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Substitution with double integral

  1. Jan 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Using transforms: [tex] u = 3x + 2y[/tex] and [tex] v = x+4y[/tex] solve:
    [tex] \iint_\textrm{R}(3x^2 + 14xy +8y^2)\,dx\,dy[/tex]
    For the region R in the first quadrant bounded by the lines:
    [tex] y = -(3/2)x +1[/tex]
    [tex] y = -(3/2)x +3[/tex]
    [tex] y = -(1/2)x [/tex]
    [tex] y = -(1/2)x +1[/tex]

    I'm itching to see where I've gone wrong on this one which is why I'm taking the time to post it:

    2. Relevant equations

    3. The attempt at a solution
    First I express [itex]x[/itex] and [itex]y[/itex] in terms of [itex]u[/itex] and [itex]v[/itex]:
    [itex]3x = u-2y = u-\frac{(v-x)}{2}[/itex], [itex]6x = 2u-v+x[/itex] therefore [itex]x = \frac{2u-v}{5}[/itex]

    [itex]4y = v-x = \frac{5v-2u+v}{5}[/itex] such that [itex] y = \frac{3v-u}{10}[/itex]

    Next I express [itex]3x^2 + 14xy +8y^2[/itex] in terms of [itex]u,v[/tex]
    [itex]=\frac{1}{25}(25uv) = uv[/tex]

    Next I use [tex]\frac{\partial(x,y)}{\partial(u,v)} = \left[ \begin {array}{cc} x_{u}&x_{v}\\\noalign{\medskip}y_{u}&y_{v}\end {array}\right]= \left[ \begin {array}{cc} \frac{2}{5}&\frac{-1}{5}\\\noalign{\medskip}\frac{-1}{10}&\frac{3}{10}\end {array}\right]=\frac{1}{10}[/tex]

    Next find my region and limits in the uv plane:
    If [itex] y = \frac{-3}{2}x+1[/tex] then [itex] \frac{3v-u}{10}=\frac{-3}{2}(\frac{2u-v}{5})+1[/tex] such that [itex] 0 = 5u + 10[/tex] ie; [itex] u = 2[/tex]
    If [itex] y = \frac{-3}{2}x+3[/tex] then using above, [itex]u=6[/tex]
    If [itex] y = \frac{-1}{2}x[/tex] then [itex] \frac{3v-u}{10}=\frac{-1}{2}(\frac{2u-v}{5})[/tex] such that [itex] 2v = -u [/tex] ie; [itex] v = \frac{-u}{2}[/tex]
    If [itex] y = \frac{-3}{2}x+1[/tex] then using above [itex]v = 5- \frac{u}{2}[/tex]

    All this gives me the following integral to solve:
    [tex]\frac{1}{10}\int_2^6 \int_{\frac{-u}{2}}^{5-\frac{u}{2}}uv \,dv\,du[/tex]
    which I work out to be 8/3. I also used my computer to solve the integral in this form and it backs me up this yet the back of my book is looking for 64/5 :confused:
    If my method is wrong or if I keep missing a blatantly obvious algabraic slip could someone please point out my error(s) as I'll be sitting an exam tomorrow (Monday) and want to reduce the number of ways I can screw myself up!
    Last edited: Jan 12, 2008
  2. jcsd
  3. Jan 12, 2008 #2

    Gib Z

    User Avatar
    Homework Helper

    When you were finding limits in the uv plane, The lines 1 and 3, did you just get u and v confused or do i See in the first line : 3v - u = 3v - 6u, u=0. and 3rd line: 3v-u = v - 2u, u=0.
  4. Jan 12, 2008 #3
    hmm...in line 1 I forgot to write +1 (thats where I get 0 = -5u + 10 from), In line 3 I wrote u/2 instead of -u/2... (though that did appear in my integral correctly)...I have fixed that post now to show my working correctly
    Last edited: Jan 13, 2008
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