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Polar Calculus - Interactive Lesson | Study Mondo
Polar Calculus - Complete Interactive Lesson Part 1: Core Concepts Polar Calculus
Part 1 of 7 — Polar Coordinates & Conversions
In polar coordinates, each point is described by ( r , θ ) (r, \theta) ( r , θ ) r r r θ \theta θ x x x
Conversion Formulas
x = r cos θ , y = r sin θ \boxed{x = r\cos\theta, \quad y = r\sin\theta} x = r cos θ , y = r sin θ
r = x 2 + y 2 , tan θ = y x \boxed{r = \sqrt{x^2+y^2}, \quad \tan\theta = \frac{y}{x}} r = x 2 + y 2
Cartesian Polar ( 1 , 1 ) (1, 1) ( 1 , 1 ) ( 2 , π / 4 ) (\sqrt{2},\, \pi/4) ( 2
Key Fact: A point ( r , θ ) (r, \theta) ( r , θ ) ( − r , θ + π ) (-r, \theta + \pi) ( − r , θ + π ) ( r , θ + 2 π ) (r, \theta + 2\pi) ( r ,
Common Polar Equations
Equation Shape r = a r = a r = a Circle of radius $ θ = c \theta = c θ = c Line through origin at angle c c c r = 2 a
Key Takeaways
Polar coordinates ( r , θ ) (r, \theta) ( r , θ )
Convert with x = r cos θ x = r\cos\theta x = r cos θ y = r sin θ y = r\sin\theta y = r
Part 2: Worked Examples Polar Calculus
Part 2 of 7 — Graphing Polar Curves & Symmetry
Graphing Strategy
To sketch r = f ( θ ) r = f(\theta) r = f ( θ )
Make a table of θ \theta θ r r r
Plot points in polar coordinates
Check for symmetry to reduce work
Note where r = 0 r = 0
Part 3: Problem-Solving Patterns Polar Calculus
Part 3 of 7 — Area in Polar Coordinates
The Polar Area Formula
The area enclosed by r = f ( θ ) r = f(\theta) r = f ( θ ) θ = α \theta = \alpha θ = α θ = β \theta = \beta θ =
Part 4: Graphs and Interpretation Polar Calculus
Part 4 of 7 — Slopes of Polar Curves
To find d y / d x dy/dx d y / d x r = f ( θ ) r = f(\theta) r = f ( θ ) θ \theta θ
x = r cos θ
Part 5: Applications Polar Calculus
Part 5 of 7 — Arc Length in Polar Coordinates
The Polar Arc Length Formula
For r = f ( θ ) r = f(\theta) r = f ( θ ) θ = α \theta = \alpha θ = α θ = β \theta = \beta θ =
Part 6: Exam Strategy Polar Calculus
Part 6 of 7 — Problem-Solving Workshop
Mixed practice covering polar coordinates, graphing, area, slopes, and arc length.
AP FRQ-Style Problem
Consider r = 2 cos θ r = 2\cos\theta r = 2 cos θ 0 ≤ θ ≤ π / 2 0 \le \theta \le \pi/2 0 ≤ θ ≤ π /2
Part 7: Mixed Review Polar Calculus
Part 7 of 7 — Comprehensive Review
Complete Polar Formula Sheet
Formula Expression Conversion x = r cos θ x = r\cos\theta x = r cos θ y = r sin θ y = r\sin\theta y = r sin θ Area
,
tan
θ
=
x y
,
π
/4
)
( 0 , 3 ) (0, 3) ( 0 , 3 ) ( 3 , π / 2 ) (3,\, \pi/2) ( 3 , π /2 )
( − 1 , 3 ) (-1, \sqrt{3}) ( − 1 , 3 ) ( 2 , 2 π / 3 ) (2,\, 2\pi/3) ( 2 , 2 π /3 )
( 3 , 0 ) (3, 0) ( 3 , 0 ) ( 3 , 0 ) (3,\, 0) ( 3 , 0 )
θ
+
2 π )
cos θ r = 2a\cos\theta r = 2 a cos θ
r = 2 a sin θ r = 2a\sin\theta r = 2 a sin θ Circle of radius $
r = a + b cos θ r = a + b\cos\theta r = a + b cos θ Lima\c{c}on (with loop if $
r = a cos ( n θ ) r = a\cos(n\theta) r = a cos ( n θ ) Rose (n n n 2 n 2n 2 n
r = 2 cos θ r = 2\cos\theta r = 2 cos θ r r r r 2 = 2 r cos θ ⟹ x 2 + y 2 = 2 x r^2 = 2r\cos\theta \implies x^2+y^2 = 2x r 2 = 2 r cos θ ⟹ x 2 + y 2 = 2 x sin
θ
r 2 = x 2 + y 2 r^2 = x^2+y^2 r 2 = x 2 + y 2 Representations are not unique: ( r , θ ) = ( − r , θ + π ) = ( r , θ + 2 π ) (r, \theta) = (-r, \theta+\pi) = (r, \theta+2\pi) ( r , θ ) = ( − r , θ + π ) = ( r , θ + 2 π )
Know the standard curves: circles, cardioids, roses, lima\c{c}ons
Next: Part 2 covers graphing polar curves and symmetry analysis.
