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GregA
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Homework Statement
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:
Homework Equations
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]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]
[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 [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
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!
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