⟹ 2 − 4 cos 2 θ + 2 cos θ = 0 2 - 2\cos^2\theta + 2\cos\theta - 2\cos^2\theta = 0 \implies 2 - 4\cos^2\theta + 2\cos\theta = 0 2 − 2 cos 2 θ + 2 cos θ − 2 cos 2 θ = 0 ⟹ 2 − 4 cos 2 θ + 2 cos θ = 0
2 cos 2 θ − cos θ − 1 = 0 ⟹ ( 2 cos θ + 1 ) ( cos θ − 1 ) = 0 2\cos^2\theta - \cos\theta - 1 = 0 \implies (2\cos\theta + 1)(\cos\theta - 1) = 0 2 cos 2 θ − cos θ − 1 = 0 ⟹ ( 2 cos θ + 1 ) ( cos θ − 1 ) = 0
cos θ = − 1 / 2 ⟹ θ = 2 π / 3 \cos\theta = -1/2 \implies \theta = 2\pi/3 cos θ = − 1/2 ⟹ θ = 2 π /3 θ = 4 π / 3 \theta = 4\pi/3 θ = 4 π /3 cos θ = 1 ⟹ θ = 0 \cos\theta = 1 \implies \theta = 0 cos θ = 1 ⟹ θ = 0
At θ = 2 π / 3 \theta = 2\pi/3 θ = 2 π /3 r = 2 ( 1 + 1 / 2 ) = 3 r = 2(1 + 1/2) = 3 r = 2 ( 1 + 1/2 ) = 3 ( − 3 / 2 , 3 3 / 2 ) (-3/2, 3\sqrt{3}/2) ( − 3/2 , 3 3 /2 ) θ = 4 π / 3 \theta = 4\pi/3 θ = 4 π /3 r = 2 ( 1 + 1 / 2 ) = 3 r = 2(1 + 1/2) = 3 r = 2 ( 1 + 1/2 ) = 3 ( − 3 / 2 , − 3 3 / 2 ) (-3/2, -3\sqrt{3}/2) ( − 3/2 , − 3 3 /2 ) θ = 0 \theta = 0 θ = 0 r = 0 r = 0 r = 0
Example 7.4: Volume of revolution in polar coordinates Problem. The region enclosed by r = 1 + cos θ r = 1 + \cos\theta r = 1 + cos θ
Solution. Using the parametric volume formula with y = r sin θ = ( 1 + cos θ ) sin θ y = r\sin\theta = (1+\cos\theta)\sin\theta y = r sin θ = ( 1 + cos θ ) sin θ d x = d x d θ d θ dx = \dfrac{dx}{d\theta}\,d\theta d x = d θ d x d θ
x = r cos θ = ( 1 + cos θ ) cos θ x = r\cos\theta = (1+\cos\theta)\cos\theta x = r cos θ = ( 1 + cos θ ) cos θ d x d θ = − sin θ − 2 cos θ sin θ = − sin θ ( 1 + 2 cos θ ) \dfrac{dx}{d\theta} = -\sin\theta - 2\cos\theta\sin\theta = -\sin\theta(1 + 2\cos\theta) d θ d x = − sin θ − 2 cos θ sin θ = − sin θ ( 1 + 2 cos θ )
By symmetry, integrate from 0 0 0 π \pi π
V = 2 π ∫ 0 π y 2 d x d θ d θ = 2 π ∫ 0 π ( 1 + cos θ ) 2 sin 2 θ ⋅ [ − sin θ ( 1 + 2 cos θ ) ] d θ V = 2\pi\int_0^{\pi} y^2\,\frac{dx}{d\theta}\,d\theta = 2\pi\int_0^{\pi}(1+\cos\theta)^2\sin^2\theta\cdot[-\sin\theta(1+2\cos\theta)]\,d\theta V = 2 π ∫ 0 π y 2 d θ d x d θ = 2 π ∫ 0 π ( 1 + cos θ ) 2 sin 2 θ ⋅ [ − sin θ ( 1 + 2 cos θ )] d θ
Let u = cos θ u = \cos\theta u = cos θ d u = − sin θ d θ du = -\sin\theta\,d\theta d u = − sin θ d θ θ = 0 \theta = 0 θ = 0 u = 1 u = 1 u = 1 θ = π \theta = \pi θ = π u = − 1 u = -1 u = − 1
V = 2 π ∫ − 1 1 ( 1 + u ) 2 ( 1 − u 2 ) ( 1 + 2 u ) d u V = 2\pi\int_{-1}^{1}(1+u)^2(1-u^2)(1+2u)\,du V = 2 π ∫ − 1 1 ( 1 + u ) 2 ( 1 − u 2 ) ( 1 + 2 u ) d u
Expanding ( 1 + u ) 2 ( 1 − u 2 ) ( 1 + 2 u ) = ( 1 + 2 u + u 2 ) ( 1 − u 2 ) ( 1 + 2 u ) (1+u)^2(1-u^2)(1+2u) = (1+2u+u^2)(1-u^2)(1+2u) ( 1 + u ) 2 ( 1 − u 2 ) ( 1 + 2 u ) = ( 1 + 2 u + u 2 ) ( 1 − u 2 ) ( 1 + 2 u )
Note: ( 1 + u ) 2 ( 1 − u 2 ) = ( 1 + u ) 2 ( 1 − u ) ( 1 + u ) = ( 1 + u ) 3 ( 1 − u ) (1+u)^2(1-u^2) = (1+u)^2(1-u)(1+u) = (1+u)^3(1-u) ( 1 + u ) 2 ( 1 − u 2 ) = ( 1 + u ) 2 ( 1 − u ) ( 1 + u ) = ( 1 + u ) 3 ( 1 − u )
So the integrand is ( 1 + u ) 4 ( 1 − u ) (1+u)^4(1-u) ( 1 + u ) 4 ( 1 − u )
Let v = 1 + u v = 1+u v = 1 + u
V = 2 π ∫ 0 2 v 4 ( 2 − v ) d v = 2 π ∫ 0 2 ( 2 v 4 − v 5 ) d v = 2 π [ 2 v 5 5 − v 6 6 ] 0 2 V = 2\pi\int_0^2 v^4(2-v)\,dv = 2\pi\int_0^2(2v^4 - v^5)\,dv = 2\pi\left[\frac{2v^5}{5} - \frac{v^6}{6}\right]_0^2 V = 2 π ∫ 0 2 v 4 ( 2 − v ) d v = 2 π ∫ 0 2 ( 2 v 4 − v 5 ) d v = 2 π [ 5 2 v 5 − 6 v 6 ] 0 2
= 2 π ( 64 5 − 64 6 ) = 2 π ⋅ 64 ( 6 − 5 ) 30 = 128 π 15 = 2\pi\left(\frac{64}{5} - \frac{64}{6}\right) = 2\pi\cdot\frac{64(6-5)}{30} = \frac{128\pi}{15} = 2 π ( 5 64 − 6 64 ) = 2 π ⋅ 30 64 ( 6 − 5 ) = 15 128 π
8. Connections to Other Topics 8.1 Polar coordinates and complex numbers The polar form of a complex number z = r e i θ z = re^{i\theta} z = r e i θ ( r , θ ) (r, \theta) ( r , θ ) r 1 e i θ 1 ⋅ r 2 e i θ 2 = r 1 r 2 e i ( θ 1 + θ 2 ) r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 e^{i(\theta_1+\theta_2)} r 1 e i θ 1 ⋅ r 2 e i θ 2 = r 1 r 2 e i ( θ 1 + θ 2 ) Complex Numbers .
