1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change of Variables for Double Integral

  1. Aug 10, 2011 #1
    Update: I figured out how to solve the problem. Nevermind.

    1. The problem statement, all variables and given/known data

    Use a suitable change of variables to evaluate the double integral:

    [itex]\int^{1}_{0}\int^{3-x}_{2x}(y-2x)e^{(x+y)^{3}}dydx[/itex]


    2. Relevant equations

    [itex]\frac{\partial(x,y)}{\partial(u,v)}=( \frac{\partial x}{\partial u} )(\frac{\partial y}{\partial v})-(\frac{\partial x}{\partial v})(\frac{\partial y}{\partial u})[/itex]

    [itex]dA=|\frac{\partial(x,y)}{\partial(u,v)}|dudv[/itex]


    3. The attempt at a solution

    I made the following change of variables:

    [itex]u=x+y[/itex]

    [itex]v=2x[/itex]

    (At first I tried u=y-2x and v=x+y but that didn't work)

    Next, I solved for x and y:

    [itex]x=\frac{v}{2}[/itex]

    [itex]y=\frac{2u-v}{2}[/itex]

    I found that the Jacobian is:

    [itex]\frac{\partial(x,y)}{\partial(u,v)}=-\frac{1}{2}[/itex]

    And therefore...

    [itex]dA=\frac{1}{2}dudv[/itex]

    Next I changed the boundaries:

    [itex]x=0\rightarrow v=0[/itex]

    [itex]x=1\rightarrow v=2[/itex]

    [itex]y=3-x\rightarrow u=3[/itex]

    [itex]y=2x\rightarrow u=\frac{3}{2}v[/itex]

    Therefore, the new region S on the uv-plane is the triangle described by:

    [itex]0\leq v\leq 2[/itex]

    [itex]\frac{3}{2}v\leq u\leq 3[/itex]

    The original integrand becomes:

    [itex](y-2x)e^{(x+y)^{3}}\rightarrow \frac{2u-3v}{2}e^{u^{3}}[/itex]

    So the new integral is:

    [itex]\frac{1}{4}\int^{2}_{0}\int^{3}_{\frac{3}{2}v}(2u-3v)e^{u^{3}}dudv[/itex]

    I used wolfram alpha to find out that the integral is equal to [itex]\frac{1}{18}(e^{27}-1)[/itex], which is the correct answer, but I am not sure how to evaluate the integral. Do I have to use another change of variables? Is there a better change of variables I could have used in the beginning?

    Thank you. Any help is much appreciated. :smile:

    Update:

    Okay, I figured out how to solve the integral. I'll post the rest of the solution in case anybody else who has this same question ever happens across this post. What I did was I switched the order of integration so that the integral becomes:

    [itex]\frac{1}{4}\int^{3}_{0}\int^{\frac{2}{3}u}_{0}(2u-3v)e^{u^{3}}dvdu[/itex]

    [itex]=\frac{1}{4}\int^{3}_{0}(2uv-\frac{3}{2}v^{2})e^{u^{3}}]^{\frac{2}{3}u}_{0}du[/itex]

    [itex]=\frac{1}{6}\int^{3}_{0}u^{2}e^{u^{3}}du[/itex]

    Use substitution with [itex]w=u^{3}[/itex] and [itex]dw=3u^{2}du\rightarrow \frac{1}{3}dw=u^{2}du[/itex]:

    [itex]\frac{1}{18}\int^{27}_{0}e^{w}dw[/itex]

    [itex]=\frac{1}{18}e^{w}]^{27}_{0}[/itex]

    [itex]=\frac{1}{18}(e^{27}-1)[/itex]
     
    Last edited: Aug 10, 2011
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Change of Variables for Double Integral
  1. Change of Variable (Replies: 0)

  2. Change of Variables (Replies: 0)

Loading...