Results 1 to 5 of 5

Math Help - variable substitution integral mmn 15 2c

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    variable substitution integral mmn 15 2c

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602
    After sketching the region



    it should be clear that you can evaluate the required area as

    \displaystyle \int_0^{\sqrt[3]{28}}{\frac{1}{2}x^2 - \frac{1}{4}x^2\,dx} + \int_{\sqrt[3]{28}}^{\sqrt[3]{112}}{\sqrt{x} - \frac{1}{4}x^2\,dx}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    coold you hekp me solve it by my way
    because on the test it not effective to solve it like this
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,411
    Thanks
    1850
    Nice graph! Transgalactic, the point of the u, v substitution is that the curves x^2= 2y, x^2= 4y and y^2= 7x, in the xy-plane, become, in the uv-plane, u= 2, u= 4, and u= 7. Of course, those don't form a bounded figure. That's because, as x and y go to 0, u and v go to infinity. So your integral will be for u= 2 to 4, v= 7 to infinity.

    Now, when you substitute for x and y, you also have to substitute for u and v: dxdy= \frac{\partial x\partial y}{\partial u\partial v} dudv. And that symbol refers to the determinant, \left|\begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}\end{array}\right|.

    In order to find that directly, you would have to solve for x and y in terms of u and v. The hint says you don't have to do that. Find, instead, \frac{\partial u\partial v}{\partial x\partial y}= \left|\begin{array}{cc}\frac{\partial u}{\partial x} & \frac{\partial v}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y}\end{array}\right| and take the reciprocal.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    i put v to be the vertical axes
    and u to be the horizontal axes
    so i dont have 7 to infinity
    but 0 to 7
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Substitution Integral
    Posted in the Calculus Forum
    Replies: 8
    Last Post: January 26th 2010, 09:04 PM
  2. Integral Substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 2nd 2009, 11:32 AM
  3. variable substitution question(diff)
    Posted in the Differential Equations Forum
    Replies: 14
    Last Post: August 17th 2009, 09:13 PM
  4. Variable substitution in a derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 25th 2009, 01:51 PM
  5. Integral Substitution
    Posted in the Calculus Forum
    Replies: 9
    Last Post: October 19th 2008, 08:15 PM

Search Tags


/mathhelpforum @mathhelpforum