Area of one loop of the rose r=cos(3theta)

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SUMMARY

The area of one loop of the rose curve defined by the polar equation r = cos(3θ) can be determined using double integrals. The limits for θ are established as -π/6 and π/6 because these values correspond to the points where r equals zero, specifically at θ = -π/6 (where 3θ = -π/2) and θ = π/6 (where 3θ = π/2). This range effectively encloses one complete loop of the curve, as r remains non-zero within these bounds. Understanding these limits is crucial for accurately calculating the area using the appropriate integral setup.

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  • Knowledge of polar coordinates and polar equations
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  • Familiarity with trigonometric functions, specifically cosine
  • Ability to interpret and analyze curves in polar form
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Homework Statement


Use double integrals to find the Area of one loop of the rose r=cos(3theta)

I know how to solve this, the only question I have is why theta is between -pi/6 and pi/6. I don't understand where those two values come from.
 
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With theta equal to -pi/6, 3theta= -pi/2 and r= cos(3theta)= cos(-pi/2)= 0. Similarly, if theta is pi/6, 3theta= pi/2 and r= cos(3theta)= cos(pi/2)= 0. The only point with r= 0 is the origin, no matter what theta is. That means that starting at -pi/6 and going to pi/6 you have closed a loop. starting at the origin and coming back to it. Since cosine in not 0 at any point between -pi/2 and pi/2, cos(3theta) is not 0 at any point between -pi/6 and pi/6 so that encloses only one loop which is what you want.,
 
finally! Now i understand it! Thank you so much! :)
 

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