∫∫_{D}( (-x-2y)*cos( (2x+2y)*(-x-2y) )dxdy; D is the area enclosed by the points (−2−1/3π,1+1/3π), (−4−1/3π,2+1/3π), (−2−1/2π,1+1/2π) and (−4−1/2π,2+1/2π).

Question

∫∫_{D}( (-x-2y)*cos( (2x+2y)*(-x-2y) )dxdy; D is the area enclosed by the points  (−2−1/3π,1+1/3π), (−4−1/3π,2+1/3π), (−2−1/2π,1+1/2π) and  (−4−1/2π,2+1/2π).

anonymous · Answer

I'm overwhelmed by the gnarly area; is there a simple substitution that I'm not seeing? Moreover, the integrand in terms of x and y is not going to be pleasant. :(

anonymous · Answer

I guess ∫∫_{E}( u*cos(uv)*|J(u,v)| )dudv; but what about E?

anonymous · Answer

u=-x-2y;v=2x+2y

ganeshie8 · Answer

|dw:1393088587498:dw|

ganeshie8 · Answer

(−2−1/3π,1+1/3π)  ->  (-1/3pi, -2 )

ganeshie8 · Answer

(−4−1/3π,2+1/3π) -> (-1/3pi, -4)

ganeshie8 · Answer

looks like it is going to be a rectangle... :)

anonymous · Answer

Is this using my above sub.?

ganeshie8 · Answer

yup !

anonymous · Answer

Man I hate my teacher devising these diabolical problems. :'(

ganeshie8 · Answer

find the other two points also, and plot the region

ganeshie8 · Answer

lol

anonymous · Answer

Thanks for clarifying that it's going to transform to a rectangle. ^^

ganeshie8 · Answer

np :)