Results 1 to 5 of 5

Thread: Evaluating integral using change of variables

  1. #1
    Newbie
    Joined
    Apr 2009
    From
    Richmond, VA
    Posts
    5

    Evaluating integral using change of variables

    Hey guys, just having a bit of trouble getting started with this.

    Have to use a change of variables to evaluate the integral
    $\displaystyle \int\int_{D}sin(x+2y)sin(3x-y)dxdy$
    where $\displaystyle D={ (x,y)\in\Re^2 | 0 \leq x+2y \leq (\Pi/2) and 0\leq 3x-y\leq (\Pi/2) }$

    I assume we use polar coordinates, I'm just a bit lost with the inequalities in D
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by MrSplashypants1 View Post
    Hey guys, just having a bit of trouble getting started with this.

    Have to use a change of variables to evaluate the integral
    $\displaystyle \int\int_{D}sin(x+2y)sin(3x-y)dxdy$
    where $\displaystyle D={ (x,y)\in\Re^2 | 0 \leq x+2y \leq (\Pi/2) and 0\leq 3x-y\leq (\Pi/2) }$

    I assume we use polar coordinates, I'm just a bit lost with the inequalities in D
    substitution is what you're supposed to do here. let $\displaystyle x+2y=u, \ 3x - y= v,$ find the Jacobian of this transformation, etc. you'll end up with a very simple integral.

    by the way that's $\displaystyle \frac{\pi}{2}$ not $\displaystyle \frac{\Pi}{2}.$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    From
    Richmond, VA
    Posts
    5
    Quote Originally Posted by NonCommAlg View Post
    substitution is what you're supposed to do here. let $\displaystyle x+2y=u, \ 3x - y= v,$ find the Jacobian of this transformation, etc. you'll end up with a very simple integral.

    by the way that's $\displaystyle \frac{\pi}{2}$ not $\displaystyle \frac{\Pi}{2}.$
    Alright, so we define
    $\displaystyle g: D^* \longrightarrow D$
    $\displaystyle (u,v) \longmapsto (x,y)$
    where u=x+2y and v=3x-y
    Then $\displaystyle D^*$ is the square region bounded by $\displaystyle 0 \leq u \leq (\pi/2) and 0\leq v \leq (\pi/2)$

    And so the original integral becomes $\displaystyle \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}sin(u)sin( v)|detJ_{g} (u,v)|dudv$

    where the determinant = $\displaystyle \frac{1}{7}$

    and so the integral comes out as just $\displaystyle \frac{1}{7}$ as well?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by MrSplashypants1 View Post
    Alright, so we define
    $\displaystyle g: D^* \longrightarrow D$
    $\displaystyle (u,v) \longmapsto (x,y)$
    where u=x+2y and v=3x-y
    Then $\displaystyle D^*$ is the square region bounded by $\displaystyle 0 \leq u \leq (\pi/2) and 0\leq v \leq (\pi/2)$

    And so the original integral becomes $\displaystyle \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}sin(u)sin( v)|detJ_{g} (u,v)|dudv$

    where the determinant = $\displaystyle \frac{1}{7}$

    and so the integral comes out as just $\displaystyle \frac{1}{7}$ as well?
    yes. $\displaystyle \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \sin u \sin v |\det J| \ du \ dv = \frac{1}{7} \left( \int_0^{\frac{\pi}{2}} \sin u \ du \right)^2=\frac{1}{7}.$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2009
    From
    Richmond, VA
    Posts
    5
    Thank you very much sir
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Double integral change of variables
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Oct 15th 2011, 07:30 PM
  2. Replies: 3
    Last Post: May 18th 2011, 07:12 AM
  3. Evaluating an integral by change of variables
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jul 1st 2010, 05:08 PM
  4. Double integral change of variables.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 19th 2009, 06:48 PM
  5. Integral Change of Variables
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Mar 19th 2009, 01:24 PM

Search Tags


/mathhelpforum @mathhelpforum