Area in polar (stuck at the intersection points)

System
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Homework Statement


find the area inside both of the curves
r = 4 cos@
r = 2+2cos@
@ = theta


Homework Equations


-------


The Attempt at a Solution


i will say 4cos@ = 2+2cos@ to find the intersection points
4cos@ = 2+2cos@
2cos@ = 2
cos@ = 1
@ = 0 !
I need the other points!
 
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System said:

The Attempt at a Solution


i will say 4cos@ = 2+2cos@ to find the intersection points
4cos@ = 2+2cos@
2cos@ = 2
cos@ = 1
@ = 0 !
I need the other points!

Yes θ=0 is one point. Now just add π/2 to your principal answer of 0 and that will give you another answer.
 
System said:

Homework Statement


find the area inside both of the curves
r = 4 cos@
r = 2+2cos@
@ = theta


Homework Equations


-------


The Attempt at a Solution


i will say 4cos@ = 2+2cos@ to find the intersection points
4cos@ = 2+2cos@
2cos@ = 2
cos@ = 1
@ = 0 !
I need the other points!

You have to be careful with this type of problem. Eyes can be deceiving. Have you drawn a plot of the two curves? (I will use t for the angle). If so you will have noted that at t = 0 the r for both equations is 4, which is the point you have found. And the graphs both touch at the origin. The trouble is, at the origin t can be anything. And these two graphs do not have r = 0 for the same value of t. That is why you are having trouble finding the other point. What this means is you can't find the area with a single integral of the form

\frac 1 2 \int_{\alpha}^\beta r_{outer}^2-r_{inner}^2\, dt

So do the areas separately, each with their correct limits and subtract the inner area from the outer one. You might notice as t goes from 0 to 2\pi, one of the curves is traced twice.
 
rock.freak667 said:
Yes θ=0 is one point. Now just add π/2 to your principal answer of 0 and that will give you another answer.

why Pi/2 and not 2Pi? I would rather have said that cos θ has two easy roots: 0 and 2Pi, and all multiples of 2Pi.
 
Quantumjump said:
why Pi/2 and not 2Pi? I would rather have said that cos θ has two easy roots: 0 and 2Pi, and all multiples of 2Pi.

Sorry, I was solving cos(t)=0 not 1. You are right it would be 0 and 2π
 

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