Answer:
a) A = 3π/16
b) A = ∑₂°° (π/n⁴) = π⁵/90 − π
Explanation:
(a) The area between the curve and the x-axis is:
A = ∫ₐᵇ y (dx/dt) dt
x = ½ (1 − cos t) cos t
dx/dt = ½ (1 − cos t) (-sin t) + ½ sin t cos t
dx/dt = -½ sin t + ½ cos t sin t + ½ sin t cos t
dx/dt = -½ sin t + sin t cos t
dx/dt = -½ sin t (1 − 2 cos t)
y dx/dt = -½ sin t (1 − 2 cos t) × ½ (1 − cos t) sin t
y dx/dt = -¼ sin²t (1 − 3 cos t + 2 cos²t)
y dx/dt = -¼ sin²t + ¾ sin²t cos t − ½ sin²t cos²t
y dx/dt = ⅛ (-2 sin²t) + ¾ sin²t cos t − ½ (sin t cos t)²
y dx/dt = ⅛ (-1 + 1 − 2 sin²t) + ¾ sin²t cos t − ½ (½ sin(2t))²
y dx/dt = ⅛ (-1 + cos(2t)) + ¾ sin²t cos t − ⅛ sin²(2t)
y dx/dt = -⅛ + ⅛ cos(2t) + ¾ sin²t cos t + ¹/₁₆ (-2 sin²(2t))
y dx/dt = -⅛ + ⅛ cos(2t) + ¾ sin²t cos t + ¹/₁₆ (-1 + 1 − 2 sin²(2t))
y dx/dt = -⅛ + ⅛ cos(2t) + ¾ sin²t cos t + ¹/₁₆ (-1 + cos(4t))
y dx/dt = -⅛ + ⅛ cos(2t) + ¾ sin²t cos t − ¹/₁₆ + ¹/₁₆ cos(4t)
y dx/dt = -³/₁₆ + ⅛ cos(2t) + ¾ sin²t cos t + ¹/₁₆ cos(4t)
A = ∫₀ᵖⁱ [-³/₁₆ + ⅛ cos(2t) + ¾ sin²t cos t + ¹/₁₆ cos(4t)] dt
A = -³/₁₆ t + ¹/₁₆ sin(2t) + ¼ sin³t + ¹/₆₄ sin(4t) |₀ᵖⁱ
A = [-³/₁₆ π + ¹/₁₆ sin(2π) + ¼ sin³π + ¹/₆₄ sin(4π)] − [0 + ¹/₁₆ sin(0) + ¼ sin³0 + ¹/₆₄ sin(0)]
A = -³/₁₆ π
We got a negative answer because the graph traces to the left (x decreases as t increases). So the area is simply ³/₁₆ π.
b) Each circle has radius 1/n², so the area is π (1/n)² = π/n⁴.
The sum of the areas from n=2 to n=∞ is:
A = ∑₂°° (π/n⁴)
A = ∑₁°° (π/n⁴) − π
A = π ∑₁°° (1/n⁴) − π
A = π (π⁴/90) − π
A = π⁵/90 − π