Question

Jacobian

Answer 1

As in Jacobian matrix?

Answer 2

\[\int\limits_{?}^{?}\int\limits_{}^{}x-3y dA\]

Answer 3

x=2u+v y=u+2v

Answer 4

vertices (0,0) (2,1) (1,2) triangle

Answer 5

i got the jacobian J(u,v)=3

Answer 6

the 3 equations of lines are y=1/2x y=2x y=-x+3

Answer 7

Im not sure what to do next to solve the ingegral

Answer 8

are you getting the jacobian in order to convert this to a polar integration?

Answer 9

no it just says to use the jacobian transformation to solve the integral, an i got the =3. now i just have to make a substitution to finish it

Answer 10

to refresh my memory, the jacobian is the determinant (discriminant?) of the matrix of partial derivatives

\[U_x~V_x\\U_y~V_y\]

Answer 11

yea that's right

Answer 12

http://tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx

im having to read up on this to remember some stuff, but this here tends to be a very readable site

Answer 13

the change of variables formula is \[\int\limits_{?}^{?}\int\limits_{?}^{?}f(x,y) dA = \int\limits_{?}^{?}\int\limits_{?}^{?}f(x(u,v),y(u,v)) J(u,v) du dv\]

Answer 14

\[\int\int~x-3y~ dA~:~x=2u+v;~y=u+2v\]
\[\int\int~(2u+v)-3(u+2v)~ dA~\]
\[\int\int~2u+v-3u-6v~ dA~\]
\[\int\int~-u-5v~ dA~\]
seems to be a start

Answer 15

dA = J du dv; soo, dA = 3 du dv

Answer 16

yes thanks one last question why did they give me the 3 points to the triangle

Answer 17

in order to convert the skewed area of xy into a nice looking right triangle of uv

x=2u+v y=u+2v

vertices (0,0) (2,1) (1,2) triangle

0=2u+v
0=u+2v

Answer 18

or so im lead to believe :)

Answer 19

yes ok i think im good ill try to work it out

Answer 20

x=2u+v y=u+2v

vertices (0,0) (2,1) (1,2) triangle

use (0,0)

0=2u+v : *-2
0=u+2v

0=-4u-2v
0= u+2v
----------
0=-3u ; therefore u = 0, and v=0




(2,1)

-4=-4u-2v
1= u+2v
----------
-3 = -3u ; u=1, v=0


(1,2)

-2=-4u-2v
2= u+2v
----------
0 = -3u ; u=0, v=1

Answer 21

thanks

Answer 22

yep, with any luck i did that right ;)