Polar coordinates and integral

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Homework Help Overview

The discussion revolves around evaluating an integral using polar coordinates. The original poster expresses confusion about converting the limits of integration from Cartesian to polar coordinates, particularly for the dx limits, while the dy limits are defined in terms of r and theta.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the transformation of Cartesian coordinates to polar coordinates, including the relationships between dx, dy, and their derivatives. There is also a focus on understanding the graphical representation of the problem, particularly the intersections of lines and circles that define the limits of integration.

Discussion Status

Some participants have offered guidance on deriving the limits of integration based on the graphical interpretation of the problem. There is an ongoing exploration of how to properly set up the integrals in polar coordinates, with no explicit consensus reached yet.

Contextual Notes

The original poster mentions a lack of information in their textbook regarding the conversion of limits, which may be influencing their understanding of the problem setup.

Ylle
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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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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).
 
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].
 
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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