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- Thread starter queenstudy
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HallsofIvy

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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

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View attachment 41642

(5pi/6)

A= 1/2 ∫ [(4 sinθ )^2-(2)^2 ]dθ

(pi/6)

4 sinθ=2 θ=π/6 ,5π/6

- #5

HallsofIvy

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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].

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Thanks again

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ok nice exercise now i got it

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HallsofIvy

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On the very next line, after integrating you have [itex]-4 sin(2\theta)+ 4\theta[/itex] which is

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