r
=
0
Note where r < 0 r < 0 r < 0
Symmetry Tests Symmetry Test Replace About x x x f ( − θ ) = f ( θ ) f(-\theta) = f(\theta) f ( − θ ) = f ( θ ) θ → − θ \theta \to -\theta θ → − θ About y y y f ( π − θ ) = f ( θ ) f(\pi - \theta) = f(\theta) f ( π − θ ) = f ( θ ) θ → π − θ \theta \to \pi - \theta θ → About origin f ( θ + π ) = f ( θ ) f(\theta + \pi) = f(\theta) f ( θ + π ) = f ( θ ) θ → θ + π \theta \to \theta + \pi θ → θ + π
Important: These are sufficient but not necessary conditions. A curve may have symmetry that the test does not detect.
Graphing Examples
Cardioid r = 1 + cos θ r = 1 + \cos\theta r = 1 + cos θ
θ \theta θ 0 0 0 π / 3 \pi/3 π /3 π / 2 \pi/2 π /2 2 π / 3 2\pi/3 2 π /3 π \pi π r r r 2 2 2 3 / 2 3/2 3/2 1 1 1 1 / 2 1/2 1/2 0 0
Symmetric about the x x x cos ( − θ ) = cos θ \cos(-\theta) = \cos\theta cos ( − θ ) = cos θ
Rose r = sin ( 3 θ ) r = \sin(3\theta) r = sin ( 3 θ )
r = 0 r = 0 r = 0 3 θ = n π 3\theta = n\pi 3 θ = nπ θ = 0 , π / 3 , 2 π / 3 , π \theta = 0, \pi/3, 2\pi/3, \pi θ = 0 , π /3 ,
Lemniscate r 2 = 4 cos ( 2 θ ) r^2 = 4\cos(2\theta) r 2 = 4 cos ( 2 θ )
Only exists when cos ( 2 θ ) ≥ 0 \cos(2\theta) \ge 0 cos ( 2 θ ) ≥ 0 − π / 4 ≤ θ ≤ π / 4 -\pi/4 \le \theta \le \pi/4 − π /4 ≤ θ ≤ π /4 3 π / 4 ≤ θ ≤ 5 π / 4 3\pi/4 \le \theta \le 5\pi/4
Key Takeaways
Graph polar curves by making a θ \theta θ r r r
Use symmetry tests to reduce graphing work
Know the shapes: circles, cardioids, lima\c{c}ons, roses, lemniscates, spirals
Negative r r r
Next: Part 3 covers polar area with the integral formula.
