Find the area of the region enclosed by one loop of the curve r = sin(2θ)

To find the area enclosed by one loop of the curve given by the polar equation r = sin(2θ), we can use the formula for area in polar coordinates:

A = (1/2) ∫ r² dθ

For the given curve, we start by figuring out the limits of integration. One loop of the curve occurs when r = 0 again. Setting sin(2θ) = 0, we can determine:

• 2θ = nπ for integers n
• θ = nπ/2

For the first loop, we can choose n = 0 leading to:

θ = 0 to θ = π/2.

Now we can set up the integral for the area:

A = (1/2) ∫ (sin(2θ))² dθ

Next, we apply the double angle identity:

(sin(2θ))² = (1 - cos(4θ))/2

Thus, the integral becomes:

A = (1/2) ∫ (1 - cos(4θ))/2 dθ

Now, integrating from 0 to π/2:

A = (1/4) ∫ (1 - cos(4θ)) dθ

This can be split into two integrals:

A = (1/4)(∫ dθ - ∫ cos(4θ) dθ)

Calculating each part:

∫ dθ = θ and ∫ cos(4θ) dθ = (1/4)sin(4θ)

Now evaluate from 0 to π/2:

A = (1/4)[θ - (1/4)sin(4θ)]_0^{π/2}

Plugging in the upper limit:

A = (1/4)[(π/2) - (1/4)sin(2π)] - (1/4)[0 - (1/4)sin(0)]

This simplifies to:

A = (1/4)(π/2) = π/8

Hence, the area enclosed by one loop of the curve r = sin(2θ) is:

π/8.