8.2 Polar area and further calculus The polar area formula 1 2 ∫ r 2 d θ \frac{1}{2}\int r^2\,d\theta 2 1 ∫ r 2 d θ Further Calculus .
8.3 Polar curves and parametric differentiation The gradient formula for polar curves is derived from parametric differentiation. The expressions For d x / d θ dx/d\theta d x / d θ d y / d θ dy/d\theta d y / d θ Further Calculus .
9. Additional Exam-Style Questions Question 11 A curve has polar equation r = a ( 1 + cos θ ) r = a(1 + \cos\theta) r = a ( 1 + cos θ ) a > 0 a > 0 a > 0
(a) Find the area enclosed by the curve.
(b) Find the equation of the tangent at θ = π / 2 \theta = \pi/2 θ = π /2
Solution (a) By symmetry:
A = 2 ⋅ 1 2 ∫ 0 π a 2 ( 1 + cos θ ) 2 d θ = a 2 ∫ 0 π ( 3 2 + 2 cos θ + cos 2 θ 2 ) d θ = 3 π a 2 2 A = 2\cdot\frac{1}{2}\int_0^{\pi}a^2(1+\cos\theta)^2\,d\theta = a^2\int_0^{\pi}\left(\frac{3}{2}+2\cos\theta+\frac{\cos 2\theta}{2}\right)d\theta = \frac{3\pi a^2}{2} A = 2 ⋅ 2 1 ∫ 0 π a 2 ( 1 + cos θ ) 2 d θ = a 2 ∫ 0 π ( 2 3 + 2 cos θ + 2 c o s 2 θ ) d θ = 2 3 π a 2
(b) At θ = π / 2 \theta = \pi/2 θ = π /2 r = a r = a r = a ( 0 , a ) (0, a) ( 0 , a )
d r / d θ = − a sin θ dr/d\theta = -a\sin\theta d r / d θ = − a sin θ d r / d θ ∣ π / 2 = − a dr/d\theta|_{\pi/2} = -a d r / d θ ∣ π /2 = − a
d y d x = ( − a ) ( 1 ) + a ( 0 ) ( − a ) ( 0 ) − a ( 1 ) = − a − a = 1 \frac{dy}{dx} = \frac{(-a)(1) + a(0)}{(-a)(0) - a(1)} = \frac{-a}{-a} = 1 d x d y = ( − a ) ( 0 ) − a ( 1 ) ( − a ) ( 1 ) + a ( 0 ) = − a − a = 1
Tangent: y − a = 1 ( x − 0 ) y - a = 1(x - 0) y − a = 1 ( x − 0 ) y = x + a y = x + a y = x + a
Question 12 Find the area of the finite region bounded by the curve r = 2 + cos θ r = 2 + \cos\theta r = 2 + cos θ θ = 0 \theta = 0 θ = 0 θ = π \theta = \pi θ = π
Solution A = 1 2 ∫ 0 π ( 2 + cos θ ) 2 d θ = 1 2 ∫ 0 π ( 4 + 4 cos θ + cos 2 θ ) d θ A = \frac{1}{2}\int_0^{\pi}(2+\cos\theta)^2\,d\theta = \frac{1}{2}\int_0^{\pi}(4 + 4\cos\theta + \cos^2\theta)\,d\theta A = 2 1 ∫ 0 π ( 2 + cos θ ) 2 d θ = 2 1 ∫ 0 π ( 4 + 4 cos θ + cos 2 θ ) d θ
= 1 2 ∫ 0 π ( 9 2 + 4 cos θ + cos 2 θ 2 ) d θ = 1 2 [ 9 θ 2 + 4 sin θ + sin 2 θ 4 ] 0 π = 9 π 4 = \frac{1}{2}\int_0^{\pi}\left(\frac{9}{2} + 4\cos\theta + \frac{\cos 2\theta}{2}\right)d\theta = \frac{1}{2}\left[\frac{9\theta}{2} + 4\sin\theta + \frac{\sin 2\theta}{4}\right]_0^{\pi} = \frac{9\pi}{4} = 2 1 ∫ 0 π ( 2 9 + 4 cos θ + 2 c o s 2 θ ) d θ = 2 1 [ 2 9 θ + 4 sin θ + 4 s i n 2 θ ] 0 π = 4 9 π
Question 13 Prove that the polar curve r = a cos θ r = \dfrac{a}{\cos\theta} r = cos θ a
Solution r = a cos θ ⟹ r cos θ = a ⟹ x = a r = \dfrac{a}{\cos\theta} \implies r\cos\theta = a \implies x = a r = cos θ a ⟹ r cos θ = a ⟹ x = a
This is the vertical line x = a x = a x = a ■ \blacksquare ■
Question 14 The curve C C C r = 4 sin 2 θ r = 4\sin 2\theta r = 4 sin 2 θ 0 ≤ θ ≤ π / 2 0 \leq \theta \leq \pi/2 0 ≤ θ ≤ π /2
(a) Find the area of one petal.
