Areas and Lengths in Polar Coordinates

Transcript

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 A 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: A θ π πr r θ Let R be the region bounded by the polar curve r fθ) 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 fθ i ). Thus from Formula we have and so an approximation to the total area A of R is A i A i [fθ i )] θ ) n i [fθ i )] θ. One can see that the approximation in ) improves as n. But the sums in ) are Riemann sums for the function gθ) [fθ)], so n b lim n [fθ i )] θ a [fθ)] 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 A b A a [fθ)] 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 A 3π/4 π/4 3π/4 3π π/4 4 π ) 4 ) π π 4 4 b) We have A 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 A 7π/4 5π/4 7π/4 5π/4 7π 4 5π 4 ) ) π π 4 4 d) We have or A π+π/4 7π/4 A 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 A 4π/3 π/3 4π/3 π/3 4π/3 π/3 4π/3 + 4 cosθ) π/ cosθ + 6 cos θ) + 8 cosθ + 8 cos θ) 4π/3 π/3 + 8 cosθ cos θ) + 8 cosθ + 8 4π/3 π/3 [ ] 4π/3 6θ + 8 sin θ + sin θ 4π π/3 ) + cos θ cosθ + 4 cos θ) EXAMPLE: Find the area enclosed by one loop of the four-leaved rose r cos θ. 3

4 Kiryl Tsishchanka EXAMPLE: Find the area enclosed by one loop of the four-leaved 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 A 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 fθ), r gθ), θ a, and θ b, where fθ) gθ) and < b a π. Then the area A of R is A b a [fθ)] [gθ)] ) 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 A 7π/6 7π/6 7π/6 3 [ 3 + sin θ) ] 5 + sinθ + 4 7π/6 ) cos θ 7π/6 5 + sin θ + 4 sin θ) 7 + sinθ cos θ) [ ] 7π/6 7θ cosθ sin θ + 4π EXAMPLE: Find the area of the region outside r 3 + sin θ and inside r. 5 + sinθ + cos θ)) 5

6 Kiryl Tsishchanka EXAMPLE: Find the area of the region outside r 3 + sin θ and inside r. Solution: We have A π/6 7π/6 [ 3 + sin θ) ] π/6 7π/6 5 sin θ 4 sin θ) π/6 7π/6 7 sin θ + cosθ) [ ] 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 fθ), a θ b, we regard θ as a parameter and write the parametric equations of the curve as x r cosθ fθ) cosθ y r sin θ fθ) 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 fθ), 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 cosθ. 8

9 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 L r + cosθ) + sin θ π π π π π cosθ + cos θ + sin θ cosθ 4 sin θ sin θ sin θ 4 cos θ ] π EXAMPLE: Find the length of the cardioid r + sin θ. 9

10 Kiryl Tsishchanka EXAMPLE: Find the length of the cardioid r + sin θ. Solution : Note that r + sin θ cos θ + π ) Therefore the graph of r + sin θ is the rotation of the graph of r cosθ. Hence the length of the cardioid r + sin θ is 8 by the previous Example. Solution : The full length of the cardioid is given by the parameter interval θ π, so Formula 5 gives π ) dr L r + π π π π 5π/ π/ + sin θ) + cos θ + sin θ + sin θ + cos θ + sin θ cos θ + π ) θ + π u d θ + π ) du du 4 sin u 5π/ du sin u du π/ π π/ π π/ 5π/ cosudu π/ sin u du + 5π/ π sin u du sin u 5π/ du sin u du ] π π ] 5π/ 4 cos u + 4 cos u π/ π 4 cosπ + 4 cos π ) + 4 cos 5π4 ) 4 4 cosπ 4 + ) + + 4) 8

11 Kiryl Tsishchanka Solution 3: 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 3π/ sin θ π/ cos θ π sin θ cosθ + π sin θ 3π/ cosθ sin θ cosθ sin θ Note that cosθ sin θ sin θ u d sin θ) du cos θ du cosθ du du u u / du u /+ / + + C u + C sin θ + C Therefore L ] π/ sin θ + ] 3π/ sin θ ] π sin θ π/ 3π/ ) sinπ/) sin + ) sin3π/) sinπ/) ) sinπ) sin3π/) ) + ) )