anonymous
  • anonymous
use the given transformation to evaluate the integral. u=x+y, v=-2x+y; integrateR: 2y dx dy where R is the parallelogram bounded by the lines y=1-x, y=4-x,y=2x+2,y=2x+5. I have already transformed the region into a square that intersects at (1,2), (4,2), (1,5), (4,5). I need help with setting up the integrals for calculation.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
after making the transformations u are asked to do i get this 4 equations corresponding to each of y=mx+b you got: u=-3/2,u=-12,v=4/3u +2,v=4/3u+5 Can you confirm befor continuing?
anonymous
  • anonymous
I am unsure of how you did that, I haven't had algebra class in quite a few years.
anonymous
  • anonymous
:) just solve u=x+y, v=-2x+y for x and y and then plug in the results in y=1-x, y=4-x,y=2x+2,y=2x+5 This is no algebra method

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anonymous
  • anonymous
I keep getting things like x=y-u and y=v-2x
anonymous
  • anonymous
plug the 1º into the secondand solve for y
anonymous
  • anonymous
so you have y=(y-u)-u
anonymous
  • anonymous
eh?? y = v-2(y-u) = v-2y+2u
anonymous
  • anonymous
but anyway i think this is wrong, becouse i got y=(v-2u)/3, x=(u-v)/3
anonymous
  • anonymous
ok I am with you so far. I still am unsure of why though.
anonymous
  • anonymous
this is for setting comfortably the integration limits
anonymous
  • anonymous
how do I put the other problems into it? would i just use y=2x+2 and put the (y-u) in for x?
anonymous
  • anonymous
|dw:1333385751096:dw| now you have this type of region
anonymous
  • anonymous
That is the region I started with.
anonymous
  • anonymous
I know. But that's what your problem whants you to do. Now you have to find this circled points that intersect horizontal lines with skewed ones. Those will be 3 intervals of integration for u|dw:1333385794920:dw|
anonymous
  • anonymous
1º going from left to right should be -21/8. 2º -51/4
anonymous
  • anonymous
i mean values of u
anonymous
  • anonymous
for that points
anonymous
  • anonymous
so new u integration values are from -51/4 to -21/8?
anonymous
  • anonymous
no
anonymous
  • anonymous
yeah i was thinking no.
anonymous
  • anonymous
you need another 2 points |dw:1333386156308:dw|
anonymous
  • anonymous
oh, so those points are brought down to the new points to show the new formation of the square?
anonymous
  • anonymous
you have to integrate now the 3 intervals|dw:1333386268902:dw|
anonymous
  • anonymous
first interval limits are v = -12, v =4/3u +5 second interval v = -12,v=-3/2 third interval v = 4/3u +2, v= -3/2
anonymous
  • anonymous
ok what do I do wih this information (sorry I sound so stupid, we just went over this in class).
anonymous
  • anonymous
now express this 2y dx dy in terms of u,v dudv and integrate
anonymous
  • anonymous
i think the rest you can do it on your own
anonymous
  • anonymous
ok i will try. Thank you.
anonymous
  • anonymous
dont forget the J matrix, :)
anonymous
  • anonymous
I have that as =3. I still can't figure out how to set up the integration

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