Converting to polar double integrals

In summary, the student is trying to find the area of a circle using polar coordinates. However, they are having trouble integrating the equation.
  • #1
cos(e)
27
0

Homework Statement


we are given the intgral from 0 to 2( the integral from 0 to sqrt(1-(x-1)^2) of ((x+y)/(x^2+y^2))dydx, so convert to polar integral and solve


Homework Equations





The Attempt at a Solution



i got integral from 0 to pi/2(integral 0 to 2cos(theta) of (x+y)/r dr d(theta), and not sure if I am right and don't know how to procede. i let x=1+cos(theta) and y=sin(theta) and r^2=x^2+y^2. sorry bout the ugly format but i don't know how to write it out in maths form.

Thanks in advanced
 
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  • #2
Strictly speaking, you have shifted the coordinate system so the origin is at the old (1, 0). That perfectly valid but might violate instruction to just change to polar coordinates.

If [itex]x= 1+ cos(\theta)[/itex] and [itex]y= sin(\theta)[/itex] then [itex]\sqrt{1- (x-1)^2}= \sqrt{1- r^2cos^2(\theta)}= sin(\theta)[/itex] so your equation becomes [itex]y= r sin(\theta)= sin(\theta)[/itex] and [itex]r= 1[/itex], a circle about the origin with radius 1 (the graph in the original coordinates is, of course, a circle about (1, 0) with radius 1).

But the fact that y is equal to the positive square root means you are only getting the upper half of the circle. The integral is from [itex]\theta= 0[/itex] to [itex]\pi[/itex], not [itex]2\pi[/itex]. And, in polar coordinates, the "differential of area", dxdy, becomes [itex]rdrd\theta[/itex].

Your integrand is [itex](x+y)/(x^2+ y^2)= [/itex]
[tex]\int_{r=0}^1\int_{\theta= 0}^\pi (r cos(\theta)+ r sin(\theta))/r^2= (1/r)(cos(\theta)+ sin(\theta)) r dr\theta[/itex]
 
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  • #3
ok i see, i thought about moving the orgin but i thought that would change my values of (x+y)/(x^2+ y^2) when integrating and screw up my integral? So in general, can i move the origin and the function in the double integral (in this case (x+y)/(x^2+ y^2) ) will always fix itself?
 
  • #4
okay i did your double integral of sin(theta) + cos(theta) drd(theta) and i got 2 yet the answers in the book have pi/2+1.

i tried letting x=1+rcos(theta) and y=rsin(theta), r^2=x^2+y^2, and i get the double integral of 1/r+cos(theta)+sin(theta), but when i antiderive with respect to r i get log|r|+... and when i sub in the terminals for r(0 and 1) i get log|0| which is undefined.

any help please?
 
  • #5
Good question Cos(e).

Worked Solution attached as Word Document. Fingers Crossed the file opens.
 

Attachments

  • coz integral.doc
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What is the purpose of converting to polar double integrals?

The purpose of converting to polar double integrals is to simplify the integration process for certain types of functions in two-dimensional space. It allows us to integrate over circular or symmetric regions by using polar coordinates, which can often yield simpler integrands.

How do you convert a double integral from rectangular to polar coordinates?

To convert a double integral from rectangular to polar coordinates, we use the following substitutions:

  • x = rcos(θ)
  • y = rsin(θ)
  • dA = rdrdθ

Then, we can rewrite the limits of integration in terms of r and θ and substitute these into the original integral.

What types of functions are best suited for conversion to polar coordinates?

Functions that have circular or symmetric regions are best suited for conversion to polar coordinates. This includes functions with equations in the form of r = f(θ) or θ = g(r), as well as functions with equations in the form of x2 + y2 = r2 or θ = arctan(y/x).

What are the benefits of converting to polar double integrals?

Converting to polar double integrals can simplify the integration process and make it easier to evaluate certain types of functions. It can also provide geometric insight into the function being integrated, as the polar coordinates represent points on a graph in terms of distance and angle from the origin.

Are there any limitations to using polar double integrals?

Yes, there are some limitations to using polar double integrals. For example, they are not suitable for integrating functions over non-circular or non-symmetric regions. They also may not be the most efficient method for evaluating certain types of functions, so it is important to consider alternative methods of integration as well.

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