Results 1 to 9 of 9

Math Help - circle transformation

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    5

    circle transformation

    Hi,

    I have been given the following transformation:

    x=u+2v+1
    y=2u+v+1

    which transforms the circle u^2+v^2<=1 to the region R in the XY field.

    I am requested to find the Volume of R.

    I tried using a double integral with polar coordinates, but couldn't exactly understand how i am supposed to use the transformation.

    Can someone help me understand what I should be doing here ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by meirgold View Post
    Hi,

    I have been given the following transformation:

    x=u+2v+1
    y=2u+v+1

    which transforms the circle u^2+v^2<=1 to the region R in the XY field.

    I am requested to find the Volume of R.

    I tried using a double integral with polar coordinates, but couldn't exactly understand how i am supposed to use the transformation.

    Can someone help me understand what I should be doing here ?
    the Jacobian of the transformation is: J=\det \begin{pmatrix} 1 & 2 \\ \ 2 & 1 \end{pmatrix} =-3. so: the area of R=\int \int_{u^2 + v^2 \leq 1} |J| \ du \ dv=3\pi. \ \ \ \square
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2008
    Posts
    5

    not sure i understood

    Hi,

    thanks for the reply, but i'm not sure i understood 2 matters:

    1. What are the numbers which you used for the double integral ? I noticed that you used: u^2+v^2<=1, but what exact values did you extract from that formula ?

    2. I need to find the volume of R and not the area. Is there a way to find the volume from the area ?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2008
    Posts
    54
    Quote Originally Posted by meirgold View Post
    Hi,

    I have been given the following transformation:

    x=u+2v+1
    y=2u+v+1

    which transforms the circle u^2+v^2<=1 to the region R in the XY field.

    I am requested to find the Volume of R.

    I tried using a double integral with polar coordinates, but couldn't exactly understand how i am supposed to use the transformation.

    Can someone help me understand what I should be doing here ?
    Huh? Are you sure that you are asked to find the volume of R? Then I'm afraid that it would be 0 since there are only 2 dimensions. Are you sure that you posted the question correctly?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2008
    Posts
    5

    oooops

    sorry, you are correct - i was refering to the area.

    Can you explain the answer to my first question ?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by meirgold View Post
    Hi,

    thanks for the reply, but i'm not sure i understood 2 matters:

    1. What are the numbers which you used for the double integral ? I noticed that you used: u^2+v^2<=1, but what exact values did you extract from that formula ?

    2. I need to find the volume of R and not the area. Is there a way to find the volume from the area ?

    Thanks
    You're given u^2+v^2<=1 in the question so there's no mystery there. It's a unit circle so the double integral clearly represents three times the area of a unit circle, so there's no mystery there either.

    I don't understand exactly where your trouble is and what exact values you're referring to ......
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jun 2008
    Posts
    5
    I'll try to be more clear:

    Each of the integrals (within the double integral) should have a lower limit and an upper limit.
    I understand that somehow I'm supposed to derive them from the equation: u^2+v^2<=1 but am not sure how to do this.

    In addition, if the answer is 3\pi, than should I be using polar coordinates ? If so, shouldn't the Jacobian be equal to the circle radius (which is r=1) ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by meirgold View Post
    I'll try to be more clear:

    Each of the integrals (within the double integral) should have a lower limit and an upper limit.
    I understand that somehow I'm supposed to derive them from the equation: u^2+v^2<=1 but am not sure how to do this.

    In addition, if the answer is 3\pi, than should I be using polar coordinates ? If so, shouldn't the Jacobian be equal to the circle radius (which is r=1) ?

    You don't need to evaluate any integral. We didn't even convert the circle region to polar coords because we simply know what a circle's area is.

    I'll try to explain it step by step.

    You first have a circle, u^2 + v^2 \leq 1. Call this region D. The area of this circle is,

    \int \int_D~du~dv.

    Transform this to x-y coordinates, call the new region R. This transformation will give,

    \int \int_R |J|~dx~dy.

    So, \int \int_D~du~dv = \int \int_R |J|~dx~dy.

    Remember, \int \int_D~du~dv represents the area of a homogenous unit circle (a circle of radius 1). We know that it's \pi r^2 = \pi.

    \pi = \int \int_R |J|~dx~dy.

    |J| will be a constant (this is not a general rule but a special case). So we can write,

     \underbrace{\int \int_R ~dx~dy}_{\text{area of R}}=\frac{\pi}{|J|}

    Note that this Jacobian is different than NonCommAlg's, because I transformed from x-y to u-v while he did it the other way.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Jun 2008
    Posts
    5
    Thanks.
    It's all clear now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Circle, tangent line, and a point not on the circle
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: March 31st 2011, 01:40 PM
  2. Replies: 6
    Last Post: July 8th 2010, 05:39 PM
  3. Replies: 7
    Last Post: March 15th 2010, 04:10 PM
  4. Replies: 2
    Last Post: February 6th 2010, 08:31 AM
  5. Replies: 0
    Last Post: October 12th 2008, 04:31 PM

Search Tags


/mathhelpforum @mathhelpforum