β
A = 1 2 ∫ α β r 2 d θ = 1 2 ∫ α β [ f ( θ ) ] 2 d θ \boxed{A = \frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta = \frac{1}{2}\int_{\alpha}^{\beta} [f(\theta)]^2\,d\theta} A = 2 1 ∫ α β r 2 d θ = 2 1 ∫ α β [ f ( θ ) ] 2 d θ
Why 1 2 r 2 \frac{1}{2}r^2 2 1 r 2 Each thin sector has area 1 2 r 2 d θ \frac{1}{2}r^2\,d\theta 2 1 r 2 d θ r d θ r\,d\theta r d θ r r r
Critical: Choose α \alpha α β \beta β
Example 1: Area of a Cardioid
Find the area enclosed by r = 1 + cos θ r = 1 + \cos\theta r = 1 + cos θ
Full curve: 0 ≤ θ ≤ 2 π 0 \le \theta \le 2\pi 0 ≤ θ ≤ 2 π x x x A = 2 ⋅ 1 2 ∫ 0 π ( 1 + cos θ ) 2 d θ A = 2 \cdot \frac{1}{2}\int_0^{\pi}(1+\cos\theta)^2\,d\theta A = 2 ⋅ 2 1 ∫ 0
= ∫ 0 π ( 1 + 2 cos θ + cos 2 θ ) d θ = ∫ 0 π ( 3 2 + 2 cos θ + cos 2 θ 2 ) d θ = \int_0^{\pi}(1 + 2\cos\theta + \cos^2\theta)\,d\theta = \int_0^{\pi}\left(\frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}\right)d\theta = ∫ 0 π ( 1 + 2
= [ 3 θ 2 + 2 sin θ + sin 2 θ 4 ] 0 π = 3 π 2 = \left[\frac{3\theta}{2} + 2\sin\theta + \frac{\sin 2\theta}{4}\right]_0^{\pi} = \frac{3\pi}{2} = [ 2 3 θ + 2 sin θ +
Example 2: Area of One Petal
One petal of r = sin ( 2 θ ) r = \sin(2\theta) r = sin ( 2 θ ) 0 ≤ θ ≤ π / 2 0 \le \theta \le \pi/2 0 ≤ θ ≤ π /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 = π 8 A = \frac{1}{2}\int_0^{\pi/2}\sin^2(2\theta)\,d\theta = \frac{1}{2}\int_0^{\pi/2}\frac{1-\cos 4\theta}{2}\,d\theta = \frac{\pi}{8} A = 2 1 ∫
Key Takeaways
Polar area: A = 1 2 ∫ α β r 2 d θ A = \frac{1}{2}\int_{\alpha}^{\beta}r^2\,d\theta A = 2 1 ∫ α β r 2 d θ
Between curves: 1 2 ∫ ( r outer 2 − r inner 2 ) d θ \frac{1}{2}\int(r_{\text{outer}}^2 - r_{\text{inner}}^2)\,d\theta 2 1 ∫ ( r outer 2
Use half-angle identities: cos 2 θ = 1 + cos 2 θ 2 \cos^2\theta = \frac{1+\cos 2\theta}{2} cos 2 θ = 2 1 + c o s 2 θ
Exploit symmetry to simplify calculations
Next: Part 4 covers slopes of polar curves (d y / d x dy/dx d y / d x
= f ( θ ) cos θ , y = r sin θ = f ( θ ) sin θ x = r\cos\theta = f(\theta)\cos\theta, \quad y = r\sin\theta = f(\theta)\sin\theta x = r cos θ = f ( θ ) cos θ , y = r sin θ = f ( θ ) sin θ
The Slope Formula 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
This comes from the product rule:
d x / d θ = f ′ cos θ − f sin θ dx/d\theta = f'\cos\theta - f\sin\theta d x / d θ = f ′ cos θ − f sin θ d y / d θ = f ′ sin θ + f cos θ dy/d\theta = f'\sin\theta + f\cos\theta d y / d θ = f ′ sin θ + f cos θ Worked Example
Find the slope of r = 1 + cos θ r = 1 + \cos\theta r = 1 + cos θ at 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 d r / d θ = − sin ( π / 3 ) = − 3 / 2 dr/d\theta = -\sin(\pi/3) = -\sqrt{3}/2 d r / d θ = − sin ( π /3 ) = − 3
d y d x = ( − 3 / 2 ) sin ( π / 3 ) + ( 3 / 2 ) cos ( π / 3 ) ( − 3 / 2 ) cos ( π / 3 ) − ( 3 / 2 ) sin ( π / 3 ) \frac{dy}{dx} = \frac{(-\sqrt{3}/2)\sin(\pi/3) + (3/2)\cos(\pi/3)}{(-\sqrt{3}/2)\cos(\pi/3) - (3/2)\sin(\pi/3)} d x d
= ( − 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{(-\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 =
The tangent is horizontal at θ = π / 3 \theta = \pi/3 θ = π /3
AP Tip: Horizontal tangents occur when d y / d θ = 0 dy/d\theta = 0 d y / d θ = 0 d x / d θ ≠ 0 dx/d\theta \ne 0 d x / d θ = 0 d x / d θ
Key Takeaways
Polar slope: d y d x = ( d r / d θ ) sin θ + r cos θ ( d r / d θ ) cos θ − r sin θ \frac{dy}{dx} = \frac{(dr/d\theta)\sin\theta + r\cos\theta}{(dr/d\theta)\cos\theta - r\sin\theta} d x d y = ( d r / d θ ) c o s θ − r s i n θ ( d r / d θ ) s i n θ + r c o s θ
Treat polar as parametric with θ \theta θ
Horizontal tangent: numerator = 0 = 0 = 0 ≠ 0 \ne 0 = 0
At the origin: tangent line has slope tan θ 0 \tan\theta_0 tan θ 0 f ( θ 0 ) = 0 f(\theta_0) = 0 f ( θ 0 ) = 0
Next: Part 5 covers arc length in polar coordinates.
