Polar coordinates

  • Thread starter queenstudy
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  • #1
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i just need to know something about the integral with polar coordinates , to know the interval of the teta angle of any domain the proffesor said that we put the pen on the x-axis and move it , i for one moment was not focusing and the doctor had to go to another class can any one explain ??? thank you
 

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  • #2
HallsofIvy
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Well, I have no clue what is meant here, or even what question you are asking. Perhaps he/she was saying that the angle, for a point (x,y) in Cartesian coordinates, is the angle the line through (0, 0) and (x, y) makes with the x-axis. That is, you measure the angle from the x-axis to that line.
 
  • #3
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i mean if you have a graph and you want to change your coordinates from cartesian to polar so that the double intergral is easier
okay here is an example a semi circle of center O(2,0) if i want to find teta in polar coordinates it will be 90 degree why
 
  • #4
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Any help is much appreciated

View attachment 41642

(5pi/6)
A= 1/2 ∫ [(4 sin⁡θ )^2-(2)^2 ]dθ
(pi/6)

4 sin⁡θ=2 θ=π/6 ,5π/6
 
  • #5
HallsofIvy
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What, exactly, are you asking? How they got those limits of integration?

You have two circles, one with center at (0, 0) and radius 2 and so equation [itex]x^2+ y^2= 4[/itex], the other with center at (0, 2) and radius 2 and so equation [itex]x^2+ (y- 2)^2= x^2+ y^2- 4y+ 4= 4[/itex]. Since those are both equal to 4, they are equal to each other: [itex]x^2+ y^2= x^2+ y^2- 4y+ 4[/itex] which reduces to 4y= 4 and y= 1. Then we have [itex]x^2+ y^2= x^2+ 1= 4[/itex] so [itex]x^2= 3[/itex] and [itex]x= \pm\sqrt{3}[/itex].
The two points of intersection are [itex](\sqrt{3}, 1)[/itex], [itex](-\sqrt{3}, 1)[/itex].

In terms of polar coordinates, [itex]\theta= tan^{-1}(y/x)[/itex] which gives [itex]\theta= tan^{-1}(1/\sqrt{3})= \pi/6[/itex] and [itex]\theta= tan^{-1}(-1/\sqrt{3})= 5\pi/6[/itex].
 
  • #6
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Thanks for replying back soo fast I appreciate it i was trying to find the shade parts are in attachment I also think I have it figured out but wouldnt mind a second pair of eyes to look over it. See attachments

Solve.png

Think_I_got_it.jpg


Thanks again
 
  • #7
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ok nice exercise now i got it
 
  • #8
HallsofIvy
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How did the "[itex]-8 cos(2\theta)+ 4[/itex]" on the fifth line become "[itex]-4 cos(2\theta)+ 4[/itex]" on the sixth line?

On the very next line, after integrating you have [itex]-4 sin(2\theta)+ 4\theta[/itex] which is correct because the integral of [itex]cos(2\theta)[/itex] is [itex](sin(2\theta))(1/2)[/itex] but there is no reason to divide by 2 before the integration.
 

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