Double integral in polar form: how do you find the boundaries?

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Homework Help Overview

The discussion revolves around using a double integral in polar coordinates to find the area of a region defined by the equation r = 3 + 3sinQ, where Q represents the angle theta. Participants are exploring how to determine the appropriate boundaries for the outer integral.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the bounds of the inner integral and question how to establish the bounds for the outer integral. There is consideration of the periodic nature of the sine function and its implications for the limits of integration.

Discussion Status

Some participants have suggested examining the periodic behavior of the sine function to identify when the value of r begins to repeat. There is an exploration of the potential symmetry in the figure, with one participant proposing to integrate from 0 to pi/2 and then double the result.

Contextual Notes

Participants are considering the implications of the periodicity of the sine function and the symmetry of the region defined by the polar equation, which may influence the choice of integration limits.

winbacker
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Hi I need to use a double integral to find the area of the region bounded by:

r = 3 + 3sinQ where Q = theta.

I know the bounds of the inner integral are from 0 to 3 + 3sinQ.

However, I do not know how to determine the bounds of the outer integral.

Any help would be greatly appreciated.
 
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The outer integral is just a typical single-variable polar integral. Play with the equation to figure out when the behavior of r begins to repeat itself as theta varies, paying particular attention to the periodic nature of sine. Perhaps a polar graph might help.
 
Last edited:
Ok, well I know that once theta = 2pi, the behavior of of sinQ will repeat itself. Should I then plug in 2pi to the equation and work with that?

I know the value of sinQ becomes zero again at pi. Does this mean the outer boundary is from 0 to pi?
 
0 to pi, although you can exploit the symmetry of the figure by doubling the value you get integrating from 0 to pi/2.
 

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