Double Integral of (x+y)x over a Quadrilateral Region R

[aks_sky]

Find the double integral of:

(x+y)x dxdy

Where R is a quadrilateral with vertices at (-4,-1), (-2,-2) (-1,1) and (-3,2)



**I have done the diagram and i know that there will be two regions R1 and R2 but i am not sure exactly how to find the limits of int. for these two regions, any suggestions?

thanx

 

[lanedance]

i drew this quickly, but looks like a rotated square to me, so you could think about a linear variable change to make the limits of integration independent & simple

 

[aks_sky]

How about taking the gradient of the 2 sides of the square? i think that might work. i will give that a try

 

[lanedance]

i'm not totally sure what you mean...

I was thinking along the lines of new variables u,v, whose level curves are lines parallel to the edges of the squares

 

[HallsofIvy]

Neither a square nor a rectangle- it is parallelogram. Two sides are of length $\sqrt{5}$ and the other two of length $\sqrt{10}$- and the sides are not perpendicular. However, it can be done as lanedance suggests: The lines through (-2,-2) and (-1,1) and through (-4,-1) and (-3,2) both have slope (1+2)/(-1+2)= 3. The first line is y= 3x+ 4 and the second y= 3x+ 7. If you let u= y- 3x, then the first line is u= -4 and the second u= -7. The lines through (-1, 1) and (-3, 2) and through (-2, -2) and (-4, -1) both have slope (2-1)/(-3+1)= -1/2. The first line is y= -(1/2)x+ 1/2 and the second is y= (-1/2)x- 1. If you let v= y+ (1/2) x, then the first line is v= 1/2 and the second is v= 1. Make that change of variables and don't forget to change the dxdy properly.

 

[aks_sky]

So basically i am solving it for dudv by making that change of variable. sweet