Answer: Area of the region enclosed by one loop of the curve r= sin(12Φ) is π/48.
Step-by-step explanation: Given the curve is r= sin(12Φ).
The range of the Φ is 0 <= Φ <= π/12.
The formula of Area (A) is given as:
A= ∫ 1/2 r^² dΦ
Inserting the limits and the r:
A = \int_0^{\dfrac{\pi}{12}} \space \dfrac{1}{2} (sin(12 \theta))^2 d\theta
A = \dfrac{1}{2} \int_0^{\dfrac{\pi}{12}} \space sin^2(12 \theta) d\theta
Using the formula:
sin^2x = \dfrac{1-cos2x}{2}
A = \dfrac{1}{2} \int_0^{\dfrac{\pi}{12}} \space \dfrac{1-cos(24 \theta)}{2} d\theta
=A = \dfrac{1}{2} \int_0^{\dfrac{\pi}{12}} \space \dfrac{1-cos(24 \theta)}{2} d\theta
= \dfrac{1}{2} \left[ \int_0^{\dfrac{\pi}{12}} \space \dfrac{1}{2} d \theta \space – \space \int_0^{\dfrac{\pi}{12}} \space \left( \dfrac{1-cos(24 \theta)}{2} \right) d\theta \right]
Integrating with respect to dΦ :
A = \dfrac{1}{2} \left[ \left( \dfrac{\theta}{2} \right)_0^{\dfrac{\pi}{12}} \space – \space \left( \dfrac{1-sin(24 \theta)}{2(24)} \right)_0^{\dfrac{\pi}{12}} \right]
= \dfrac{1}{2} \left[ \left( \dfrac{\pi/12}{2} – \dfrac{0}{2} \right) \space – \space \left( \dfrac{1-sin(24 \dfrac{\pi}{12})}{48} \space – \space \dfrac{1-sin(24 (0))}{48} \right) \right]
= \dfrac{1}{2} \left[ \left( \dfrac{\pi}{24} \right) \space – \space \left( \dfrac{\pi}{24} – \dfrac{\pi}{24} \right) \right]
= \dfrac{1}{2} \left[ \left( \dfrac{\pi}{24} \right) \right]
A = \dfrac{\pi}{48}
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