Results 1 to 7 of 7

Math Help - Cartesian Polar Integral Converstion

  1. #1
    Newbie
    Joined
    Jun 2010
    Posts
    10

    Cartesian Polar Integral Converstion

    PLZ HELP! appreciations

    change the cartesian integral into polar integral then evaluate the polar integral
    Cartesian Polar Integral Converstion-math.jpg 0<y<6 0<x<y xdxdy
    Attached Thumbnails Attached Thumbnails Cartesian Polar Integral Converstion-math.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,407
    Thanks
    1294
    I don't see any reason to change this integral to a polar integral...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    I agree with Prove It. But if this is just your professor being annoying and wants you to use Polar coordinates even when it is unnecessary, you can recall that for Polar coordinates

    x \mapsto r \cos \theta

    y \mapsto r \sin \theta

    and

    dx~dy \mapsto r ~dr~d \theta

    Hopefully you can use this to figure it out.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2010
    Posts
    10
    it is required bythe book ..........i wouldnt want to change it ether if i wasnt required to
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2010
    Posts
    10
    can u atleast show me the bounds of integrations pleasae? i was wondering if i can have 0<r<root72 sinthata or 0<r<rsin thata
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Hopefully you can see that \frac {\pi}4 \le \theta \le \frac {\pi}2

    Now, you know that 0 < y < 6

    Switch to polar:

    0 < r \sin \theta < 6

    which means

    0 < r < 6 \csc \theta
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jun 2010
    Posts
    10

    Thumbs up

    Quote Originally Posted by jhevon View Post
    hopefully you can see that \frac {\pi}4 \le \theta \le \frac {\pi}2

    now, you know that 0 < y < 6

    switch to polar:

    0 < r \sin \theta < 6

    which means

    0 < r < 6 \csc \theta
    thank you!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: September 24th 2010, 04:33 AM
  2. Replies: 5
    Last Post: August 10th 2010, 11:15 PM
  3. Replies: 6
    Last Post: April 30th 2010, 01:19 PM
  4. Polar to cartesian
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: May 28th 2008, 10:17 AM
  5. cartesian and polar
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 3rd 2006, 08:02 AM

Search Tags


/mathhelpforum @mathhelpforum