45–60. Areas of regions Find the area of the following regions.
The region inside the curve r = √(cos θ) and inside the circle r = 1/√2 in the first quadrant
45–60. Areas of regions Find the area of the following regions.
The region inside the curve r = √(cos θ) and inside the circle r = 1/√2 in the first quadrant
45–60. Areas of regions Find the area of the following regions.
The region inside one leaf of the rose r = cos 5θ
45–60. Areas of regions Find the area of the following regions.
The region inside the rose r = 4 sin 2θ and inside the circle r = 2
45–60. Areas of regions Find the area of the following regions.
The region common to the circles r = 2 sin θ and r = 1
45–60. Areas of regions Find the area of the following regions.
The region inside the outer loop but outside the inner loop of the limaçon r = 3 - 6 sin θ
45–60. Areas of regions Find the area of the following regions.
The region inside the lemniscate r² = 6 sin 2θ
45–60. Areas of regions Find the area of the following regions.
The region inside the limaçon r = 4 - 2 cos θ
63–74. Arc length of polar curves Find the length of the following polar curves.
The complete circle r = a sin θ, where a > 0
63–74. Arc length of polar curves Find the length of the following polar curves.
The spiral r = θ², for 0 ≤ θ ≤ 2π
63–74. Arc length of polar curves Find the length of the following polar curves.
The complete cardioid r = 4 + 4 sin θ
63–74. Arc length of polar curves Find the length of the following polar curves.
The curve r = sin³(θ/3), for 0 ≤ θ ≤ π/2
63–74. Arc length of polar curves Find the length of the following polar curves.
{Use of Tech} The complete limaçon r=4−2cosθ
63–74. Arc length of polar curves Find the length of the following polar curves.
{Use of Tech} The complete limaçon r=4−2cosθ
Spiral arc length Consider the spiral r=4θ, for θ≥0.
a. Use a trigonometric substitution to find the length of the spiral, for 0≤θ≤√8.
Spiral arc length Consider the spiral r=4θ, for θ≥0.
c. Show that L′(θ)>0. Is L″(θ) positive or negative? Interpret your answer.