β
L = ∫ α β r 2 + ( d r d θ ) 2 d θ \boxed{L = \int_{\alpha}^{\beta}\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta} L = ∫ α β r 2 + ( d θ d r ) 2 d θ
Derivation: From the parametric formula with x = r cos θ x = r\cos\theta x = r cos θ y = r sin θ y = r\sin\theta y = r sin θ
( d x / d θ ) 2 + ( d y / d θ ) 2 = ( r ′ ) 2 cos 2 θ − 2 r r ′ cos θ sin θ + r 2 sin 2 θ (dx/d\theta)^2 + (dy/d\theta)^2 = (r')^2\cos^2\theta - 2rr'\cos\theta\sin\theta + r^2\sin^2\theta ( d x / d θ ) 2 + ( d y / d θ ) 2 = ( r ′ ) 2 cos 2 θ − 2 r r ′ cos θ sin θ + r 2 sin 2 θ + ( r ′ ) 2 sin 2 θ + 2 r r ′ sin θ cos θ + r 2 cos 2 θ = ( r ′ ) 2 + r 2 + (r')^2\sin^2\theta + 2rr'\sin\theta\cos\theta + r^2\cos^2\theta = (r')^2 + r^2 + ( r ′ ) 2 sin 2
Key Fact: The polar arc length formula is simpler than the parametric one because the cross terms cancel!
Example 1: Arc Length of a Circle
r = a r = a r = a d r / d θ = 0 dr/d\theta = 0 d r / d θ = 0
L = ∫ 0 2 π a 2 + 0 d θ = a ⋅ 2 π = 2 π a ✓ L = \int_0^{2\pi}\sqrt{a^2+0}\,d\theta = a \cdot 2\pi = 2\pi a \checkmark L = ∫ 0 2 π a 2 + 0 d a ⋅ 2 π = 2 πa ✓
Example 2: Cardioid r = 1 + cos θ r = 1 + \cos\theta r = 1 + cos θ
d r / d θ = − sin θ dr/d\theta = -\sin\theta d r / d θ = − sin θ
L = ∫ 0 2 π ( 1 + cos θ ) 2 + sin 2 θ d θ L = \int_0^{2\pi}\sqrt{(1+\cos\theta)^2 + \sin^2\theta}\,d\theta L = ∫ 0 2 π ( 1 + cos θ
= ∫ 0 2 π 2 + 2 cos θ d θ = ∫ 0 2 π 4 cos 2 ( θ / 2 ) d θ = ∫ 0 2 π 2 ∣ cos ( θ / 2 ) ∣ d θ = \int_0^{2\pi}\sqrt{2 + 2\cos\theta}\,d\theta = \int_0^{2\pi}\sqrt{4\cos^2(\theta/2)}\,d\theta = \int_0^{2\pi}2|\cos(\theta/2)|\,d\theta = ∫ 0 2 π
Using symmetry: = 4 ∫ 0 π cos ( θ / 2 ) d θ = 4 [ 2 sin ( θ / 2 ) ] 0 π = 8 = 4\int_0^{\pi}\cos(\theta/2)\,d\theta = 4[2\sin(\theta/2)]_0^{\pi} = 8 = 4 ∫ 0 π cos ( θ /2 ) d θ =
AP Note: The half-angle identity 1 + cos θ = 2 cos 2 ( θ / 2 ) 1 + \cos\theta = 2\cos^2(\theta/2) 1 + cos θ = 2 cos 2 ( θ /2 )
Key Takeaways
Polar arc length: L = ∫ r 2 + ( d r / d θ ) 2 d θ L = \int\sqrt{r^2 + (dr/d\theta)^2}\,d\theta L = ∫ r 2 + ( d r / d θ ) 2 d θ
For a circle r = a r = a r = a [ α , β ] [\alpha, \beta] [ α , β ] a ( β − α ) a(\beta - \alpha) a ( β − α )
Half-angle identities simplify cardioid integrals
Logarithmic spiral r = e θ r = e^\theta r = e θ e θ 2 e^\theta\sqrt{2} e θ 2
Next: Part 6 is a Problem-Solving Workshop with mixed polar problems.
