PP 11.3-11.4

# D dx ds 2 2 cos sin sin cos dx dr r d d dy dr r d d

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d dx ds 2 2 + = cos sin sin cos dx dr r d d dy dr r d d θ θ θ θ θ θ θ θ = - = + From before:

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Since cos 2 θ + sin 2 θ = 1. 2 2 2 2 2 2 2 2 2 2 2 2 cos 2 cos sin sin sin 2 sin cos cos dx dy d d dr dr r r d d dr dr r r d d dr r d θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ + = - + + + + = +
ARC LENGTH IN POLAR COORDINATES Therefore, the length of a curve with polar equation r = f ( θ ), a θ b, is: 2 2 b a dr L r d d θ θ = +

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Find the length of the polar curve r = 2cosθ Example Previously we graphed the curve, so we know the interval of integration is [0, π]
2 2 b a dr L r d d θ θ = + θ θ θ π d - + = 0 2 2 ) sin 2 ( ) cos 2 ( = + = π π θ θ θ θ 0 0 2 2 2 sin 4 cos 4 d d π θ π 2 2 0 = =

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Area in Polar Coordinates Sector: a slice of a circle determine by a central angle. radians in , 2 2 1 θ θ r A = Area of a sector:
Using Riemann sums and the above formula when working with a polar region:

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We see that each polar rectangle has area [ ] θ θ = 2 2 1 ) ( i i f A So the area of the entire region is θ θ β α d f A A n i i n 2 2 1 1 )] ( [ lim = = =
Example Find the area enclosed by the polar curve r = 2cosθ From previous work, we know that the interval needed to trace the entire curve is [0, π] θ θ π d A 2 0 2 1 ] cos 2 [ = θ θ π d = 0 2 cos 2 θ θ π d + = 0 2 ) 2 cos( 1 2 π θ θ π = + = 0 2 1 ) 2 sin(
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