Calculating Area of Polar Function: Spiral r = 5(e^.1θ)

klarge
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1. Problem: Use the spiral r = 5(e^.1θ). Find the area of the region in Quadrant I that is outside the second revolution of the spiral and inside the third revolution.



2. Homework Equations :
39d48006ae0953cf0cc5bdec86aa9332.png




3. Attempt at solution:

My problem with finding the area of a polar graph is determining the bounds, so to get the right bounds for this graph do I set the equation equal to zero? I am really at a loss as to how to set up the bounds. Any hints would be helpful.
 
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So... the second revolution is between which values of \theta and the third is between which values of \theta? \theta = 0 is the start of the first.
 
2nd between -2pi and 2pi and the third between -4pi and 4pi?
 
One full revolution is 2\pi radians. If revolution 1 is for 0 < \theta < 2\pi and the second revolution starts at \theta = 2\pi radians and spans 2\pi subsequent radians, the second revolution has what range of \theta?.
 
2\pi < \theta < 4\pi
and the third would be 4\pi < \theta < 6\pi.
 
Correct.
 
So the bounds would be between 2\pi < \theta < 4\pi?
 
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