Find Region Enclosed by Function r=2sin2θ

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Homework Help Overview

The discussion revolves around finding the area enclosed by one leaf of the polar function r = 2 sin(2θ). Participants are exploring how to set up the integration necessary to determine this area.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand how to integrate to find the area, mentioning the use of cylindrical coordinates and the Jacobian. Some participants suggest using the formula for area in polar coordinates, while others question the specific integration setup.

Discussion Status

Participants are actively discussing the integration process, with some suggesting a formula for area while others are questioning its application. There is no explicit consensus on the correct approach yet, but the dialogue is focused on clarifying the necessary steps.

Contextual Notes

There is uncertainty regarding the specific function to integrate and whether the factor of one-half in the area formula is appropriate for the problem at hand. Participants are also considering the implications of finding the area of just one half of the leaf.

aks_sky
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Find the region enclosed by one leaf of the function r=2 sin2θ


Now i know how the diagram looks like for this function.
The diagram can be obtained by using :

x= r cosθ = 2 sin2θ cosθ
y= r sinθ = 2 sin2θ sinθ

Now i have to find the region enclosed by one of the leaves which i am not able to do.
I know i have to use the jacobian and also that i can use the above as cylindrical coordinates but what do i integrate? i don't know which function to integrate.

Any Suggestions for that??

thank you
 
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[tex]\frac{1}{2}\int r^{2} .d\theta[/tex]

You can use that formula
 
rootX said:
[tex]\frac{1}{2}\int r^{2} .d\theta[/tex]

You can use that formula

Wouldn't it be [tex]\int r d\theta[/tex]

I could be missing something of course...
 
Is the half for finding each half of the region of the leaf?
 

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