Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the area of a sector of a circle 1 A = r2θ 2 (1) where r is the radius and θ is the radian measure of the central angle. Formula 1 follows from the fact that the area of a sector is proportional to its central angle: A= θ 1 · πr 2 = r 2 θ 2π 2 Let R be the region bounded by the polar curve r = f (θ) and by the rays θ = a and θ = b, where f is a positive continuous function and where 0 < b − a ≤ 2π. We divide the interval [a, b] into subintervals with endpoints θ0 , θ1 , θ2 , . . . , θn and equal width ∆θ. The rays θ = θi then divide R into n smaller regions with central angle ∆θ = θi − θi−1 . If we choose θi∗ in the ith subinterval [θi−1 , θi ], then the area ∆Ai of the ith region is approximated by the area of the sector of a circle with central angle ∆θ and radius f (θi∗ ). Thus from Formula 1 we have 1 ∆Ai ≈ [f (θi∗ )]2 ∆θ 2 and so an approximation to the total area A of R is A ≈ n X 1 i=1 2 (2) [f (θi∗ )]2 ∆θ. One can see that the approxima- 1 tion in (2) improves as n → ∞. But the sums in (2) are Riemann sums for the function g(θ) = [f (θ)]2 , 2 so Z n b X1 1 lim [f (θi∗ )]2 ∆θ = [f (θ)]2 dθ n→∞ 2 a 2 i=1 It therefore appears plausible (and can in fact be proved) that the formula for the area A of the polar region R is Z b 1 (3) A= [f (θ)]2 dθ 2 a This formula is often written as A= Z b a 1 1 2 r dθ 2 (4) Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka EXAMPLE: Find the area of each of the following regions: (a) (b) (c) (d) Solution: (a) We have A= Z 3π/4 π/4 1 2 1 · 1 dθ = 2 2 Z 3π/4 π/4 1 dθ = 2 3π π − 4 4 1 = 2 2π 4 = π 4 (b) We have A= Z 2π+π/4 3π/4 1 1 2 · 1 dθ = 2 2 Z 2π+π/4 3π/4 1 dθ = 2 π 3π 2π + − 4 4 1 = 2 2π π 3π 2π − =π− = 4 4 4 (c) We have A= Z 7π/4 5π/4 1 1 2 · 1 dθ = 2 2 Z 7π/4 5π/4 1 dθ = 2 7π 5π − 4 4 1 = 2 2π 4 = π 4 (d) We have A= Z 2π+π/4 7π/4 1 2 1 · 1 dθ = 2 2 or A= Z π/4 −π/4 Z 2π+π/4 7π/4 1 1 2 · 1 dθ = 2 2 1 dθ = 2 Z π 7π 2π + − 4 4 π/4 dθ = −π/4 1 = 2 6π 3π π 2π − =π− = 4 4 4 1 π π 1 π π π = = − − + 2 4 4 2 4 4 4 EXAMPLE: Find the area of the inner loop of r = 2 + 4 cos θ. 2 Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka EXAMPLE: Find the area of the inner loop of r = 2 + 4 cos θ. Solution: We first find a and b: 2 + 4 cos θ = 0 Therefore the area is Z A = 4π/3 2π/3 = Z 4π/3 Z 4π/3 =⇒ 1 (2 + 4 cos θ)2 dθ = 2 cos θ = − Z 2 =⇒ θ= 2π 4π , 3 3 1 (4 + 16 cos θ + 16 cos2 θ)dθ 2 2π/3 (2 + 8 cos θ + 8 cos θ)dθ = 2π/3 = 4π/3 1 2 Z 4π/3 2π/3 (2 + 8 cos θ + 4(1 + cos 2θ)dθ = 2π/3 1 + cos 2θ dθ 2 + 8 cos θ + 8 · 2 Z 4π/3 (6 + 8 cos θ + 4 cos 2θ)dθ 2π/3 h i4π/3 √ = 6θ + 8 sin θ + 2 sin 2θ = 4π − 6 3 ≈ 2.174 2π/3 EXAMPLE: Find the area enclosed by one loop of the four-leaved rose r = cos 2θ. 3 Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka EXAMPLE: Find the area enclosed by one loop of the four-leaved rose r = cos 2θ. Solution: Notice that the region enclosed by the right loop is swept out by a ray that rotates from θ = −π/4 to θ = π/4. Therefore, Formula 4 gives A = Z π/4 Z π/4 −π/4 = 0 1 2 1 r dθ = 2 2 Z π/4 2 cos 2θdθ = −π/4 Z π/4 cos2 2θdθ 0 π/4 π 1 1 1 = θ + sin 4θ (1 + cos 4θ)dθ = 2 2 4 8 0 Let R be the region bounded by curves with polar equations r = f (θ), r = g(θ), θ = a, and θ = b, where f (θ) ≥ g(θ) ≥ 0 and 0 < b − a ≤ 2π. Then the area A of R is A= Z b a 1 [f (θ)]2 − [g(θ)]2 dθ 2 EXAMPLE: Find the area that lies inside r = 3 + 2 sin θ and outside r = 2. 