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

Solution:

Given, r = sin(2θ)

When, r = 0

sin(2θ) = 0

2θ = 0 or π

θ = 0 or π/2

Thus, the limit lies in the interval 0 to + π/2.

Area of the polar region = \(\frac{1}{2}\int_{a}^{b}r^{2}\, d\theta\)

A = \(\frac{1}{2}\int_{0}^{\frac{\pi }{2}}(sin(2\theta ))^{2}\, d\theta\)

A = \(\frac{1}{2}\int_{0}^{\frac{\pi }{2}}sin^{2}(2\theta )\, d\theta\) --- (1)

We know, cos²(2x) = 1 - 2sin²(x)

So, cos(4θ) = 1 - 2sin²(2θ)

On rearranging,

2sin²(2θ) = 1 - cos(4θ)

sin²(2θ) = 1/2 - cos(4θ)/2

Substitute the value of sin²(2θ) in (1)

A = \(\frac{1}{2}\int_{0}^{\frac{\pi }{2}}(\frac{1}{2}-\frac{cos(4\theta )}{2})\, d\theta\)

= \(\frac{1}{2}[(\frac{1}{2}\theta) -\frac{1}{2}(\frac{1}{4}sin(4\theta))]_{0}^{\frac{\pi }{2}}\)

= \(\frac{1}{2}[(\frac{1}{2}\theta) -(\frac{1}{8}sin(4\theta))]_{0}^{\frac{\pi }{2}}\)

= \(\frac{1}{2}[(\frac{1}{2}\frac{\pi }{2}) -(\frac{1}{8}sin(4(\frac{\pi }{2})))-0]\)

= \(\frac{1}{2}[\frac{\pi }{4} -(\frac{1}{8}sin(2\pi))-0]\)

= \(\frac{1}{2}[\frac{\pi }{4} -\frac{1}{8}(0)]\)

= \(\frac{1}{2}[\frac{\pi }{4}]\)

= \(\frac{\pi }{8}\)

Therefore, the area is \(\frac{\pi }{8}\) square units.

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

Summary:

The area of the region enclosed by one loop of the curve r = sin(2θ) is \(\frac{\pi }{8}\)square units.