(a)
A = 1 2 ∫ 0 π / 2 4 cos 2 θ d θ = 2 ∫ 0 π / 2 1 + cos 2 θ 2 d θ = [ θ + sin 2 θ 2 ] 0 π / 2 = π 2 A = \frac{1}{2}\int_0^{\pi/2}4\cos^2\theta\,d\theta = 2\int_0^{\pi/2}\frac{1+\cos 2\theta}{2}\,d\theta = [\theta + \frac{\sin 2\theta}{2}]_0^{\pi/2} = \frac{\pi}{2} A = 2 1 ∫ 0 π /2 4 cos 2 θ d θ = 2 ∫ 0 π /2 2 1 + c o s 2 θ d θ = [ θ + 2 s i n 2 θ ] 0 π /2 = 2 π
(b) Find d y / d x dy/dx d y / d x θ = π / 6 \theta = \pi/6 θ = π /6
r = 3 r = \sqrt{3} r = 3 d r / d θ = − 2 sin ( π / 6 ) = − 1 dr/d\theta = -2\sin(\pi/6) = -1 d r / d θ = − 2 sin ( π /6 ) = − 1
d y d x = ( − 1 ) ( 1 / 2 ) + 3 ( 3 / 2 ) ( − 1 ) ( 3 / 2 ) − 3 ( 1 / 2 ) = − 1 / 2 + 3 / 2 − 3 / 2 − 3 / 2 = 1 − 3 = − 1 3 \frac{dy}{dx} = \frac{(-1)(1/2) + \sqrt{3}(\sqrt{3}/2)}{(-1)(\sqrt{3}/2) - \sqrt{3}(1/2)} = \frac{-1/2 + 3/2}{-\sqrt{3}/2 - \sqrt{3}/2} = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}} d x d y = ( − 1 ) ( 3 /2 ) − 3 − 3 /2 − 3 − 3 1 = − 3 1
L = ∫ 0 π / 2 4 cos 2 θ + 4 sin 2 θ d θ = ∫ 0 π / 2 2 d θ = π L = \int_0^{\pi/2}\sqrt{4\cos^2\theta + 4\sin^2\theta}\,d\theta = \int_0^{\pi/2}2\,d\theta = \pi L = ∫ 0 π /2 4 cos 2 θ + 4 sin 2 θ d θ = ∫ 0 π /2 2 d θ = π
Workshop Recap
Polar Problem Checklist:
Identify the curve type (circle, cardioid, rose, etc.)
Determine symmetry and appropriate limits
Set up the correct integral (area: 1 2 r 2 \frac{1}{2}r^2 2 1 r 2 r 2 + ( r ′ ) 2 \sqrt{r^2+(r')^2} r 2 + ( r ′ ) 2
Use trig identities to evaluate
Check your answer against geometric intuition
Coming Up: Part 7 is the Comprehensive Review of polar calculus.
