Answer :
The area of the region enclosed by one loop of the curve r = 3/2 sin(2θ) is 9π/16
How to find the area of and enclosed by equation?
When we need to find the area bounded by a single loop of the polar curve, we use the same formula which was used to find area inside the polar curve in general.
[tex]A = \int\limits^a_b {1/2 r^2} \, d\theta\\[/tex]
where [a,b] is the interval and r is the given curve equation
According to the given question:
Equation of the given curve is r = 3/2 sin(2θ)
When r = 0, θ = π/2 which corresponds to one loop
Therefore area of the region enclosed by this curve is
A = 9/4 * 1/2 int 0 to π/2 sin²2θ dθ
A = 9/8 int 0 to π/2 (1 - cos4θ)/2 dθ
A = 9/8 * π/2
A = 9π/16
Area of the required loop is 9π/16
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