Results 1 to 6 of 6

Math Help - changing order of integration

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    6

    changing order of integration

    ∫∫x^3 y/ 1-y^2 dy dx

    x= 0
    x=1
    y= x^2
    y=x

    By changing the order of integration show that the value of the integral is 1/16

    please could anyone help on this one please!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    \int_{0}^{1}\int_{y}^{\sqrt{y}}\left(\frac{x^{3}y}  {1-y^{2}}\right)dxdy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2007
    Posts
    6
    dude the y integral is not the square root of x. it is x squared!

    could u please help me solve this?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by stuwy View Post
    dude the y integral is not the square root of x. it is x squared!
    stop shouting.

    y = x^2 \implies x = \pm \sqrt{y}

    after sketching the region, galactus realize he needed to take the positive square root
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    645
    Hello, duude!

    \int^1_0\int^x_{x^2}\frac{x^3y}{1-y^2}\,dy\,dx

    By changing the order of integration, show that the value of the integral is \frac{1}{16}
    The limits are: . \begin{Bmatrix}x = 1 \\ x=0\end{Bmatrix} \text{ and }\begin{Bmatrix}y = x \\ y = x^2\end{Bmatrix}

    The region looks like this:
    Code:
            |
           1+ - - - - - *
            |         **:
            |       *:: :
            |     *:::* :
            |   *::::*  :
            | *::::*    :
          - **- - - - - + - -
            |           1

    Reversing the limits, we have: . \begin{Bmatrix}y = 1 \\ y = 0\end{Bmatrix}\text{ and }\begin{Bmatrix}x = \sqrt{y} \\ x = y\end{Bmatrix}

    And the integral is: . \int^1_0\int^{\sqrt{y}}_y\frac{x^3y}{1-y^2}\,dx\,dy


    The inner integral is: . \int^{\sqrt{y}}_y\frac{x^3y}{1-y^2}\,dx \;\;=\;\;\frac{1}{4}\cdot\frac{x^4y}{1-y^2}\,\bigg|^{\sqrt{y}}_y \;\;=\; \;\frac{1}{4}\,\left[\frac{(\sqrt{y})^4y}{1-y^2} - \frac{y^4\cdot y}{1-y^2}\right]

    . . = \;\;\frac{1}{4}\left[\frac{y^3}{1-y^2} - \frac{y^5}{1-y^2}\right] \;\;=\;\;\frac{1}{4}\left[\frac{y^3(1-y^2)}{1-y^2}\right] \;\;=\;\;\frac{1}{4}y^3


    Then we have: . \int^1_0\frac{1}{4}y^3\,dy \;\;=\;\;\frac{1}{16}y^4\,\bigg|^1_0 \;\;=\;\;\frac{1}{16}

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2007
    Posts
    6

    Talking

    Thanks dude!
    That was great!

    Cheers!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Changing order or integration
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 27th 2011, 03:33 PM
  2. changing the order of integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 16th 2009, 04:31 AM
  3. Changing the Order of Integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 5th 2009, 03:06 PM
  4. Changing order of integration?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 20th 2008, 01:39 PM
  5. Changing order of integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 27th 2008, 06:13 AM

Search Tags


/mathhelpforum @mathhelpforum