Results 1 to 5 of 5

Math Help - difficult integral

  1. #1
    Member kezman's Avatar
    Joined
    Jul 2006
    Posts
    88
    Thanks
    2

    difficult integral

    Find the area bounded by the curve given by the equation:
    (x^2 + y^2)^2 = 2a^2(x^2 - y^2)
    How do you transform the equation to the integral?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,664
    Thanks
    298
    Awards
    1
    Quote Originally Posted by kezman View Post
    Find the area bounded by the curve given by the equation:
    (x^2 + y^2)^2 = 2a^2(x^2 - y^2)
    How do you transform the equation to the integral?
    Solve for y:
    (x^2 + y^2)^2 = 2a^2(x^2 - y^2)

    x^4 + y^4 + 2x^2y^2 = 2a^2x^2 - 2a^2y^2

    y^4 + 2(x^2 + a^2)y^2+ x^2(x^2 - 2a^2) = 0

    y^2 = \frac{-2(x^2 + a^2) \pm \sqrt{[2(x^2 + a^2)]^2 - 4x^2(x^2 - 2a^2)}}{2}

    We need to keep the "+" solution else y isn't real. So:
    y^2 = \frac{-2(x^2 + a^2) + \sqrt{16x^2a^2 +  4a^4}}{2}

    y^2 = \frac{-2(x^2 + a^2) + 2a\sqrt{4x^2 +  a^2}}{2}

    y^2 = -(x^2 + a^2) + a\sqrt{4x^2 +  a^2}

    Thus
    y = \pm \sqrt{-(x^2 + a^2) + a\sqrt{4x^2 +  a^2}}

    The area you are after will be between the "+" and "-" curves, and you know by symmetry they cross on the x-axis so you should be able to find these fairly easily. The integral you will be doing will be of the form:
    \int_a^b \sqrt{-(x^2 + a^2) + a\sqrt{4x^2 +  a^2}}~dx
    which isn't going to be at all pretty. You'll likely have to do a numerical estimate. (Unless Krizalid or someone else pulls a trick out of their hat.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    If a>0 this is the Leminscate of Bernoulli.

    Express this curve in polar coordinates for simplification.

    This is Mine 69th Post!!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hellow, kezman!

    Find the area bounded by the curve given by the equation:
    . . (x^2 + y^2)^2 \:= \:2a^2(x^2 - y^2)
    I would convert to polar coordinates . . .

    Conversions: . x \: =\: r\cos\theta,\quad y \:=\: r\sin\theta,\quad x^2+y^2 \:=\: r^2


    Then we have: . \left(r^2\right)^2 \;=\;2a^2\left(r^2\cos^2\!\theta - r^2\sin^2\!\theta\right)\quad\Rightarrow\quad r^4\;=\;2a^2r^2\left(\cos^2\!\theta - \sin^2\!\theta\right)

    . . which simplifies to: . r^2\;=\;2a^2\cos2\theta

    This looks like a large Infinite-sign: \infty

    . . It is symmetric to both the "x-axis" and the "y-axis".


    We can integrate from 0 to \frac{\pi}{4} and multiply by 4.

    . . Area \;=\;4 \times \frac{1}{2}\int^{\frac{\pi}{4}}_0 2a^2\cos2\theta\,d\theta  \;=\;4a^2\int^{\frac{\pi}{4}}_0\cos2\theta\,d\thet  a


    Got it?

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member kezman's Avatar
    Joined
    Jul 2006
    Posts
    88
    Thanks
    2
    Yes! excelent all the responses. Thank you very much for the help-
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Difficult integral
    Posted in the Calculus Forum
    Replies: 10
    Last Post: April 15th 2010, 11:33 AM
  2. Difficult integral!?!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 20th 2010, 10:33 AM
  3. Difficult Integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 16th 2010, 02:20 PM
  4. Difficult Integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 22nd 2009, 10:08 AM
  5. Difficult integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 30th 2008, 09:27 AM

Search Tags


/mathhelpforum @mathhelpforum