(b) Find the angle at which the tangent to C C C
Solution (a) One petal of r = 4 sin 2 θ r = 4\sin 2\theta r = 4 sin 2 θ θ = 0 \theta = 0 θ = 0 θ = π / 2 \theta = \pi/2 θ = π /2
A = 1 2 ∫ 0 π / 2 16 sin 2 2 θ d θ = 8 ∫ 0 π / 2 1 − cos 4 θ 2 d θ = 4 [ θ − sin 4 θ 4 ] 0 π / 2 = 2 π A = \frac{1}{2}\int_0^{\pi/2}16\sin^2 2\theta\,d\theta = 8\int_0^{\pi/2}\frac{1-\cos 4\theta}{2}\,d\theta = 4\left[\theta - \frac{\sin 4\theta}{4}\right]_0^{\pi/2} = 2\pi A = 2 1 ∫ 0 π /2 16 sin 2 2 θ d θ = 8 ∫ 0 π /2 2 1 − c o s 4 θ d θ = 4 [ θ − 4 s i n 4 θ ] 0 π /2 = 2 π
(b) Tangent parallel to the initial line means d y / d θ = 0 dy/d\theta = 0 d y / d θ = 0
r = 4 sin 2 θ r = 4\sin 2\theta r = 4 sin 2 θ d r / d θ = 8 cos 2 θ dr/d\theta = 8\cos 2\theta d r / d θ = 8 cos 2 θ
d y d θ = 8 cos 2 θ sin θ + 4 sin 2 θ cos θ = 8 cos 2 θ sin θ + 8 sin θ cos 2 θ \dfrac{dy}{d\theta} = 8\cos 2\theta\sin\theta + 4\sin 2\theta\cos\theta = 8\cos 2\theta\sin\theta + 8\sin\theta\cos^2\theta d θ d y = 8 cos 2 θ sin θ + 4 sin 2 θ cos θ = 8 cos 2 θ sin θ + 8 sin θ cos 2 θ
= 8 sin θ ( cos 2 θ + cos 2 θ ) = 8 sin θ ( 2 cos 2 θ − 1 + cos 2 θ ) = 8 sin θ ( 3 cos 2 θ − 1 ) = 8\sin\theta(\cos 2\theta + \cos^2\theta) = 8\sin\theta(2\cos^2\theta - 1 + \cos^2\theta) = 8\sin\theta(3\cos^2\theta - 1) = 8 sin θ ( cos 2 θ + cos 2 θ ) = 8 sin θ ( 2 cos 2 θ − 1 + cos 2 θ ) = 8 sin θ ( 3 cos 2 θ − 1 )
= 0 = 0 = 0 sin θ = 0 \sin\theta = 0 sin θ = 0 θ = 0 \theta = 0 θ = 0 r = 0 r = 0 r = 0 cos 2 θ = 1 / 3 \cos^2\theta = 1/3 cos 2 θ = 1/3 cos θ = ± 1 / 3 \cos\theta = \pm 1/\sqrt{3} cos θ = ± 1/ 3
For 0 ≤ θ ≤ π / 2 0 \leq \theta \leq \pi/2 0 ≤ θ ≤ π /2 θ = arccos ( 1 / 3 ) \theta = \arccos(1/\sqrt{3}) θ = arccos ( 1/ 3 )
Question 15 Find the maximum distance from the origin to any point on the curve r = 2 + 3 sin θ r = 2 + 3\sin\theta r = 2 + 3 sin θ
Solution The distance from the origin is ∣ r ∣ |r| ∣ r ∣ 2 + 3 sin θ ≥ 2 − 3 = − 1 2 + 3\sin\theta \geq 2 - 3 = -1 2 + 3 sin θ ≥ 2 − 3 = − 1 ∣ r ∣ |r| ∣ r ∣ r r r r r r
d r / d θ = 3 cos θ = 0 ⟹ θ = π / 2 dr/d\theta = 3\cos\theta = 0 \implies \theta = \pi/2 d r / d θ = 3 cos θ = 0 ⟹ θ = π /2 θ = 3 π / 2 \theta = 3\pi/2 θ = 3 π /2
At θ = π / 2 \theta = \pi/2 θ = π /2 r = 5 r = 5 r = 5 θ = 3 π / 2 \theta = 3\pi/2 θ = 3 π /2 r = − 1 r = -1 r = − 1
∣ r ∣ = 5 |r| = 5 ∣ r ∣ = 5 θ = π / 2 \theta = \pi/2 θ = π /2 ∣ r ∣ = 1 |r| = 1 ∣ r ∣ = 1 θ = 3 π / 2 \theta = 3\pi/2 θ = 3 π /2
The maximum distance is 5 \boxed{5} 5
8. Advanced Worked Examples Example 8.1: Area enclosed by a limacon Problem. Find the area enclosed by the limacon r = 2 + cos θ r = 2 + \cos\theta r = 2 + cos θ
Solution. Since r = 2 + cos θ > 0 r = 2 + \cos\theta > 0 r = 2 + cos θ > 0 θ \theta θ
A = 1 2 ∫ 0 2 π ( 2 + cos θ ) 2 d θ = 1 2 ∫ 0 2 π ( 4 + 4 cos θ + cos 2 θ ) d θ A = \frac{1}{2}\int_0^{2\pi} (2+\cos\theta)^2\,d\theta = \frac{1}{2}\int_0^{2\pi} (4 + 4\cos\theta + \cos^2\theta)\,d\theta A = 2 1 ∫ 0 2 π ( 2 + cos θ ) 2 d θ = 2 1 ∫ 0 2 π ( 4 + 4 cos θ + cos 2 θ ) d θ
= 1 2 ∫ 0 2 π ( 4 + 4 cos θ + 1 + cos 2 θ 2 ) d θ = 1 2 ∫ 0 2 π ( 9 2 + 4 cos θ + cos 2 θ 2 ) d θ = \frac{1}{2}\int_0^{2\pi} \!\left(4 + 4\cos\theta + \frac{1+\cos 2\theta}{2}\right)d\theta = \frac{1}{2}\int_0^{2\pi} \!\left(\frac{9}{2} + 4\cos\theta + \frac{\cos 2\theta}{2}\right)d\theta = 2 1 ∫ 0 2 π ( 4 + 4 cos θ + 2 1 + c o s 2 θ ) d θ = 2 1 ∫ 0 2 π ( 2 9 + 4 cos θ + 2 c o s 2 θ ) d θ
