Substitution with double integral

[GregA]

Homework Statement

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:

Homework Equations

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]<br /> 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<br /> =\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]&lt;br /&gt; =\frac{1}{25}(12u^2-12uv+3v^2+42uv-21v^2-14u^2+7uv+18v^2-12uv+2u^2)[/tex]&amp;lt;br /&amp;gt; =\frac{1}{25}(25uv) = uv[/tex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; Next I use \frac{\partial(x,y)}{\partial(u,v)} = \left[ \begin {array}{cc} x_{u}&amp;amp;amp;amp;amp;x_{v}\\\noalign{\medskip}y_{u}&amp;amp;amp;amp;amp;y_{v}\end {array}\right]= \left[ \begin {array}{cc} \frac{2}{5}&amp;amp;amp;amp;amp;\frac{-1}{5}\\\noalign{\medskip}\frac{-1}{10}&amp;amp;amp;amp;amp;\frac{3}{10}\end {array}\right]=\frac{1}{10}&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; Next find my region and limits in the uv plane:&amp;amp;lt;br /&amp;amp;gt; If y = \frac{-3}{2}x+1[/tex] then \frac{3v-u}{10}=\frac{-3}{2}(\frac{2u-v}{5})+1[/tex] such that 0 = 5u + 10[/tex] ie; u = 2[/tex]&amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;gt; If y = \frac{-3}{2}x+3[/tex] then using above, u=6[/tex]&amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;gt; If y = \frac{-1}{2}x[/tex] then \frac{3v-u}{10}=\frac{-1}{2}(\frac{2u-v}{5})[/tex] such that 2v = -u [/tex] ie; v = \frac{-u}{2}[/tex]&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; If y = \frac{-3}{2}x+1[/tex] then using above v = 5- \frac{u}{2}[/tex]&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; All this gives me the following integral to solve:&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; \frac{1}{10}\int_2^6 \int_{\frac{-u}{2}}^{5-\frac{u}{2}}uv \,dv\,du &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; 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 &amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;img src=&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot;https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f615.png&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot; class=&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot;smilie smilie--emoji&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot; loading=&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot;lazy&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot; width=&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot;64&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot; height=&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot;64&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot; alt=&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot;:confused:&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot; title=&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot;Confused :confused:&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot; data-smilie=&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot;5&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot;data-shortname=&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot;:confused:&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;quot; /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt;&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; 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&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;#039;ll be sitting an exam tomorrow (Monday) and want to reduce the number of ways I can screw myself up!

 

[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.

 

[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