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Some questions about double integral and vector calculus

  1. Feb 14, 2009 #1
    I've written all the questions in the PDF file ......

    Please help me !

    Attached Files:

  2. jcsd
  3. Feb 14, 2009 #2


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    If you can't be bothered to
    1) type the problem itself here
    2) Do anything at all at all on the problem

    I don't see why I should bother.
  4. Feb 14, 2009 #3

    I'm sorry for that.

    I scanned the problem I had just because they are "hard" to type ......I concede that my computer ability is not good.... So I made manuscript instead.

    And I did not discribed the problem I have .....is my false .

    I'm now typing the problem I have.....

    For Q1:

    I tried change of variable and fail.....

    I could not figure out the craftsmanship solving the problem...

    Give me a hint please....

    For Q2:

    I was considering expand the formula by assuming

    A=(A1)i+(A2)j+(A3)k {also B}

    substitude the equations....

    No doubt ........It is quite terrible !

    So I asked for help!

    Sorry again for ignoring the "detail problem" :shy:
  5. Feb 14, 2009 #4


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    Hi abcdefg10645! That's better! :smile:

    (have a del: ∇ and a dot: · and an integral: ∫ :wink:)

    For 1, make the obvious substitution …

    u = 3x + y, v = x - 2y, dxdy = … ?

    For 2, just write each of the coordinates out in full …

    try a. first …

    what do you get? :smile:
  6. Feb 14, 2009 #5

    Thaks for your help....

    But for question 1......I still have a little problem....

    Could you give me a hint for changing the upper and lower limit?

    And even tell me the "principle" resoving the limits .....

  7. Feb 15, 2009 #6


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    I don't understand :confused:

    just convert x and y to u and v …

    what do you get? :smile:
  8. Feb 15, 2009 #7


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    The region in problem 1 has boundaries 2x+ y= 0, 3x+ y= 0, x- 2y= 1 and x- 2y= 2.

    tiny-tim suggested the substitution u = 3x + y, v = x - 2y.

    Solve for x and y in terms of u and v: Multiplying the first equation by 2 and adding to the second equation gives 2u+ v= 6x+ 2y+ x- 2y= 7x so x= (2u+ v)/7. Multiplying the second equation by 3 and subtracting the first equation gives 3v- u= 3x-6y- 3x- y= -7y so y= (u- 3v)/7.

    Convert the boundary to uv-coordinates by substituting those for x and y:
    2x+ y= 2(2u+v)/7+ (u-3v)/7= (4u+ 2v+ u- 3v)/7= (5u- v)/7= 0 so 5u- v= 0

    It should be obvious that the last three equations become u= 0, v= 1, and v= 2, but as a demonstration, x- 2y= (2u+ v)/7- 2(u- 3v)/7= (2u+ v- 2u+ 6v)/7= 7v/7= v= 1.

    So you want to integrate on the region bounded by the lines u= 0, u= v/5, v= 1, and v= 2. I recommend taking the "outer integral" to be with respect to v, from 1 to 2, and the "inner integral" with respect to u, from 0 to v/5. Be careful about dudv. It is, of course, the Jacobian.
  9. Feb 15, 2009 #8


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    try sketching your boundaries
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