= 1 2 [ 9 θ 2 + 4 sin θ + sin 2 θ 4 ] 0 2 π = 1 2 ⋅ 9 π = 9 π 2 = \frac{1}{2}\left[\frac{9\theta}{2} + 4\sin\theta + \frac{\sin 2\theta}{4}\right]_0^{2\pi} = \frac{1}{2} \cdot 9\pi = \boxed{\frac{9\pi}{2}} = 2 1 [ 2 9 θ + 4 sin θ + 4 s i n 2 θ ] 0 2 π = 2 1 ⋅ 9 π = 2 9 π
Example 8.2: Tangents to a polar curve Problem. Find the angle ψ \psi ψ r = a ( 1 + cos θ ) r = a(1+\cos\theta) r = a ( 1 + cos θ ) θ = π / 2 \theta = \pi/2 θ = π /2
Solution. tan ψ = r d r / d θ \tan\psi = \dfrac{r}{dr/d\theta} tan ψ = d r / d θ r
d r / d θ = − a sin θ dr/d\theta = -a\sin\theta d r / d θ = − a sin θ θ = π / 2 \theta = \pi/2 θ = π /2 r = a r = a r = a d r / d θ = − a dr/d\theta = -a d r / d θ = − a
tan ψ = a − a = − 1 ⟹ ψ = 3 π 4 \tan\psi = \dfrac{a}{-a} = -1 \implies \psi = \dfrac{3\pi}{4} tan ψ = − a a = − 1 ⟹ ψ = 4 3 π 135 ° 135° 135°
The tangent makes an angle of 135 ° 135° 135°
Example 8.3: Cartesian equation of a spiral Problem. Convert the spiral r = e 2 θ r = e^{2\theta} r = e 2 θ
Solution. r = e 2 θ ⟹ ln r = 2 θ ⟹ θ = 1 2 ln r r = e^{2\theta} \implies \ln r = 2\theta \implies \theta = \dfrac{1}{2}\ln r r = e 2 θ ⟹ ln r = 2 θ ⟹ θ = 2 1 ln r
Since θ = arctan ( y / x ) \theta = \arctan(y/x) θ = arctan ( y / x ) r = x 2 + y 2 r = \sqrt{x^2+y^2} r = x 2 + y 2
arctan ( y x ) = 1 2 ln ( x 2 + y 2 ) \arctan\!\left(\frac{y}{x}\right) = \frac{1}{2}\ln(x^2+y^2) arctan ( x y ) = 2 1 ln ( x 2 + y 2 )
y x = exp ( 1 2 ln ( x 2 + y 2 ) ) = x 2 + y 2 \frac{y}{x} = \exp\!\left(\frac{1}{2}\ln(x^2+y^2)\right) = \sqrt{x^2+y^2} x y = exp ( 2 1 ln ( x 2 + y 2 ) ) = x 2 + y 2
y 2 x 2 = x 2 + y 2 ⟹ y 2 = x 2 ( x 2 + y 2 ) \frac{y^2}{x^2} = x^2 + y^2 \implies y^2 = x^2(x^2+y^2) x 2 y 2 = x 2 + y 2 ⟹ y 2 = x 2 ( x 2 + y 2 )
Example 8.4: Area between two polar curves Problem. Find the area inside r = 3 cos θ r = 3\cos\theta r = 3 cos θ r = 1 + cos θ r = 1 + \cos\theta r = 1 + cos θ
Solution. First find intersection points: 3 cos θ = 1 + cos θ ⟹ 2 cos θ = 1 ⟹ θ = ± π / 3 3\cos\theta = 1 + \cos\theta \implies 2\cos\theta = 1 \implies \theta = \pm\pi/3 3 cos θ = 1 + cos θ ⟹ 2 cos θ = 1 ⟹ θ = ± π /3
A = 1 2 ∫ − π / 3 π / 3 [ ( 3 cos θ ) 2 − ( 1 + cos θ ) 2 ] d θ A = \frac{1}{2}\int_{-\pi/3}^{\pi/3} \!\left[(3\cos\theta)^2 - (1+\cos\theta)^2\right]\,d\theta A = 2 1 ∫ − π /3 π /3 [ ( 3 cos θ ) 2 − ( 1 + cos θ ) 2 ] d θ
= 1 2 ∫ − π / 3 π / 3 ( 9 cos 2 θ − 1 − 2 cos θ − cos 2 θ ) d θ = 1 2 ∫ − π / 3 π / 3 ( 8 cos 2 θ − 2 cos θ − 1 ) d θ = \frac{1}{2}\int_{-\pi/3}^{\pi/3} (9\cos^2\theta - 1 - 2\cos\theta - \cos^2\theta)\,d\theta = \frac{1}{2}\int_{-\pi/3}^{\pi/3} (8\cos^2\theta - 2\cos\theta - 1)\,d\theta = 2 1 ∫ − π /3 π /3 ( 9 cos 2 θ − 1 − 2 cos θ − cos 2 θ ) d θ = 2 1 ∫ − π /3 π /3 ( 8 cos 2 θ − 2 cos θ − 1 ) d θ
= 1 2 ∫ − π / 3 π / 3 ( 4 + 4 cos 2 θ − 2 cos θ − 1 ) d θ = 1 2 ∫ − π / 3 π / 3 ( 3 + 4 cos 2 θ − 2 cos θ ) d θ = \frac{1}{2}\int_{-\pi/3}^{\pi/3} \!\left(4 + 4\cos 2\theta - 2\cos\theta - 1\right)d\theta = \frac{1}{2}\int_{-\pi/3}^{\pi/3} (3 + 4\cos 2\theta - 2\cos\theta)\,d\theta = 2 1 ∫ − π /3 π /3 ( 4 + 4 cos 2 θ − 2 cos θ − 1 ) d θ = 2 1 ∫ − π /3 π /3 ( 3 + 4 cos 2 θ − 2 cos θ ) d θ
