Find the area of the region enclosed by one loop of the curve. r = 4 cos(3θ).
Solution:
Given, r = 4 cos(3θ)
When, r = 0
⇒ 4 cos(3θ) = 0
⇒ cos(3θ) = 0
⇒ 3θ = π/2 + nπ
⇒ θ = π/6 + nπ/3
Thus, the limit lies in the interval -π/6 to + π/6
Area of polar region, A = \(\int_{a}^{b}\frac{1}{2}r^{2}dθ\)
Substituting the values
\(A=\int_{-\pi /6}^{\pi /6}\frac{1}{2}(4cos3θ )^{2}dθ\)
Here,
\(\\A=2\int_{0}^{\pi /6}\frac{1}{2}(4cos3θ )^{2}dθ \\ \\A=2\int_{0}^{\pi /6}\frac{1}{2}16cos^{2}3θ dθ \\ \\A=2\int_{0}^{\pi /6}8cos^{2}3θ dθ \\ \\A=16\int_{0}^{\pi /6}cos^{2}3θ dθ\)
It can be written as
\(\\A=16\int_{0}^{\pi /6}\frac{1}{2}(1+cos6θ )dθ \\ \\A=8\int_{0}^{\pi /6}(1+cos6θ )dθ\)
A = 8 [θ + 1/6 sin6θ] in the limit 0 to π/6
A = 8[(π/6 + 0) - (0 + 0)]
A = 4π/3
Therefore, area of the region is 4π/3.
Find the area of the region enclosed by one loop of the curve. r = 4 cos(3θ).
Summary:
The area of the region enclosed by one loop of the curve r = 4 cos(3θ) is 4π/3.
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