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

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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 [itex]\theta[/itex] and the third is between which values of [itex]\theta[/itex]? [itex]\theta = 0[/itex] is the start of the first.
 
2nd between -2pi and 2pi and the third between -4pi and 4pi?
 
One full revolution is [itex]2\pi[/itex] radians. If revolution 1 is for [itex]0 < \theta < 2\pi[/itex] and the second revolution starts at [itex]\theta = 2\pi[/itex] radians and spans [itex]2\pi[/itex] subsequent radians, the second revolution has what range of [itex]\theta[/itex]?.
 
[itex]2\pi < \theta < 4\pi[/itex]
and the third would be [itex]4\pi < \theta < 6\pi[/itex].
 
So the bounds would be between [itex]2\pi < \theta < 4\pi[/itex]?