= 1 2 [ 3 θ + 2 sin 2 θ − 2 sin θ ] − π / 3 π / 3 = \frac{1}{2}\left[3\theta + 2\sin 2\theta - 2\sin\theta\right]_{-\pi/3}^{\pi/3} = 2 1 [ 3 θ + 2 sin 2 θ − 2 sin θ ] − π /3 π /3
= 1 2 [ π + 2 sin 2 π 3 − 2 sin π 3 − ( − π − 2 sin 2 π 3 + 2 sin π 3 ) ] = \frac{1}{2}\left[\pi + 2\sin\frac{2\pi}{3} - 2\sin\frac{\pi}{3} - \left(-\pi - 2\sin\frac{2\pi}{3} + 2\sin\frac{\pi}{3}\right)\right] = 2 1 [ π + 2 sin 3 2 π − 2 sin 3 π − ( − π − 2 sin 3 2 π + 2 sin 3 π ) ]
= 1 2 [ 2 π + 2 3 − 3 + 2 3 − 3 ] = 1 2 ( 2 π + 2 3 ) = π + 3 = \frac{1}{2}\left[2\pi + 2\sqrt{3} - \sqrt{3} + 2\sqrt{3} - \sqrt{3}\right] = \frac{1}{2}(2\pi + 2\sqrt{3}) = \boxed{\pi + \sqrt{3}} = 2 1 [ 2 π + 2 3 − 3 + 2 3 − 3 ] = 2 1 ( 2 π + 2 3 ) = π + 3
Example 8.5: Converting a parametric curve to polar Problem. The curve x = 2 t 1 + t 2 x = \dfrac{2t}{1+t^2} x = 1 + t 2 2 t y = 1 − t 2 1 + t 2 y = \dfrac{1-t^2}{1+t^2} y = 1 + t 2 1 − t 2
Solution. x 2 + y 2 = 4 t 2 + ( 1 − t 2 ) 2 ( 1 + t 2 ) 2 = 4 t 2 + 1 − 2 t 2 + t 4 ( 1 + t 2 ) 2 = ( 1 + t 2 ) 2 ( 1 + t 2 ) 2 = 1 x^2 + y^2 = \dfrac{4t^2 + (1-t^2)^2}{(1+t^2)^2} = \dfrac{4t^2 + 1 - 2t^2 + t^4}{(1+t^2)^2} = \dfrac{(1+t^2)^2}{(1+t^2)^2} = 1 x 2 + y 2 = ( 1 + t 2 ) 2 4 t 2 + ( 1 − t 2 ) 2 = ( 1 + t 2 ) 2 4 t 2 + 1 − 2 t 2 + t 4 = ( 1 + t 2 ) 2 ( 1 + t 2 ) 2 = 1
So r = 1 r = 1 r = 1 t t t
cos θ = x r = 2 t 1 + t 2 \cos\theta = \dfrac{x}{r} = \dfrac{2t}{1+t^2} cos θ = r x = 1 + t 2 2 t sin θ = 1 − t 2 1 + t 2 \sin\theta = \dfrac{1-t^2}{1+t^2} sin θ = 1 + t 2 1 − t 2 t = tan ( θ / 2 ) t = \tan(\theta/2) t = tan ( θ /2 )
cos θ = cos θ \cos\theta = \cos\theta cos θ = cos θ sin θ = sin θ \sin\theta = \sin\theta sin θ = sin θ
Example 8.6: Arc length of a cardioid Problem. Find the total arc length of the cardioid r = a ( 1 + cos θ ) r = a(1 + \cos\theta) r = a ( 1 + cos θ )
Solution. s = ∫ 0 2 π r 2 + ( d r d θ ) 2 d θ s = \displaystyle\int_0^{2\pi} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta s = ∫ 0 2 π r 2 + ( d θ d r ) 2 d θ
r = a ( 1 + cos θ ) r = a(1+\cos\theta) r = a ( 1 + cos θ ) d r / d θ = − a sin θ dr/d\theta = -a\sin\theta d r / d θ = − a sin θ
r 2 + ( d r / d θ ) 2 = a 2 ( 1 + cos θ ) 2 + a 2 sin 2 θ = a 2 ( 1 + 2 cos θ + cos 2 θ + sin 2 θ ) = 2 a 2 ( 1 + cos θ ) = 4 a 2 cos 2 ( θ / 2 ) r^2 + (dr/d\theta)^2 = a^2(1+\cos\theta)^2 + a^2\sin^2\theta = a^2(1+2\cos\theta+\cos^2\theta+\sin^2\theta) = 2a^2(1+\cos\theta) = 4a^2\cos^2(\theta/2) r 2 + ( d r / d θ ) 2 = a 2 ( 1 + cos θ ) 2 + a 2 sin 2 θ = a 2 ( 1 + 2 cos θ + cos 2 θ + sin 2 θ ) = 2 a 2 ( 1 + cos θ ) = 4 a 2 cos 2 ( θ /2 )
s = ∫ 0 2 π 2 a ∣ cos ( θ / 2 ) ∣ d θ s = \int_0^{2\pi} 2a|\cos(\theta/2)|\,d\theta s = ∫ 0 2 π 2 a ∣ cos ( θ /2 ) ∣ d θ
For 0 ≤ θ ≤ 2 π 0 \leq \theta \leq 2\pi 0 ≤ θ ≤ 2 π cos ( θ / 2 ) ≥ 0 \cos(\theta/2) \geq 0 cos ( θ /2 ) ≥ 0 0 ≤ θ ≤ π 0 \leq \theta \leq \pi 0 ≤ θ ≤ π ≤ 0 \leq 0 ≤ 0 π ≤ θ ≤ 2 π \pi \leq \theta \leq 2\pi π ≤ θ ≤ 2 π
s = 2 a [ ∫ 0 π cos ( θ / 2 ) d θ + ∫ π 2 π ( − cos ( θ / 2 ) ) d θ ] = 2 a [ 2 + 2 ] = 8 a s = 2a\!\left[\int_0^{\pi} \cos(\theta/2)\,d\theta + \int_{\pi}^{2\pi} (-\cos(\theta/2))\,d\theta\right] = 2a[2+2] = \boxed{8a} s = 2 a [ ∫ 0 π cos ( θ /2 ) d θ + ∫ π 2 π ( − cos ( θ /2 )) d θ ] = 2 a [ 2 + 2 ] = 8 a
