Jacobian Transformation, Substitution of varibles

[Idealistic]

integral[(x + y)e[sup:d6ae0qa7](x^2 - y^2)[/sup:d6ae0qa7]dA], bounded by the lines:

y = -x; y = x - 2; y = 3 - x; y = x;

These four lines form a parellelogram whose verticies are (0, 0), (1, -1), (3/2, 3/2), (5/2, 1/2)

If I let u = (x + y) and v = (x[sup:d6ae0qa7]2[/sup:d6ae0qa7] - y[sup:d6ae0qa7]2[/sup:d6ae0qa7]) ..to make the double integral above easier to evaluate..

I get a new Region, a triangle, with verticies (0,0), (3, 0), and (3, 6).

I get this from pluging in the original four verticies into my u and v equations to form the new verticies with coordintes (u[sub:d6ae0qa7]i[/sub:d6ae0qa7], v[sub:d6ae0qa7]i[/sub:d6ae0qa7]).. i.e:

(u = x+ y; v = x[sup:d6ae0qa7]2[/sup:d6ae0qa7] - y[sup:d6ae0qa7]2[/sup:d6ae0qa7]) by plugging in the first verticy with coordinates (x, y) = (0, 0) from above, I get:

u = (0) + (0) = 0; v = (0)[sup:d6ae0qa7]2[/sup:d6ae0qa7] - (0)[sup:d6ae0qa7]2[/sup:d6ae0qa7] = 0 ..so my coordinate (u, v) = (0,0) ..I continue to do this for the remaing verticies to get another (0, 0), (3, 0) and (3, 6),

My only problem is that to find the new integral I need to find the Jacobian. And to find the Jacobian I need to find an equation for x and y in terms of u and v. (so I can calculate the det(d(x,y)/d(u,v))

How do I write x and/or y strictly in terms of u and v in this particular question?

.

 

[BigGlenntheHeavy]

\(\displaystyle \int_{R}\int(x+y)e^{x^{2}-y^{2}}dA, \ find \ Volume.\)

\(\displaystyle Given: \ y \ = \ -x, \ y \ = \ x, \ y \ = \ x-2, \ and \ \ y=3-x\)

\(\displaystyle y+x \ = \ 0 \ \ \ y-x \ = \ 0\)
\(\displaystyle y+x \ = \ 3 \ \ \ y-x \ = \ -2, \ Let \ u \ = \ y+x \ and \ v \ = \ y-x\)

\(\displaystyle Then \ x \ = \ \frac{u-v}{2} \ and \ y \ = \ \frac{u+v}{2}\)

\(\displaystyle Note \ also:\)
\(\displaystyle y+x \ = \ 0 \ \implies \ u \ = \ 0\)
\(\displaystyle y+x \ = \ 3 \ \implies \ u \ = \ 3\)
\(\displaystyle y-x \ = \ 0 \ \implies \ v \ = \ 0\)
\(\displaystyle y-x \ = \ -2 \ \implies \ v \ = \ -2\)

\(\displaystyle \frac{\partial(x,y)}{\partial(u,v)} \ = \ \bigg|\frac{(\partial x)/(\partial u) \ (\partial x)/(\partial v)}{(\partial y)/(\partial u) \ (\partial y)/(\partial v)}\bigg| \ = \ \bigg|\frac{1/2 \ -1/2}{1/2 \ \ 1/2} \ \bigg| \ = \ \frac{1}{4}+\frac{1}{4} \ = \ \frac{1}{2}\)

\(\displaystyle Hence, \ V \ = \ \frac{1}{2}\int_{0}^{3}\int_{-2}^{0}ue^{-uv}dvdu \ = \ \frac{e^{6}-7}{4}\)

 

[Idealistic]

Thanks for the clarification.

 

[BigGlenntheHeavy]

You're welcome Idealistic. I'm a little rusty on Carl, but I think I got it right.