Question

Double integrals: given trapezoid corner points (1,0),(2,0),(0,-2) and (0,-1) and have to use these to find my boundaries for the double integral of exp((x+y)/(x-y) dA. I know that x can be between 0 to 2. Same for y. But that would be the wrong area. So not sure what integral boundaries to use here...

Answer 1

it looks like they expect you to use a change of variables.
notice u= x+y will be parallel to the line connecting (0,-1) to (1,0)
ditto (0,-2) and (2,0)

Answer 2

yes, I noticed that. But wasn't sure what to do with it

Answer 3

Still not sure now :-)

Answer 4

so I have to use another set of variables instead of x and y. The integrals of the function in there is a pain. But the sphere variables don't apply.

Answer 5

u= x+y simplifies things
I am wonder what's a good choice for v

Answer 6

it looks like v= x-y is what you want to do
you now need the Jacobian, and the new limits.

Answer 7

Ok, I'll work out the Jacobian

Answer 8

so I have
\[\left[\begin{matrix}1 & 1 \\ 1 & -1\end{matrix}\right]\]

Answer 9

and you want the determinate of that

Answer 10

Which is -2, and I'll have to use the absolute, so that makes 2

Answer 11

So now I have
\[\int\limits_{}^{}\int\limits_{K}^{}2dudv\]

Answer 12

my notes say
du dv = |J| dx dy
so dx dy = 1/|J| du dv

Answer 13

Trying to figure out where to draw the v axis

Answer 14

You are correct. The roles of the u,v and x,y are reversed here in contrast to those of my course

Answer 15

the v axis is perpendicular to the u

Answer 16

That would make for the limits for u:-2,2 and for v:0,2?

Answer 17

Let's first clear up a detail.
I originally said
u= x+y

but the line connecting (0,-1) to (1,0) which has slope 1
in other words, we want u to be parallel to the line y=x or y-x=0
so let u = y-x

and v= x+y

Answer 18

Is it ok if I just do u=x+y and v=y-x?

Answer 19

ok, but then you have to relabel your picture

Answer 20

Isn't the one I drew x+y?

Answer 21

no, that is the mistake I made.
the line we want must have positive slope (= +1) based on change in y / change in x
the line x+y=0 --> y=-x (with y intercept of 0) has the wrong slope
y-x=0 --> y=x (with y intercept of 0) has the correct slope

Answer 22

Yeah, you're right, just checked with the slope for myself. And the perpendicular one would be y=-x => y+x

Answer 23

Ok then I'll have u=y-x and v=x+y

Answer 24

and the Jocobian is still the same

Answer 25

yup

Answer 26

But my chosen limits are correct no?

Answer 27

When I use -2<u<2 and 0<v<2 for my limits I find as a solution 4 for the double integral

Answer 28

u goes from x=0 (the starting left leg) to y=0 (the right leg) of the figure
we need to find x and y in terms of u and v

Answer 29

So u has (0,0) as limits?

Answer 30

v would have (1,2)

Answer 31

the limits of u will be a function of v

u= y -x
v= y+x
u+v = 2y
or y= (u+v)/2

also
v-u = 2x
x = (v-u)/2

Answer 32

I based my limits just based on the drawing

Answer 33

If I use the equations you put forward for
let's v=1 and y=0 then u=-1
let's v=2 and y=0 then u=-2

Answer 34

Well I think I've unconfused myself

if we use u= y-x and v= y+x

then that means the constant level curves for u are