9. Common Pitfalls Pitfall Correct Approach Forgetting the 1 2 \frac{1}{2} 2 1 A = 1 2 ∫ r 2 d θ A = \dfrac{1}{2}\displaystyle\int r^2\,d\theta A = 2 1 ∫ r 2 d θ ∫ r 2 d θ \int r^2\,d\theta ∫ r 2 d θ Not checking if r r r If r < 0 r < 0 r < 0 Confusing the angle ψ \psi ψ θ \theta θ tan ψ = r / ( d r / d θ ) \tan\psi = r / (dr/d\theta) tan ψ = r / ( d r / d θ ) θ + ψ \theta + \psi θ + ψ Using the wrong limits for symmetric curves Exploit symmetry: if the curve is symmetric about θ = 0 \theta = 0 θ = 0 0 0 0 π \pi π
10. Additional Exam-Style Questions Question 8 Find the area of the region enclosed by one loop of the curve r 2 = 4 cos 2 θ r^2 = 4\cos 2\theta r 2 = 4 cos 2 θ
Solution This is a lemniscate. One loop is traced for − π / 4 ≤ θ ≤ π / 4 -\pi/4 \leq \theta \leq \pi/4 − π /4 ≤ θ ≤ π /4
A = 1 2 ∫ − π / 4 π / 4 4 cos 2 θ d θ = 2 [ sin 2 θ 2 ] − π / 4 π / 4 = 2 ( 1 − ( − 1 ) ) = 4 ... wait A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} 4\cos 2\theta\,d\theta = 2\!\left[\frac{\sin 2\theta}{2}\right]_{-\pi/4}^{\pi/4} = 2(1-(-1)) = 4 \text{... wait} A = 2 1 ∫ − π /4 π /4 4 cos 2 θ d θ = 2 [ 2 s i n 2 θ ] − π /4 π /4 = 2 ( 1 − ( − 1 )) = 4 ... wait
A = ∫ − π / 4 π / 4 2 cos 2 θ d θ = [ sin 2 θ ] − π / 4 π / 4 = 1 − ( − 1 ) = 2 A = \int_{-\pi/4}^{\pi/4} 2\cos 2\theta\,d\theta = [\sin 2\theta]_{-\pi/4}^{\pi/4} = 1 - (-1) = 2 A = ∫ − π /4 π /4 2 cos 2 θ d θ = [ sin 2 θ ] − π /4 π /4 = 1 − ( − 1 ) = 2
A = 1 2 ∫ r 2 d θ = 1 2 ∫ − π / 4 π / 4 4 cos 2 θ d θ = 2 [ sin 2 θ ] − π / 4 π / 4 = 2 × 2 = 4 A = \dfrac{1}{2}\displaystyle\int r^2\,d\theta = \dfrac{1}{2}\int_{-\pi/4}^{\pi/4} 4\cos 2\theta\,d\theta = 2[\sin 2\theta]_{-\pi/4}^{\pi/4} = 2 \times 2 = \boxed{4} A = 2 1 ∫ r 2 d θ = 2 1 ∫ − π /4 π /4 4 cos 2 θ d θ = 2 [ sin 2 θ ] − π /4 π /4 = 2 × 2 = 4
Question 9 Prove that the tangent to r = a sec θ r = a\sec\theta r = a sec θ
Solution r = a sec θ ⟹ d r / d θ = a sec θ tan θ r = a\sec\theta \implies dr/d\theta = a\sec\theta\tan\theta r = a sec θ ⟹ d r / d θ = a sec θ tan θ
tan ψ = r d r / d θ = a sec θ a sec θ tan θ = cot θ \tan\psi = \dfrac{r}{dr/d\theta} = \dfrac{a\sec\theta}{a\sec\theta\tan\theta} = \cot\theta tan ψ = d r / d θ r = a sec θ tan θ a sec θ = cot θ
So ψ = π / 2 − θ \psi = \pi/2 - \theta ψ = π /2 − θ θ + ψ = π / 2 \theta + \psi = \pi/2 θ + ψ = π /2 ■ \blacksquare ■
Question 10 Find the Cartesian equation of the curve r = 2 a cos θ + 2 b sin θ r = 2a\cos\theta + 2b\sin\theta r = 2 a cos θ + 2 b sin θ
Solution r = 2 a cos θ + 2 b sin θ ⟹ r 2 = 2 a r cos θ + 2 b r sin θ r = 2a\cos\theta + 2b\sin\theta \implies r^2 = 2ar\cos\theta + 2br\sin\theta r = 2 a cos θ + 2 b sin θ ⟹ r 2 = 2 a r cos θ + 2 b r sin θ
x 2 + y 2 = 2 a x + 2 b y ⟹ ( x − a ) 2 + ( y − b ) 2 = a 2 + b 2 x^2 + y^2 = 2ax + 2by \implies (x-a)^2 + (y-b)^2 = a^2 + b^2 x 2 + y 2 = 2 a x + 2 b y ⟹ ( x − a ) 2 + ( y − b ) 2 = a 2 + b 2
This is a circle with centre ( a , b ) (a, b) ( a , b ) a 2 + b 2 \sqrt{a^2+b^2} a 2 + b 2
11. Connections to Other Topics 11.1 Polar coordinates and complex numbers The polar form z = r ( cos θ + i sin θ ) z = r(\cos\theta+i\sin\theta) z = r ( cos θ + i sin θ ) ( r , θ ) (r,\theta) ( r , θ ) Complex Numbers .
11.2 Polar curves and calculus Finding tangents, areas, and arc lengths in polar coordinates requires differentiation and Integration. See Further Calculus .
