(adsbygoogle = window.adsbygoogle || []).push({}); 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 - 7/2v^{2}

And after plugging in 9 and 5 I get -345.333

which is wrong. So what am I doing wrong?

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# Change of Variables

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