4 Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka EXAMPLE: Find the area that lies inside r = 3 + 2 sin θ and outside r = 2. Solution: We first find a and b: 3 + 2 sin θ = 2 =⇒ 1 sin θ = − 2 =⇒ 7π π θ= , − 6 6 11π 6 Therefore the area is Z 7π/6 Z 7π/6 1 1 2 2 A = (3 + 2 sin θ) − 2 dθ = (5 + 12 sin θ + 4 sin2 θ)dθ −π/6 2 −π/6 2 = = Z Z 7π/6 −π/6 7π/6 −π/6 1 2 Z 7π/6 1 − cos 2θ 1 5 + 12 sin θ + 4 · dθ = (5 + 12 sin θ + 2(1 − cos 2θ))dθ 2 −π/6 2 i7π/6 1h 1 (7 + 12 sin θ − 2 cos 2θ)dθ = 7θ − 12 cos θ − sin 2θ −π/6 2 2 √ 11 3 14π = + ≈ 24.187 2 3 EXAMPLE: Find the area of the region outside r = 3 + 2 sin θ and inside r = 2. 5 Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka EXAMPLE: Find the area of the region outside r = 3 + 2 sin θ and inside r = 2. Solution: We have Z 11π/6 1 2 2 − (3 + 2 sin θ)2 dθ A = 2 7π/6 = Z 11π/6 Z 11π/6 7π/6 = 7π/6 1 (−5 − 12 sin θ − 4 sin2 θ)dθ 2 i11π/6 11√3 7π 1h 1 = − 7θ + 12 cos θ + sin 2θ (−7 − 12 sin θ + 2 cos 2θ)dθ = − ≈ 2.196 7π/6 2 2 2 3 1 EXAMPLE: Find all points of intersection of the curves r = cos 2θ and r = . 2 1 1 Solution: If we solve the equations r = cos 2θ and r = , we get cos 2θ = and, therefore, 2 2 2θ = π/3, 5π/3, 7π/3, 11π/3 Thus the values of θ between 0 and 2π that satisfy both equations are θ = π/6, 5π/6, 7π/6, 11π/6 We have found four points of intersection: 1 1 1 1 , π/6 , , 5π/6 , , 7π/6 , and , 11π/6 2 2 2 2 However, you can see from the above figure that the curves have four other points of intersection — namely, 1 1 1 1 , π/3 , , 2π/3 , , 4π/3 , and , 5π/3 2 2 2 2 1 These can be found using symmetry or by noticing that another equation of the circle is r = − and then 2 1 solving the equations r = cos 2θ and r = − . 2 6 Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka Arc Length To find the length of a polar curve r = f (θ), a ≤ θ ≤ b, we regard θ as a parameter and write the parametric equations of the curve as x = r cos θ = f (θ) cos θ y = r sin θ = f (θ) sin θ Using the Product Rule and differentiating with respect to θ, we obtain dx dr = cos θ − r sin θ dθ dθ dy dr = sin θ + r cos θ dθ dθ So, using cos2 θ + sin2 θ = 1, we have 2 2 2 dx dy dr dr + = cos2 θ − 2r cos θ sin θ + r 2 sin2 θ dθ dθ dθ dθ + dr dθ 2 dr sin θ + 2r sin θ cos θ + r 2 cos2 θ = dθ 2 dr dθ 2 + r2 Assuming that f ′ is continuous, we can use one of the formulas from Section 9.2 to write the arc length as s Z b 2 2 dy dx L= + dθ dθ dθ a Therefore, the length of a curve with polar equation r = f (θ), a ≤ θ ≤ b, is L= Z b a s r2 + dr dθ EXAMPLE: Find the length of the curve r = θ, 0 ≤ θ ≤ 1. 