# Substitution with double integral

1. Jan 12, 2008

### GregA

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

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 $x$ and $y$ in terms of $u$ and $v$:
$3x = u-2y = u-\frac{(v-x)}{2}$, $6x = 2u-v+x$ therefore $x = \frac{2u-v}{5}$

$4y = v-x = \frac{5v-2u+v}{5}$ such that $y = \frac{3v-u}{10}$

Next I express $3x^2 + 14xy +8y^2$ in terms of $u,v[/tex] [itex]3x^2+14xy+8y^2=3(\frac{2u-v}{5})^2+14(\frac{2u-v}{5})(\frac{3v-u}{10})+8(\frac{3v-u}{10})^2$
[itex]=\frac{3}{25}(4u^2-4uv+v^2)+\frac{7}{25}(6uv-3v^2-2u^2+uv)+\frac{2}{25}(9v^2-6uv+u^2)[/tex]
[itex]=\frac{1}{25}(12u^2-12uv+3v^2+42uv-21v^2-14u^2+7uv+18v^2-12uv+2u^2)[/tex]
[itex]=\frac{1}{25}(25uv) = uv[/tex]

Next I use \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}

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:
$$\frac{1}{10}\int_2^6 \int_{\frac{-u}{2}}^{5-\frac{u}{2}}uv \,dv\,du$$
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
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. Jan 12, 2008

### Gib Z

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.

3. Jan 12, 2008

### GregA

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