Calculus 3 double integral

Question

Find ∫∫R(2x+3y)dA where R is the parallelogram with vertices (0,0), (1,3), (-3,5), and (-2,8).Use the transformation x=u−3v, y=3u+5v

Answer 1

Using the x,y points one can find the equations of the sides of the parallelogram. Side one from (-3,5) to (-2,8) is 3x-y=-14, Side 2 from (-2,8) to (1,3) is 5x/3 +y +14/3 side 3 from (1,3) to (0,0) is 3x-y=0 and side 4 from (0,0) to (-3,5) is 5x/3+y=0> These can now be used to find the mapping into (u,v) by substituting x= u - 3v and y=3u + 5v into the equations for the lines and find u ranges from 0 to 1 and v ranges from -1 to 0. The Jacobian can be found from the x and y equations in terms of u and v and gives J= 14. So the integral becomes as follows:

∫_-1⁰∫₀¹(2(u-3v) +3(3u +5v))14dudv. Doing the integration gives 14 as the area.

Answer 2

∫∫R(2x+3y) d A.

Solving for u and v we get u = [ 5x +3y ] / 14 and v= [y-3x] / 14

Then the given points in the xy- plane correspond to the points in the uv-plane

xy- plane uv- plane

( 0,0 ) (0,0)

( 1, 3) (1, -3/7)

( -3, 5) ( 0,1)

( -2, 8) (1,1)

Let us find the Jacobian ∂x/∂u= 1 ∂x/∂v=-3

∂y/∂u =3 ∂y/∂v= 5

Then ∂(x,y) / ∂(u,v) = 14

∫∫R(2x+3y) d A. = ∫^u=1_u=0 ∫^v=1_{v=- 3/7u}[2u - 6v +3u+15v] 14 dv du =

=14 ∫^u=1_u=0 ∫^v=1_{v=- 3/7u}[5u+9v]dv du = 14 ∫^u=1_u=0[5uv +9v²/2]|¹_-3u/7du=

= 14 ∫^u=1_u=0[ 5u+9/2 + 24u²/49]du

= 14 { 5u²/2 + 9u/2 +8u³/ 49 } |¹₀

= 14 {5/2 +9/2 +8/49 } =14 {7+ 8/49}



= 702 /7