11.3 Polar coordinates and parametric equations Polar curves are a special case of parametric equations with x = r ( θ ) cos θ x = r(\theta)\cos\theta x = r ( θ ) cos θ y = r ( θ ) sin θ y = r(\theta)\sin\theta y = r ( θ ) sin θ
12. Key Results Summary Quantity Formula Cartesian from polar x = r cos θ x = r\cos\theta x = r cos θ y = r sin θ y = r\sin\theta y = r sin θ Polar from Cartesian r = x 2 + y 2 r = \sqrt{x^2+y^2} r = x 2 + y 2 θ = arctan ( y / x ) \theta = \arctan(y/x) θ = arctan ( y / x ) Polar area A = 1 2 ∫ α β r 2 d θ A = \dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta A = 2 1 ∫ α β r 2 d θ Polar arc length s = ∫ α β r 2 + ( d r d θ ) 2 d θ s = \displaystyle\int_\alpha^\beta \sqrt{r^2+\left(\dfrac{dr}{d\theta}\right)^2}\,d\theta s = ∫ α β r 2 + ( d θ d r ) 2 d θ Tangent-radius angle tan ψ = r d r / d θ \tan\psi = \dfrac{r}{dr/d\theta} tan ψ = d r / d θ r Tangent to x x x d y d x = r ′ sin θ + r cos θ r ′ cos θ − r sin θ \dfrac{dy}{dx} = \dfrac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta} d x d y = r ′ cos θ − r sin θ r ′ sin θ + r cos θ
13. Further Exam-Style Questions Question 11 A curve has polar equation r = a ( 1 + cos θ ) r = a(1+\cos\theta) r = a ( 1 + cos θ )
Solution Since r > 0 r > 0 r > 0 θ \theta θ 0 0 0 2 π 2\pi 2 π
A = 1 2 ∫ 0 2 π a 2 ( 1 + cos θ ) 2 d θ = a 2 2 ∫ 0 2 π ( 3 2 + 2 cos θ + cos 2 θ 2 ) d θ A = \dfrac{1}{2}\displaystyle\int_0^{2\pi} a^2(1+\cos\theta)^2\,d\theta = \dfrac{a^2}{2}\displaystyle\int_0^{2\pi} \!\left(\dfrac{3}{2}+2\cos\theta+\dfrac{\cos 2\theta}{2}\right)d\theta A = 2 1 ∫ 0 2 π a 2 ( 1 + cos θ ) 2 d θ = 2 a 2 ∫ 0 2 π ( 2 3 + 2 cos θ + 2 cos 2 θ ) d θ
= a 2 2 [ 3 θ 2 + 2 sin θ + sin 2 θ 4 ] 0 2 π = a 2 2 ⋅ 3 π = 3 π a 2 2 = \dfrac{a^2}{2}\!\left[\dfrac{3\theta}{2}+2\sin\theta+\dfrac{\sin 2\theta}{4}\right]_0^{2\pi} = \dfrac{a^2}{2}\cdot 3\pi = \boxed{\dfrac{3\pi a^2}{2}} = 2 a 2 [ 2 3 θ + 2 sin θ + 4 sin 2 θ ] 0 2 π = 2 a 2 ⋅ 3 π = 2 3 π a 2
Question 12 Prove that the curve r = 2 a cos θ r = 2a\cos\theta r = 2 a cos θ a a a ( a , 0 ) (a, 0) ( a , 0 )
Solution r = 2 a cos θ ⟹ r 2 = 2 a r cos θ ⟹ x 2 + y 2 = 2 a x ⟹ ( x − a ) 2 + y 2 = a 2 r = 2a\cos\theta \implies r^2 = 2ar\cos\theta \implies x^2+y^2 = 2ax \implies (x-a)^2+y^2 = a^2 r = 2 a cos θ ⟹ r 2 = 2 a r cos θ ⟹ x 2 + y 2 = 2 a x ⟹ ( x − a ) 2 + y 2 = a 2
This is a circle with centre ( a , 0 ) (a,0) ( a , 0 ) a a a ■ \blacksquare ■
14. Advanced Topics 14.1 The pedal equation The pedal equation of a curve gives the distance p p p r r r
p = r sin ψ = r 2 r 2 + ( d r / d θ ) 2 p = r\sin\psi = \frac{r^2}{\sqrt{r^2+(dr/d\theta)^2}} p = r sin ψ = r 2 + ( d r / d θ ) 2 r 2
14.2 The p − r p-r p − r For a conic with focus at the origin and directrix at distance d d d r = e d 1 + e cos θ r = \dfrac{ed}{1+e\cos\theta} r = 1 + e cos θ e d e e e
e < 1 e < 1 e < 1 e = 1 e = 1 e = 1 e > 1 e > 1 e > 1 14.3 Rose curves Curves of the form r = a cos ( n θ ) r = a\cos(n\theta) r = a cos ( n θ ) r = a sin ( n θ ) r = a\sin(n\theta) r = a sin ( n θ )
If n n n n n n If n n n 2 n 2n 2 n 14.4 Limacons r = a + b cos θ r = a + b\cos\theta r = a + b cos θ
a > b a > b a > b a = b a = b a = b a < b a < b a < b 15. Further Exam-Style Questions Question 13 Sketch the curve r = 1 + 2 cos θ r = 1 + 2\cos\theta r = 1 + 2 cos θ
Solution Since 1 + 2 cos θ = 0 1 + 2\cos\theta = 0 1 + 2 cos θ = 0 cos θ = − 1 / 2 \cos\theta = -1/2 cos θ = − 1/2 θ = 2 π / 3 \theta = 2\pi/3 θ = 2 π /3 θ = 4 π / 3 \theta = 4\pi/3 θ = 4 π /3
Area of inner loop: A = 1 2 ∫ 2 π / 3 4 π / 3 ( 1 + 2 cos θ ) 2 d θ A = \dfrac{1}{2}\displaystyle\int_{2\pi/3}^{4\pi/3} (1+2\cos\theta)^2\,d\theta A = 2 1 ∫ 2 π /3 4 π /3 ( 1 + 2 cos θ ) 2 d θ
= 1 2 ∫ 2 π / 3 4 π / 3 ( 1 + 4 cos θ + 4 cos 2 θ ) d θ = 1 2 ∫ 2 π / 3 4 π / 3 ( 3 + 4 cos θ + 2 cos 2 θ ) d θ = \dfrac{1}{2}\displaystyle\int_{2\pi/3}^{4\pi/3} (1+4\cos\theta+4\cos^2\theta)\,d\theta = \dfrac{1}{2}\displaystyle\int_{2\pi/3}^{4\pi/3} (3+4\cos\theta+2\cos 2\theta)\,d\theta = 2 1 ∫ 2 π /3 4 π /3 ( 1 + 4 cos θ + 4 cos 2 θ ) d θ = 2 1 ∫ 2 π /3 4 π /3 ( 3 + 4 cos θ + 2 cos 2 θ ) d θ
= 1 2 [ 3 θ + 4 sin θ + sin 2 θ ] 2 π / 3 4 π / 3 = \dfrac{1}{2}\!\left[3\theta+4\sin\theta+\sin 2\theta\right]_{2\pi/3}^{4\pi/3} = 2 1 [ 3 θ + 4 sin θ + sin 2 θ ] 2 π /3 4 π /3
= 1 2 [ ( 4 π − 2 π ) + 4 ( − 3 2 − 3 2 ) + ( 3 2 − 3 2 ) ] = 1 2 ( 2 π − 4 3 ) = π − 2 3 = \dfrac{1}{2}\!\left[(4\pi-2\pi)+4\!\left(-\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}\right)+\!\left(\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}\right)\right] = \dfrac{1}{2}(2\pi-4\sqrt{3}) = \boxed{\pi-2\sqrt{3}} = 2 1 [ ( 4 π − 2 π ) + 4 ( − 2 3 − 2 3 ) + ( 2 3 − 2 3 ) ] = 2 1 ( 2 π − 4 3 ) = π − 2 3
