Polar coordinates and integral

  • Thread starter Ylle
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  • #1
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Homework Statement


I have a problem I hope you guys can help me with.
It's quite simple I think, but there is one thing that I can't figure out.


Homework Equations


I have to use polar coordinates to evaluate this integral:
See image


The Attempt at a Solution


I really don't have an attempt, because I can't remember, and it doesn't say so in my book, how I turn the dx limits into polar coordinates (theta).
The dy limits are:
x = r*cos(theta)
y=r*sin(theta)

, where you just insert them at the x and y. But it's the dx limits that annoy me. Do I just leave them as they are (Which sound unlogical to me), or do I change them into angles ? If yes, how ?


Hope you can help.


- Ylle
 

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Answers and Replies

  • #2
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Hi,

If you let [tex]x=r*cos(\theta) \ and \ y=r*sin(\theta)[/tex]

then [tex]\frac{dx}{d\theta}=-r*sin(\theta) and \frac{dy}{d\theta}=r*cos(\theta)[/tex]

You can then rearrange these to get [tex]dx=-r*sin(\theta)*d\theta \ and \ dy=r*cos(\theta)*d\theta[/tex]

Your limits on the integals can be found by subbing the x and y values into the new expressions for x and y. Remember that the inside integral is with respect to y and the outside integral is with respect to x. Also r=sqrt(x^2+y^2).
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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Have you draw a graph of this?

You have two circles, both with center (0,0), radius 1 and 2, and the straight line y= x.

the line y= x crosses the circle x+ y2= 1 at [itex]x= 1/\sqrt{2}[/itex] and y= 0 crosses the circle at x= 1. That's the reason for the limits of integration in the first integral. That line y= x crosses the circle x2+ y2= 4 at [itex]x= \sqrt{2}[/itex]. That's where the limits on the second integral come from. The line y= 2 crosses the circle at y= 2 which is where the final limits come from. Look closely at your graph and you should be able to see why it is divided into 3 integrals.

To change the order of integration, look at your graph. What are the least and greatest values of y? The least is, of course, y= 0. The greatest occurs where the line y= x crosses the larger circle which is [itex]\sqrt{2}[/itex]. Those will be the limits on the outer, dy, integral: y= 0 and [itex]y= \sqrt{2}[/itex]. Now, the limits on the inner, dx, integral for each y will be the left and right hand x-values of the horizontal line at y. You should be able to see that there will be two distinct integrals because the left-hand curve changes where y= x intersects the smaller circle: at [itex]y= 1/\sqrt{2}[/itex].
 
  • #4
79
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But when I have the limits, do I then just integrate x*y with the limits I found, and then replace x and with x=r*cos(theta) and y=r*sin(theta) for dr dtheta ?
 

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