A = 1 2 ∫ α β r 2 d θ A = \frac{1}{2}\int_{\alpha}^{\beta}r^2\,d\theta A = 2 1 ∫ α β r 2 d θ
Area between curves 1 2 ∫ ( r out 2 − r in 2 ) d θ \frac{1}{2}\int(r_{\text{out}}^2 - r_{\text{in}}^2)\,d\theta 2 1 ∫ ( r out 2 − r in 2 ) d θ
Slope d y d x = r ′ sin θ + r cos θ r ′ cos θ − r sin θ \frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta} d x d y = r ′ c o s θ − r s i n θ r ′ s i n θ + r c o s θ
Arc length L = ∫ r 2 + ( r ′ ) 2 d θ L = \int\sqrt{r^2+(r')^2}\,d\theta L = ∫ r 2 + ( r ′ ) 2 d θ
Horizontal tangent d y / d θ = 0 dy/d\theta = 0 d y / d θ = 0 d x / d θ ≠ 0 dx/d\theta \ne 0 d x / d θ = 0
Vertical tangent d x / d θ = 0 dx/d\theta = 0 d x / d θ = 0 d y / d θ ≠ 0 dy/d\theta \ne 0 d y / d θ = 0
Curve Classification Summary
Equation Type Petals/Loops r = a r = a r = a Circle (origin-centered) — r = 2 a cos θ r = 2a\cos\theta r = 2 a cos θ Circle through origin — r = a + a cos θ r = a + a\cos\theta r = a + a cos θ Cardioid 0 loops r = a + b cos θ r = a + b\cos\theta r = a + b cos θ a > r = a + b cos θ r = a + b\cos\theta r = a + b cos θ a < r = a cos ( n θ ) r = a\cos(n\theta) r = a cos ( n θ ) n n n Rose n n n r = a cos ( n θ ) r = a\cos(n\theta) r = a cos ( n θ ) n n n Rose 2 n 2n 2 n r 2 = a cos ( 2 θ ) r^2 = a\cos(2\theta) r 2 = a cos ( 2 θ ) Lemniscate 2 loops
Polar Calculus Complete!
You have mastered:
Polar coordinates and conversions
Graphing polar curves and symmetry analysis
Polar area formula and area between curves
Slopes of polar curves via parametric conversion
Arc length in polar coordinates
AP Exam Note: The polar/parametric FRQ is one of the most predictable on the BC exam. Practice setting up area and arc length integrals with correct limits. Always check for intersections at the origin separately!
π
−
θ
0
2
π
/3
,
π
3 3 3 π / 3 \pi/3 π /3 First petal: 0 ≤ θ ≤ π / 3 0 \le \theta \le \pi/3 0 ≤ θ ≤ π /3 r = 1 r = 1 r = 1 θ = π / 6 \theta = \pi/6 θ = π /6 3 π /4 ≤ θ ≤ 5 π /4
Figure-eight shape through the origin
π
(
1
+
cos θ ) 2 d θ
cos
θ
+
cos 2 θ ) d θ =
∫ 0 π ( 2 3 + 2 cos θ + 2 c o s 2 θ ) d θ
4 s i n 2 θ
]
0 π
=
2 3 π
0 π /2
sin 2
(
2
θ
)
d
θ
=
2 1 ∫ 0 π /2 2 1 − c o s 4 θ d θ =
8 π
−
r inner 2 ) d θ
sin 2 θ = 1 − cos 2 θ 2 \sin^2\theta = \frac{1-\cos 2\theta}{2} sin 2 θ = 2 1 − c o s 2 θ
d r
sin
θ
+
r
cos
θ
/2
y
=
( − 3 /2 ) c o s ( π /3 ) − ( 3/2 ) s i n ( π /3 ) ( − 3 /2 ) s i n ( π /3 ) + ( 3/2 ) c o s ( π /3 )
=
0
= 0 dx/d\theta = 0 d x / d θ = 0
d y / d θ ≠ 0 dy/d\theta \ne 0 d y / d θ = 0
θ
+
2 r r ′ sin θ cos θ +
r 2 cos 2 θ =
( r ′ ) 2 +
r 2
θ
=
) 2
+
sin 2
θ
d
θ
d
θ
=
∫ 0 2 π 2∣ cos ( θ /2 ) ∣ d θ
4 [ 2 sin ( θ /2 ) ] 0 π =
8
(
1/2
)
=
/2
− 1/2 + 3/2
=