7 2 dθ (5) Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka EXAMPLE: Find the length of the curve r = θ, 0 ≤ θ ≤ 1. Solution: We have L = = Z Z 1 √ 0 =⇒ √ θ2 + 1 = √ tan2 x + 1 = θ2 + 1dθ = dθ = d tan x √ sec2 x = | sec x| = sec x dθ = sec2 xdx π/4 1 sec xdx = (sec x tan x + ln | sec x + tan x|) 2 3 0 θ = tan x EXAMPLE: Find the length of the cardioid r = 1 − cos θ. 8 π/4 0 √ 1 √ = ( 2 + ln(1 + 2)) 2 Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka EXAMPLE: Find the length of the cardioid r = 1 − cos θ. Solution: The full length of the cardioid is given by the parameter interval 0 ≤ θ ≤ 2π, so Formula 5 gives s 2 Z 2π Z 2π q dr L= dθ = r2 + (1 − cos θ)2 + sin2 θdθ dθ 0 0 = = Z Z 2π 0 2π 0 2π p √ 1 − 2 cos θ + cos2 θ + sin2 θdθ 2 − 2 cos θdθ r θ 4 sin2 dθ 2 0 Z 2π θ = 2 sin dθ 2 0 Z 2π θ = 2 sin dθ 2 0 2π θ =4+4=8 = −4 cos 2 0 = Z EXAMPLE: Find the length of the cardioid r = 1 + sin θ. 9 Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka EXAMPLE: Find the length of the cardioid r = 1 + sin θ. Solution 1: Note that π r = 1 + sin θ = 1 − cos θ + 2 Therefore the graph of r = 1 + sin θ is the rotation of the graph of r = 1 − cos θ. Hence the length of the cardioid r = 1 + sin θ is 8 by the previous Example. Solution 2: The full length of the cardioid is given by the parameter interval 0 ≤ θ ≤ 2π, so Formula 5 gives s 2 Z 2π dr 2 L= dθ r + dθ 0 Z 2π p = (1 + sin θ)2 + cos2 θdθ 0 2π = Z 2π = Z 0 p 1 + 2 sin θ + sin2 θ + cos2 θdθ √ 2 + 2 sin θdθ 0 π θ+ =u r Z 5π/2 2π 2 π √ π = = dθ = 2 − 2 cos θ + 2 − 2 cos udu d θ + = du 2 0 π/2 2 dθ = du Z 2π Z 5π/2 Z 5π/2 Z 5π/2 r u u u 2 u 2 sin du + 2 sin du 2 sin du = = 4 sin du = 2 2 2 2 π/2 2π π/2 π/2 Z 2π Z 5π/2 u u = 2 sin du − 2 sin du 2 2 π/2 2π u i5π/2 u i2π + 4 cos = −4 cos 2 π/2 2 2π 5π π + 4 cos − 4 cos π = −4 cos π + 4 cos 4 4 √ √ = (4 + 2 2) + (−2 2 + 4) = 8 Z 10 Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka Solution 3: The full length of the cardioid is given by the parameter interval 0 ≤ θ ≤ 2π, so Formula 5 gives s 2 Z 2π Z 2π p Z 2π p dr 2 2 2 L = dθ = r + (1 + sin θ) + cos θdθ = 1 + 2 sin θ + sin2 θ + cos2 θdθ dθ 0 0 0 √ √ Z 2π √ √ Z 2π √ √ Z 2π 1 + sin θ 1 − sin θ √ 2 + 2 sin θdθ = 2 1 + sin θdθ = 2 = dθ 1 − sin θ 0 0 0 √ Z 2π p √ Z 2π √ Z 2π | cos θ| √ 1 − sin2 θ cos2 θ √ √ √ dθ = 2 dθ = 2 dθ = 2 1 − sin θ 1 − sin θ 1 − sin θ 0 0 0 Z π/2 √ Z 3π/2 √ Z 2π √ cos θ cos θ cos θ √ √ √ = 2 dθ − 2 dθ + 2 dθ 1 − sin θ 1 − sin θ 1 − sin θ 0 π/2 3π/2 Note that Z 1 − sin θ = u d(1 − sin θ) = du cos θ √ dθ = − cos θdθ = du 1 − sin θ cos θdθ = −du Z Z −1/2+1 du = − √ = − u−1/2 du = − u +C u −1/2 + 1 √ = −2 u + C √ = −2 1 − sin θ + C Therefore iπ/2 i3π/2 i2π √ √ √ √ √ √ + 2 2 1 − sin θ − 2 2 1 − sin θ L = −2 2 1 − sin θ π/2 0 √ p √ 1 − sin(π/2) − 1 − sin 0 = −2 2 p √ p 1 − sin(3π/2) − 1 − sin(π/2) +2 2 p √ p −2 2 1 − sin(2π) − 1 − sin(3π/2) √ √ √ √ √ = −2 2 (0 − 1) + 2 2 2 − 0 − 2 2 1 − 2 √ √ =2 2+4−2 2+4 =8 11 3π/2

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