Evaluate $$\int\limits\int\limits_{R}\frac{x+2y}{\cos(x-y)}dA $$
where R is the parallelogram bounded by the lines y=x, y=x-1, x+2y=0, and x+2y=2

I could use some help setting this up.

Question

Evaluate $$\int\limits\int\limits_{R}\frac{x+2y}{\cos(x-y)}dA $$
where R is the parallelogram bounded by the lines y=x, y=x-1, x+2y=0, and x+2y=2

I could use some help setting this up.

zarkon · Answer

looks like you will have two double integrals

chriss · Answer

I've got my four corner points for the parallelogram, but I get so confused setting up my limits of integration, and figuring out what order to integrate in.

My corners are (0,0), (2/3,2/3), (2/3,-1/3), (4/3,1/3)

zarkon · Answer

so one will have x=0 to 2/3

the other will have x=2/3 to 4/3

chriss · Answer

Two double integrals?  one for above the x axis and one for below?

zarkon · Answer

|dw:1336572645343:dw|

chriss · Answer

Ok, I see, so x will be my outside variable for both.  But I still don't really get how to get my y-limits for the inside part.

zarkon · Answer

the bounds for y are the equations of the lines

zarkon · Answer

for the first one -x/2<y<x

experimentx · Answer

http://www4c.wolframalpha.com/Calculate/MSP/MSP4451a1a5h962f8ag51700005a138998cah189bh?MSPStoreType=image/gif&s=30&w=386&h=299&cdf=RangeControl $$ \int_{0}^{2/3}\int_{x}^{-2x}f(x)dxdy + \int_{2/3}^{4/3}\int_{x-1}^{(2-x)/2}f(x)dxdy$$

zarkon · Answer

@experimentX 

it is x+2y=0, and x+2y=2

not 2x+y=0 and 2x+y=2

experimentx · Answer

sorry ... i was just giving it a try!!

zarkon · Answer

trying is good :)

experimentx · Answer

|dw:1336573162656:dw|
I haven't done this kinda question before!!