# Change to polar coordinates integration Problem

1. Aug 4, 2012

### Delphi27

1. The problem statement, all variables and given/known data

Integrate y/(x^2+y^2) for x^2+y^2<1 and y> 1/2 ; use change of variables to polar coordinates

2. Relevant equations

THe above

3. The attempt at a solution

the variables transform as

y=rsinz
x=rcosz, where z is an angle between pi/6 and 5*pi/6 = which is the ArcSin of 1/2

That is sinz > 1/2 means pi/6 < z < 5*pi/6,

The Integral in polar coordinates becomes

Integral of [ r*sinz / { (r*sinz)^2 + (r*cosz)^2 } ] * r drdz

This simplifies to

Integral of [r*sinz / { r^2 * 1 } ]*r drdz, or

=Integral [ sinz drdz ]

Now y > 1/2 means that r must be between [ (1/2)^ + (cosz)^2 ] ^ (1/2) and 1

by the Pythagorean Theorum.

So i end up with the double integral

Integral rsinz dz evaluated at r = 1, minus same evaluated at
r = [ (1/2)^ + (cosz)^2 ] ^ (1/2)

So I then get Integral [ 1 - [ (1/2)^ + (cosz)^2 ] ^ (1/2) ] * sinz dz.

The book gives the answer (sqrt 3 ) - pi/3, that is after evaluating what I would think would be the above from z = pi/6 to 5*pi/6.

The integral I came up with after substituting the range of r stated as functions of z, does not seem to integrate and must be wrong -

Can anyone tell me where I went off track ?

Would be much appreciated - hours on this, and cannot get it.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 4, 2012

### SammyS

Staff Emeritus
.

Hello Delphi27. Welcome to PF !

Why not express this in terms of sin(z), rather than cos(z) ?

You know that y = r sin(z), right?

So y = 1/2 becomes r sin(z) = 1/2.

3. Aug 4, 2012

### voko

$y > \frac {1} {2}$ means $r sin z > \frac {1} {2}$

4. Aug 4, 2012

### Delphi27

Volko and SammyS --- thanks so much - you guys got me on the right track !!

5. Aug 4, 2012

### SammyS

Staff Emeritus
You're very welcome !