How Do You Calculate the Area Bounded by \( r = 8\cos(10\Theta) \)?

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SUMMARY

The area bounded by the polar curve \( r = 8\cos(10\Theta) \) is calculated using the integral \( A = \int(1/2)64\cos^2(10\Theta) d\Theta \). The correct bounds for the integral are \( \Theta = \pi/20 \) and \( \Theta = 3\pi/20 \). To find the total area of the rose curve, the result of the integral must be multiplied by 20, as the curve consists of 20 petals. The integral can be simplified using the identity \( \cos^2{nx} = \frac{1+\cos{2nx}}{2} \) for easier evaluation.

PREREQUISITES
  • Understanding of polar coordinates and polar curves
  • Knowledge of integral calculus, specifically integration techniques
  • Familiarity with trigonometric identities, particularly \( \cos^2{nx} \)
  • Ability to evaluate definite integrals
NEXT STEPS
  • Study polar coordinate systems and their applications in calculus
  • Learn advanced integration techniques, including integration by parts and trigonometric substitution
  • Explore the properties of rose curves and their geometric interpretations
  • Practice evaluating integrals involving trigonometric functions and identities
USEFUL FOR

Students studying calculus, particularly those focusing on polar coordinates and integral calculus, as well as educators seeking to enhance their teaching methods in these areas.

Cici2006
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Homework Statement


Find the area of the region bounded by r=8cos10\Theta


Homework Equations





The Attempt at a Solution



I set r=0 to find \Theta, which i used for my bounds
\Theta=pi/20, 3pi/20
A= \int(1/2)64cos^2(10\Theta) d\Theta
 
Last edited:
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What you need to do is multiply your answer by 20 since you found the area of one of the 20 petals of the rose curve.
 
Are you having trouble finding the correct solution since yours is too small? or because you don't know how to evaluate the integral?

If you need help evaluating the integral, use the fact that

\cos^2{nx} = \frac{1+\cos{2nx}}{2}, n\in\mathbb{N}
 
Okay, let me state what i did in more detail:
A=(1/2)integral 64(cos^2(10theta)) d(theta)
=32 integral (1/2)(1+cos20theta) (theta)
=16[theta-(1/20)sin20theta]
did i do it correct so far?
then i just plug in my bounds which are pi/20 to 3pi/20 right?
now should i just multiply my answer by 20?
 
thanks for the help i solved it
 

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