Double integral ( and checking)

[vintwc]

Homework Statement

Let Ω ⊂ R^2 be the parallelogram with vertices at (1,0), (3,-1), (4,0) and (2,1). Evaluate ∫∫_Ω e^x dxdy.

Hint: It may be helpful to transform the integral by a suitable (affine) linear change of variables.

Homework Equations

The Attempt at a Solution

Ok here is what I have done:
From the sketch of the parallelogram, I have found the limits to be x=y+1 to (4-2y) and y=-1 to 1. With this, I am able to determine the solution of the integral which is 3/2*e^2 -1 -1/2*e^6.

Could anyone please verify this for me? Also, if my solution turns out to be right, how would I approach the hint to find the solution of this double integral? Thanks.

 

[Roni1985]

From what I understand, you need to separate the parallelogram into two areas when you calculate this double integral.
What you did was that you took only the upper part of the parallelogram .

you have 4 different sides which you have to take into account.

I think this is how you need to do it...

 

[vintwc]

Right, so should my integral look like as follows?

∫∫ e^x dxdy with x=4-2y to y+1, y=0 to 1 + ∫∫ e^x dxdy with x=1-2y to y+4, y=-1 to 0

 

[Roni1985]

Right, so should my integral look like as follows?

∫∫ e^x dxdy with x=4-2y to y+1, y=0 to 1 + ∫∫ e^x dxdy with x=1-2y to y+4, y=-1 to 0

I think it should be this:

∫∫ e^x dxdy with x=y+1 to 4-2y, y=0 to 1 + ∫∫ e^x dxdy with x=1-2y to 3y+4, y=-1 to 0

 

[vintwc]

I think it should be this:

∫∫ e^x dxdy with x=y+1 to 4-2y, y=0 to 1 + ∫∫ e^x dxdy with x=1-2y to 3y+4, y=-1 to 0

You are right for the limits of x but I think the y limits should be the way I first wrote it? Did you calculate the gradient wrongly? Btw, thanks for your help so far.