Parametric equations and polar coordinates

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SUMMARY

The discussion focuses on finding the area enclosed by the inner loop of the polar curve defined by the equation r = 1 - 3sin(θ). The relevant formula for calculating the area is A = 0.5 ∫ r² dθ. Participants highlight the challenge of determining the correct integration interval, specifically identifying when r = 0, which leads to the value sin⁻¹(1/3). The conversation emphasizes the need to clarify whether this expression refers to the inverse sine function or its reciprocal.

PREREQUISITES
  • Understanding of polar coordinates and their representation
  • Familiarity with integration techniques in calculus
  • Knowledge of trigonometric functions and their inverses
  • Experience with area calculations in polar coordinates
NEXT STEPS
  • Study the derivation of area formulas in polar coordinates
  • Learn about the properties of inverse trigonometric functions
  • Explore examples of polar curves with inner loops
  • Practice solving integrals involving trigonometric identities
USEFUL FOR

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

maiad
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Homework Statement


Find the area enclosed by the inner loop of the curve r=1-3sinθ


Homework Equations


A=o.5\int r^2 dθ


The Attempt at a Solution


I found the integral but i don't know how to find the interval at which i will be integrating from. I tried finding when r=0 and it turns out to be sin^-1(1/3) but it's not a special angle so it would be messy if i plugged that interval in cos(θ).Any hints will be appreciated
 
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\sin \theta = \frac{1}{3}
\cos^2 \theta = 1 - \sin^2 \theta = \frac {8}{9}
\cos \theta = \pm \frac {2 \sqrt 2}{3}
So it's not a problem. You have to figure out whether it's + or -, depending on theta.
 
the answer had a sin(1/3)^-1 so i think you have to use that as one of your intervals I am assuming
 
maiad said:
the answer had a sin(1/3)^-1 so i think you have to use that as one of your intervals I am assuming
Do you mean (sin(1/3))^{-1}= 1/sin(1/3)) or sin^{-1}(1/3)= arcsin(1/3)?
 
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