Results 1 to 3 of 3

Math Help - Double integral and polar coordinates

  1. #1
    Member
    Joined
    Jul 2005
    Posts
    187

    Double integral and polar coordinates

    I don't understand how to do these transformations as depicted on the picture.
    There's 2R = pi/3 and R/2=0 How to get these pi/3 and 0?
    Attached Thumbnails Attached Thumbnails Double integral and polar coordinates-polaar.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by totalnewbie View Post
    I don't understand how to do these transformations as depicted on the picture.
    There's 2R = pi/3 and R/2=0 How to get these pi/3 and 0?
    There are a number of typographical errors in your post (at a glance, there's two dx's and a missing term in the transformation Jacobian).

    From what I can tell, the region of integration is the upper half of the circle (x - R)^2 + y^2 = R^2 between the line x = R/2 and the y-axis. Do you realise this?

    A transformation is then made to polar coordinates. Do you realise that? Are you familiar with them?

    Do you know how to describe the region of integration using polar coordinates? In particular:

    1. Do you see how to express the line x = R/2 and the circle in polar coordinates.

    2. Do you see how to get the integral terminals for the angle?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    There are a number of typographical errors in your post (at a glance, there's two dx's and a missing term in the transformation Jacobian).

    From what I can tell, the region of integration is the upper half of the circle (x - R)^2 + y^2 = R^2 between the line x = R/2 and the y-axis. Do you realise this?

    A transformation is then made to polar coordinates. Do you realise that? Are you familiar with them?

    Do you know how to describe the region of integration using polar coordinates? In particular:

    1. Do you see how to express the line x = R/2 and the circle in polar coordinates.

    2. Do you see how to get the integral terminals for the angle?
    x = \frac{R}{2} \Rightarrow r \cos \phi = \frac{R}{2} \Rightarrow r = \frac{R}{2 \cos \phi}.

    y = \sqrt{2Rx - x^2} \Rightarrow y^2 = 2 R x - x^2

    \Rightarrow r^2 \sin^2 \phi =  2 R r \cos \phi - r^2 \cos^2 \phi \Rightarrow r = 2 R \cos \phi. (r = 0 is rejected - why?)


    At the point (2R, 0) \phi = 0.

    The line x = \frac{R}{2} cuts the semi-circle at \left( \frac{R}{2}, ~ \frac{R \sqrt{3}}{2}\right). At this point, \tan \phi = \frac{R \sqrt{3}/2}{R/2} = \sqrt{3} \Rightarrow \phi = \frac{\pi}{3}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Double Integral with Polar Coordinates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 8th 2011, 08:43 AM
  2. Double integral with polar coordinates
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 20th 2011, 03:44 PM
  3. Double integral in polar coordinates
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 16th 2010, 08:24 AM
  4. Double integral in polar coordinates 2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 24th 2008, 07:07 PM
  5. Double Integral Over Polar Coordinates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 9th 2007, 07:26 AM

Search Tags


/mathhelpforum @mathhelpforum