Areas and Lengths in Polar Coordinates

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SUMMARY

The area enclosed by one loop of the polar curve defined by r = sin(10θ) is calculated using the formula A = (1/2) ∫ (r^2) dθ. The correct limits of integration are from 0 to π/10, resulting in an area of A = π/8. The solution involves applying the half-angle formula for sine, specifically (sin(θ))^2 = (1/2)(1 - cos(2θ)), to simplify the integral. The final evaluation confirms the area as π/8.

PREREQUISITES
  • Understanding of polar coordinates and curves
  • Familiarity with integration techniques in calculus
  • Knowledge of the half-angle formula for trigonometric functions
  • Experience with definite integrals and their applications
NEXT STEPS
  • Study the properties of polar curves and their areas
  • Learn advanced integration techniques, including trigonometric identities
  • Explore the application of polar coordinates in real-world scenarios
  • Investigate the behavior of other polar functions, such as r = sin(nθ) for various n
USEFUL FOR

Students studying calculus, particularly those focusing on polar coordinates and integration, as well as educators seeking to enhance their teaching of these concepts.

JSGhost
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Homework Statement


Find the area of the region enclosed by one loop of the curve.
r = sin(10θ)

I can't seem to get the correct answer...I checked every step. I was not sure what to integrate from but the polar graph of sin(10θ) should be similar to polar graph of sin(2θ). From pi/2 to 0?

Homework Equations


A = integral(b to a) (1/2)r^2 dθ

half angle formula
(sinθ)^2 = (1/2)(1-cos2θ)dθ

The Attempt at a Solution


A = integral(b to a) (1/2)r^2 dθ = integral(pi/2 to 0) (1/2)r^2 dθ

A = (1/2) integral(pi/2 to 0) (sin(10θ))^2 dθ
A = (1/2) integral(pi/2 to 0) (1/2)(1-cos(20θ))dθ
A = (1/4) integral(pi/2 to 0) (1-cos(20θ))dθ
A = (1/4) [(θ-(1/20)sin(20θ)] (pi/2 to 0)
A = (1/4) [(pi/2-(1/20)sin(20*(pi/2)) - (0 - 0)]
A = (1/4)* (pi/2) = pi/8
 
Last edited:
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Nevermind. I figured it out. Integrating from pi/10 to 0. Thanks.
 

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