Polar equation problems

  • Thread starter rjs123
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  • #1
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Homework Statement




[tex] r = \theta + sin(2\theta)[/tex] for [tex] 0 \le \theta \le \pi [/tex]

a. Find the area bounded by the curve and the x-axis

b. Find the angle [tex]\theta[/tex] that corresponds to the point on the curve with x-coordinate -2

c. For [tex]\frac{\pi}{3} < \theta <\frac{2\pi}{3} , \frac{dr}{d\theta}[/tex] is negative. What does this fact say about r? What does this fact say about the curve?

d. Find the value of [tex]\theta[/tex] in the interval [tex] 0 \le \theta \le \frac{\pi}{2}[/tex] that corresponds to the point on the curve in the first quadrant with the greatest distance from the origin.


The Attempt at a Solution




a. [tex]1/2\int_{0}^{\pi }(\theta + sin2\theta)^2 d\theta}[/tex]


When I foil I end up with [tex] \theta^2 + 2\theta\sin2\theta + sin^2(2\theta)[/tex]

this is the part where I believe I made the mistake. I know the area is 4.9348 by the use of my calculator...if it is indeed correct...I can easily integrate the 1st and 3rd term...but i think the second term needs to be done by parts if im not mistaken.
 
Last edited:

Answers and Replies

  • #2
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Parts should be ok to solve that. It's pretty straight forward. If you're familiar with the "ultra-violet voodoo" pneumonic, then let the [tex]\theta[/tex] term be the "doo" (du) part.
 
  • #3
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i seem to be having trouble doing the integration by parts

[tex] 2\theta\sin2\theta[/tex]

using uv - [tex]\int[/tex] vdu

u = 2[tex]\theta[/tex] dv = [tex]sin2\theta[/tex]
du = 2[tex]d\theta[/tex] v = [tex]\frac{-1}{2}cos2\theta[/tex]

[tex]-\theta\cos2\theta + 1/2sin2\theta[/tex]



the final integral from [tex] \theta^2 + 2\theta\sin2\theta + sin^2(2\theta)[/tex] becomes:


[tex]\frac{1}{3}\theta^3-\theta\cos2\theta+ 1/2sin2\theta + \frac{1 - cos4\theta}{2}[/tex]

from 0 to [tex]\pi[/tex]
 
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  • #4
33
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Try

[tex]u=2\theta[/tex]
[tex]\Rightarrow du=2d\theta[/tex]
[tex]v=\sin 2\theta[/tex]
[tex]\Rightarrow dv=-2\cos 2\theta[/tex]

Then plug into your expression of

[tex]uv-\int v du[/tex]

Don't forget to evaluate that first part with limits.
 
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  • #5
90
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Try

[tex]u=2\theta[/tex]
[tex]\Rightarrow du=2d\theta[/tex]
[tex]v=\sin 2\theta[/tex]
[tex]\Rightarrow dv=-2\cos 2\theta[/tex]

Then plug into your expression of

[tex]uv-\int v du[/tex]

Don't forget to evaluate that first part with limits.

i'm pretty sure dv = sin2[tex]\theta[/tex]

and its integral v = -1/2cos2[tex]\theta[/tex]
 
  • #6
33
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Yes, you're correct. But it is of no consequence because we don't need the differential of v.
 
  • #7
33
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Wait. Hang on. No.

[tex]dv=-(1/2)\cos 2\theta d\theta[/tex]
 
  • #8
90
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Wait. Hang on. No.

[tex]dv=-(1/2)\cos 2\theta d\theta[/tex]

you set u to one part, and dv to another

du is the differential of u,

v is the integral of dv.

I'll work on this tomorrow...ahhhh so frustrating.
 
  • #9
90
0
any help at integrating this?


[tex] 2\theta\sin2\theta[/tex]
 

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