Find area between 2 polar curves

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SUMMARY

The area between the polar curves defined by r=4sin(θ) and r=2 is calculated using the formula for the area under a polar curve. The correct integral setup involves integrating from the intersection points of the curves, specifically from π/6 to 5π/6 for the outer curve and from 0 to π/2 for the inner curve. The final area is determined to be (10/3)π after evaluating the integrals. The discussion highlights the importance of correctly identifying the limits of integration based on the intersection points of the polar curves.

PREREQUISITES
  • Understanding of polar coordinates and polar curves
  • Familiarity with integral calculus, specifically integration techniques
  • Knowledge of how to find intersection points of polar curves
  • Ability to evaluate definite integrals
NEXT STEPS
  • Learn how to find intersection points of polar curves
  • Study the application of the area under polar curves formula
  • Practice evaluating definite integrals involving trigonometric functions
  • Explore advanced topics in polar coordinates, such as area and arc length
USEFUL FOR

Students studying calculus, particularly those focusing on polar coordinates and integration techniques, as well as educators looking for examples of polar area calculations.

bavman
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Homework Statement



Fine the area of the region inside the polar curve r=4sin(theta) and outside the polar curve r=2.

Homework Equations



area under polar curve = 1/2 integral (a,b) r^2 d\Theta

The Attempt at a Solution



I set up the integral like follows:

integral (0,pi/2) (4sin(\Theta)^2 d\Theta - integral (pi/2, pi/6) 2 d\Theta

Then from plugging those into my calculator i got 4pi - (2/3)pi = (10/3) pi

Sorry i can't get the integral latex code to work for some reason.
 
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bavman said:
integral (0,pi/2) (4sin(\Theta)^2 d\Theta - integral (pi/2, pi/6) 2 d\Theta

Why do u think that the integral should be from 0,\frac{\pi}{2}?
Don't you think its the intersection(s) of the 2 polar curves?
 
hmm..well i really don't know how i messed that up.

so it should be :

integral (pi/6, 5pi/6) r1^2 - r2^2 dtheta ?
 
That should be fine! :)
 

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