Evaluate double integral of x^2 + y^2 dx dy over the square with corners (-1,0),(0,1),(1,0),and (0,-1). ans. 2/3?
Evaluate in two ways: directly, and using x = (u + v)/2, y = (u-v)/2.
I have done this and always get 8/3 for direct integration. I used the [-1,1]-dx and [-1,1]-dy. For using the substitution, I got 4/3. I used the [-1,1]--du and [-1,1]--dv.
Is there something wrong with my boundaries or anything?
- rotchmLv 74 months ago
∫∫ x² + y² dy dx , x=-1 to 0 & y = -x-1 to x+1. PLUS
∫∫ x² + y² dy dx, x=0 to 1 & y = x-1 to -x+1
Evaluating indeed gives 2/3. You do see why the ranges...?
Via the trans. x = (u + v)/2, y = (u-v)/2. don't forget that dxdy becomes
J(u,v)*dudv. Evaluating J gives J = 1/2. This is what you probably forgot. So your integral is now
∫∫ ... * (1/2) * du dv, u=-1 to 1 & v = -1 to 1. Evaluating gives again 2/3.
Do you "see" what the transformations does graphically?
- Anonymous4 months ago
Sorry, but your professor wants to see your answer, not ours. Cheating = Epic fail.