Solving Change of Variables in Parallelogram | Homework

[shards5]

Homework Statement

Suppose D is the parallelogram enclosed by the lines 2x-3y = 0, 2x-3y = 2, 3x-y = 0 and 3x-y = 1.
\int\int^{}_{D} [(2x-3y) e^(3x-y) dA

Homework Equations

The Attempt at a Solution

Set u to be equal to 2x-3y -> x = (u+3y)/2
Set v to be equal to 3x-y -> v = 3/2(u+3y)-y -> 7/2y = v - 3/2u
Solve for x gives me x = -1/7u + 3/7v
Solve for y gives me y = 2/7v-37u
Using the above I get the following derivatives for Jacobian
du/dx = -1/7 dv/dx = 3/7
du/dy = 3/7 dv/dy = 2/7
which gives me -11/49 and the formula requires the absolute value so I will use 11/49 as the Jacobian
Now I need the intervals of integration so I plug in the x and y into the boundaries.
2x-3y = 0 -> 2(-1/7u+3/7v) - 3(2/7v-3/7u) = 0 -> -2/7u + 6/7v -6/7v +9/7u = 0 -> u = 0
2x-3y = 2 -> since this is in the same format as the above I get u = 2
3x-y = 0 -> 3(-1/7u+3/7v)-(2/7v-3/7u) = 0 -> -3/7u + 9/7v - 2/7v +3/7u = 0 -> v = 0
3x-y = 1 -> since this is similar to the above I get v = 1
Plugging in the values of integration and the new equations for x and y I get the following.
\int^{1}_{0}\int^{2}_{0} 2*(-1/7u + 3/7v) - 3(2/7v - 3/7u) e^(3(-1/7u+3/7v) - (2/7v - 3/7u) (11/49) dudv
Simplifying we get.
\int^{1}_{0}\int^{2}_{0} (11/49) ue^v dudv
After the first integration I get:
\int^{1}_{0} (u^2/2)e^v (11/49) dv
Plugging in I get
\int^{1}_{0} 22/49e^v dv
Integrating for v I get:
22/49e^v
Plugging in 1 and 0 I get
22/49*e¹-22/49*e⁰

And this is wrong. I am not exactly sure what I am doing wrong.

 

[LCKurtz]

Your integrand and limits look OK. I didn't check your actual integration steps, but one place you have an error is in calculating the Jacobian. You should be calculating

\frac{\partial(x,y)}{\partial(u,v)}<br /> =\left | \begin{array}{cc}<br /> x_u &amp; x_v\\<br /> y_u &amp; y_v<br /> \end{array}\right |

and using its absolute value.

 

[shards5]

I think I calculated my Jacobian correctly since I am taking the correct partial derivatives of my functions.
dx/du -1/7u + 3/7v = -1/7
dx/dv -1/7u + 3/7v = 3/7
dy/du 2/7u - 3/7v = 2/7
dy/dv 2/7u - 3/7v = 3/7
-1/7*2*7-3/7*3/7 = -2/49-9/49 = -11/49 and absolute value gives me 11/49

 

[LCKurtz]

Read it a little more carefully. You did not calculate dx/du etc. Or maybe you called it du/dx.

[Edit] Shouldn't y_v= -3/7?

 

[shards5]

Oh I am so sorry about that, I completely mistyped that. It was supposed to be dx/du, dx/dv, dy/du, dy/dv. Also thanks you pointed out where I got it wrong. That helped a lot.