Change of Variables


by shards5
Tags: variables
shards5
shards5 is offline
#1
Mar25-11, 05:25 PM
P: 38
1. The problem statement, all variables and given/known data
Suppose D is the parallelogram in the xy-plane with vertices P(-1,5), Q(1,-5), R(5,-1), S(3,9)
[tex] \int\int ^{}_{D} (6x+12y) dA [/tex]
HINT: Use transformation x = [tex]\frac{1}{6}[/tex](u+v) and y = [tex]\frac{1}{6}[/tex] (-5u+v).

2. Relevant equations



3. The attempt at a solution
Calculating the Jacobian I get
dx/du = 1/6 dx/dv = 1/6
dy/du = -5/6 dy/dv = 1/6
which gives me
1/36 - (-5/36) = 1/6
From there I need the new intervals of integration.
P-Q
[tex]\frac{5-(-5)}{-1-(1)}[/tex] = 10/-2 = -5
y = 5x + 5 -> plugging in . . .
-5/6u+1/6v = -5/6u - 5/6v + 5 -> u cancels out and then we get 6/6v = 5 which means v = 5
S-R
[tex]\frac{9-(-1)}{3-5}[/tex] = -5
y = -5x + 9
-5/6u+1/6v = -5/6u+5/6v + 9 -> v = 9
R-Q
[tex]\frac{(-1)-(-5)}{5-1}[/tex] = 1
y = x-1 -> -5/6u + 1/6v = 1/6u +1/6v -1 -> u = 1
S-P
[tex]\frac{(9-(5)}{3-(-1)}[/tex] = 1
y = x-9 -> -5/6u +1/6v = 1/6u + 1/6v - 9 -> u = 9

Plugging in the x and y values and the Jacobian into the integral we get the following new integral.
[tex] \int^{9}_{5}\int^{9}_{1} (u+v + (-10u+2v)) * 1/6 dudv [/tex]
which simplifies to . . .
[tex] \int^{9}_{5}\int^{9}_{1} (-4/3u+1/2v) dudv [/tex]
After the first integration I get . . .
[tex] \int^{9}_{5} (-2/3u^2+1/2v*u) dudv [/tex]
Plugging in 9 and 1 I get
-54-9/2v+50/3-5/2v = -37.33333 - 7v
Integrating what I got I get
-37.33333v - 7/2v2
And after plugging in 9 and 5 I get -345.333
which is wrong. So what am I doing wrong?
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