Question 14 Prove that the area enclosed by one petal of r = a cos ( 3 θ ) r = a\cos(3\theta) r = a cos ( 3 θ ) π a 2 12 \dfrac{\pi a^2}{12} 12 π a 2
Solution One petal is traced for − π / 6 ≤ θ ≤ π / 6 -\pi/6 \leq \theta \leq \pi/6 − π /6 ≤ θ ≤ π /6
A = 1 2 ∫ − π / 6 π / 6 a 2 cos 2 ( 3 θ ) d θ = a 2 2 ∫ − π / 6 π / 6 1 + cos 6 θ 2 d θ A = \dfrac{1}{2}\displaystyle\int_{-\pi/6}^{\pi/6} a^2\cos^2(3\theta)\,d\theta = \dfrac{a^2}{2}\displaystyle\int_{-\pi/6}^{\pi/6} \frac{1+\cos 6\theta}{2}\,d\theta A = 2 1 ∫ − π /6 π /6 a 2 cos 2 ( 3 θ ) d θ = 2 a 2 ∫ − π /6 π /6 2 1 + cos 6 θ d θ
= a 2 4 [ θ + sin 6 θ 6 ] − π / 6 π / 6 = a 2 4 ( π 3 + 0 ) = π a 2 12 = \dfrac{a^2}{4}\!\left[\theta+\dfrac{\sin 6\theta}{6}\right]_{-\pi/6}^{\pi/6} = \dfrac{a^2}{4}\!\left(\dfrac{\pi}{3}+0\right) = \boxed{\dfrac{\pi a^2}{12}} = 4 a 2 [ θ + 6 sin 6 θ ] − π /6 π /6 = 4 a 2 ( 3 π + 0 ) = 12 π a 2 ■ \blacksquare ■
16. Further Advanced Topics Using the focus-directrix definition, all conics with a focus at the origin have polar equation:
r = e d 1 + e cos θ r = \frac{ed}{1+e\cos\theta} r = 1 + e c o s θ e d
Where e e e d d d
e = 0 e = 0 e = 0 r = d r = d r = d 0 < e < 1 0 < e < 1 0 < e < 1 e = 1 e = 1 e = 1 e > 1 e > 1 e > 1 16.2 Spirals Archimedean spiral: r = a θ r = a\theta r = a θ Logarithmic spiral: r = a e b θ r = ae^{b\theta} r = a e b θ Hyperbolic spiral: r = a / θ r = a/\theta r = a / θ The logarithmic spiral appears in nature (nautilus shells, hurricanes, galaxies).
16.3 Tangents at the pole If r = 0 r = 0 r = 0 θ = θ 0 \theta = \theta_0 θ = θ 0 θ = θ 0 + π 2 \theta = \theta_0 + \dfrac{\pi}{2} θ = θ 0 + 2 π
16.4 Converting parametric curves to polar Many parametric curves can be simplified in polar form. The cardioid, limacon, and rose curves are Most expressed in polar coordinates.
17. Further Exam-Style Questions Question 15 Find the area inside r = 1 + sin θ r = 1 + \sin\theta r = 1 + sin θ r = 1 r = 1 r = 1
Solution 1 + sin θ = 1 1 + \sin\theta = 1 1 + sin θ = 1 sin θ = 0 \sin\theta = 0 sin θ = 0 θ = 0 , π \theta = 0, \pi θ = 0 , π
The curve r = 1 + sin θ r = 1 + \sin\theta r = 1 + sin θ r = 1 r = 1 r = 1
The required area is:
A = 1 2 ∫ 0 2 π [ ( 1 + sin θ ) 2 − 1 ] d θ = 1 2 ∫ 0 2 π ( 2 sin θ + sin 2 θ ) d θ A = \dfrac{1}{2}\displaystyle\int_0^{2\pi} [(1+\sin\theta)^2 - 1]\,d\theta = \dfrac{1}{2}\displaystyle\int_0^{2\pi} (2\sin\theta + \sin^2\theta)\,d\theta A = 2 1 ∫ 0 2 π [( 1 + sin θ ) 2 − 1 ] d θ = 2 1 ∫ 0 2 π ( 2 sin θ + sin 2 θ ) d θ
= 1 2 ∫ 0 2 π ( 2 sin θ + 1 − cos 2 θ 2 ) d θ = 1 2 [ − 2 cos θ + θ 2 − sin 2 θ 4 ] 0 2 π = 1 2 ⋅ π 2 = π 4 = \dfrac{1}{2}\displaystyle\int_0^{2\pi} \!\left(2\sin\theta + \frac{1-\cos 2\theta}{2}\right)d\theta = \dfrac{1}{2}\!\left[-2\cos\theta + \frac{\theta}{2} - \frac{\sin 2\theta}{4}\right]_0^{2\pi} = \dfrac{1}{2}\cdot\dfrac{\pi}{2} = \boxed{\dfrac{\pi}{4}} = 2 1 ∫ 0 2 π ( 2 sin θ + 2 1 − cos 2 θ ) d θ = 2 1 [ − 2 cos θ + 2 θ − 4 sin 2 θ ] 0 2 π = 2 1 ⋅ 2 π = 4 π
Question 16 Prove that the spiral r = e a θ r = e^{a\theta} r = e a θ θ = θ 0 \theta = \theta_0 θ = θ 0
Solution At θ = θ 0 \theta = \theta_0 θ = θ 0 r = e a θ 0 r = e^{a\theta_0} r = e a θ 0
For a given θ 0 \theta_0 θ 0 r r r ■ \blacksquare ■
Intuition Polar coordinates describe points by distance and direction rather than horizontal and vertical displacement. Imagine standing at the origin and pointing a compass: the angle tells you which direction to look, and the radius tells you how far to walk. Many curves that are complicated in Cartesian coordinates become simple in polar form because their symmetry aligns with radial distance from a center point. A cardioid is shaped like a heart because its radius varies smoothly with angle, while rose curves create petal patterns when the radius oscillates between zero and its maximum. The polar area formula works by summing infinitesimal wedges, like slicing a pie into infinitely thin pieces and adding their areas.
Cross-References Complex Numbers — The polar form z = re^{i theta} is identical to polar coordinates, linking complex multiplication to combining angles and moduli.Further Calculus — The polar area formula and tangent gradients are direct applications of integration and differentiation techniques.Matrices — Polar representations of rotation matrices connect coordinate transformations to trigonometric parameterisation.Further Algebra — De Moivre’s theorem uses polar form to derive multiple angles and roots of complex numbers.