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Stitz-Zeager_College_Algebra_e-book

# 18 and the ensuing discussion with theorem 83 in

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Unformatted text preview: θ) and r = 2 − 2 sin(θ) intersect when θ = π . Hence, for the 6 ﬁrst region, (r, θ) : 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π , we are shading the region between the origin 6 (r = 0) out to the circle (r = 2 sin(θ)) as θ ranges from 0 to π , which is the angle of intersection 6 of the two curves. For the second region, (r, θ) : 0 ≤ r ≤ 2 − 2 sin(θ), π ≤ θ ≤ π , θ picks 6 2 up where it left oﬀ at π and continues to π . In this case, however, we are shading from the 6 2 origin (r = 0) out to the cardioid r = 2 − 2 sin(θ) which pulls into the origin at θ = π . Putting 2 these two regions together gives us our ﬁnal answer. y y 1 θ= 1 π 6 1 r = 2 − 2 sin(θ) and r = 2 sin(θ ) x 1 x (r, θ) : 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π ∪ 6 (r, θ) : 0 ≤ r ≤ 2 − 2 sin(θ), π ≤ θ ≤ π 6 2 816 Applications of Trigonometry 11.5.1 Exercises 1. Plot the graphs of the following polar equations by hand. Carefully label your graphs. (a) Circle: r = 6 sin(θ) (j) Cardioid: r = 5 + 5 sin(θ) (b) Circle: r = 2 cos(θ) (k) Cardioid: r = 2 + 2 c...
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