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?

2 Answers

  • rotchm
    Lv 7
    4 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? 

  • Anonymous
    4 months ago

    Sorry, but your professor wants to see your answer, not ours. Cheating = Epic fail. 

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