Areas and Lengths in Polar Coordinates


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1 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 r θ () where r is the radius and θ is the radian measure of the central angle. Formula follows from the fact that the area of a sector is proportional to its central angle: θ π πr r θ 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 < b a π. We divide the interval [a, b] into subintervals with endpoints θ, θ, θ,...,θ n and equal width θ. The rays θ θ i then divide R into n smaller regions with central angle θ θ i θ i. If we choose θ i in the ith subinterval [θ i, θ i ], then the area A i of the ith region is approximated by the area of the sector of a circle with central angle θ and radius f(θ i ). Thus from Formula we have and so an approximation to the total area A of R is A i A i [f(θ i )] θ () n i [f(θ i )] θ. One can see that the approximation in () improves as n. But the sums in () are Riemann sums for the function g(θ) [f(θ)], so n b lim n [f(θ i )] θ a [f(θ)] It therefore appears plausible (and can in fact be proved) that the formula for the area A of the polar region R is This formula is often written as b a [f(θ)] (3) b a r (4)
2 Kiryl Tsishchanka EXAMPLE: Find the area of each of the following regions: (a) (b) (c) (d) Solution: (a) We have 3π/4 π/4 3π/4 ( 3π π/4 4 π ) 4 ( ) π π 4 4 (b) We have π+π/4 3π/4 π+π/4 3π/4 ( π + π 4 3π 4 ) ( π π 4 ) π π 4 3π 4 (c) We have 7π/4 5π/4 7π/4 5π/4 ( 7π 4 5π 4 ) ( ) π π 4 4 (d) We have or π+π/4 7π/4 π/4 π/4 π+π/4 7π/4 ( π + π 4 7π 4 ) ( π 6π 4 π/4 ( π ( π/4 4 π )) ( π π 4) π 4 EXAMPLE: Find the area of the inner loop of r + 4 cosθ. ) π 3π 4 π 4
3 Kiryl Tsishchanka EXAMPLE: Find the area of the inner loop of r + 4 cosθ. Solution: We first find a and b: Therefore the area is + 4 cosθ cosθ θ π 3, 4π 3 4π/3 π/3 4π/3 π/3 4π/3 ( + 4 cosθ) π/3 (4 + 6 cosθ + 6 cos θ) ( + 8 cosθ + 4( + cos(θ)) 4π/3 π/3 [ ] 4π/3 6θ + 8 sin θ + sin(θ) 4π π/3 (6 + 8 cosθ + 4 cos(θ)) EXAMPLE: Find the area enclosed by one loop of the fourleaved rose r cos θ. 3
4 Kiryl Tsishchanka EXAMPLE: Find the area enclosed by one loop of the fourleaved rose r cos θ. 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 π/4 π/4 π/4 r π/4 π/4 cos θ cos θ π/4 ( + cos 4θ) [θ + 4 ] π/4 sin 4θ π 8 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) and < b a π. Then the area A of R is b a ( [f(θ)] [g(θ)] ) EXAMPLE: Find the area that lies inside r 3 + sinθ and outside r. 4
5 Kiryl Tsishchanka EXAMPLE: Find the area that lies inside r 3 + sinθ and outside r. Solution: We first find a and b: Therefore the area is 3 + sin θ sin θ θ 7π 6, π 6 ( ) π 6 7π/6 π/6 7π/6 π/6 3 [ (3 + sin θ) ] 7π/6 π/6 (5 + sinθ + 4 sin θ) (7 + sin θ cos(θ)) [ ] 7π/6 7θ cosθ sin(θ) π/6 + 4π EXAMPLE: Find the area of the region outside r 3 + sin θ and inside r. 5
6 Kiryl Tsishchanka EXAMPLE: Find the area of the region outside r 3 + sin θ and inside r. Solution: We have π/6 7π/6 [ (3 + sinθ) ] π/6 7π/6 ( 5 sinθ 4 sin θ) π/6 7π/6 ( 7 sinθ + cos(θ)) [ ] π/6 7θ + cosθ + sin(θ) 3 7π 7π/ EXAMPLE: Find all points of intersection of the curves r cos θ and r. Solution: If we solve the equations r cos θ and r, we get cos θ and, therefore, θ π/3, 5π/3, 7π/3, π/3 Thus the values of θ between and π that satisfy both equations are We have found four points of intersection: ( ) ( ), π/6,, 5π/6, θ π/6, 5π/6, 7π/6, π/6 ( ), 7π/6, and ( ), π/6 However, you can see from the above figure that the curves have four other points of intersection namely, ( ) ( ) ( ) ( ), π/3,, π/3,, 4π/3, and, 5π/3 These can be found using symmetry or by noticing that another equation of the circle is r and then solving the equations r cos θ and r. 6
7 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 cosθ f(θ) cosθ y r sin θ f(θ) sin θ Using the Product Rule and differentiating with respect to θ, we obtain dx dr dy cos θ r sin θ dr sin θ + r cos θ So, using cos θ + sin θ, we have ( ) dx + ( ) dy ( ) dr cos θ r dr cosθ sin θ + r sin θ + ( ) dr sin θ + r dr sin θ cos θ + r cos θ ( ) dr + r Assuming that f is continuous, we can use one of the formulas from Section 9. to write the arc length as b (dx ) ( ) dy L + Therefore, the length of a curve with polar equation r f(θ), a θ b, is a L b a r + EXAMPLE: Find the length of the curve r θ, θ. ( ) dr (5) 7
8 Kiryl Tsishchanka EXAMPLE: Find the length of the curve r θ, θ. Solution: We have L θ + θ tanx θ + tan x + sec x sec x sec x d tanx sec xdx π/4 sec 3 xdx [ ] π/4 (sec x tan x + ln sec x + tanx ) ( + ln( + )) EXAMPLE: Find the length of the cardioid r + sin θ. 8
9 Kiryl Tsishchanka EXAMPLE: Find the length of the cardioid r + sin θ. Solution: The full length of the cardioid is given by the parameter interval θ π, so Formula 5 gives π ( ) dr π π L r + ( + sin θ) + cos θ + sinθ + sin θ + cos θ π π π + sin θ + sin θ + sin θ sin θ sin θ π π/ sin θ π sin θ cosθ π/+π sin θ π/ cosθ du cos θ π sin θ cosθ + π sin θ π/+π Note that sin θ u cosθ d( sin θ) du du sin θ cos θ du u cosθ sin θ cosθ sin θ u / du u /+ / + + C u + C sin θ + C Therefore L ] π/ sin θ + ] π/+π sin θ ] π sin θ π/ π/+π ( ) sin(π/) sin + ( ) sin(π/ + π) sin(π/) ( ) sin(π) sin(π/ + π) ( ) + ( ) ( ) EXAMPLE: Find the length of the cardioid r cosθ. 9
10 Kiryl Tsishchanka EXAMPLE: Find the length of the cardioid r cosθ. Solution: The full length of the cardioid is given by the parameter interval θ π, so Formula 5 gives π ( ) dr π π L r + ( cosθ) + sin θ cosθ + cos θ + sin θ π π cosθ 4 sin θ π sin θ π sin θ 4 cos θ ] π
Areas and Lengths in Polar Coordinates
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
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