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anonymous

  • 5 years ago

I am working on a double integral of 3xy where R is the region in the first quadrant bounded by x^2+y^2=9, y=4x, y=0. I drew a picture and I have the inner integral from 4x to sqrt(9-x^2) and the outer integral from 0 to 3-- of 3xy dy dx

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    I need to get a pen and paper for this, good practice for me too :-) next week exam about this. 2min and I am back

  3. anonymous
    • 5 years ago
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    ok- I know I have a problem since my answer is negative (and I'm in the first quadrant)

  4. anonymous
    • 5 years ago
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    ok I have a picture now

  5. anonymous
    • 5 years ago
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    y goes from 0 to 3

  6. anonymous
    • 5 years ago
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    x goes from 1/4y to sqrt(9-y^2)

  7. anonymous
    • 5 years ago
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    what do you think?

  8. anonymous
    • 5 years ago
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    wouldn't it also be the same to have x goes from 0 to 3 and y goes from 4x to sqrt(9-x^2)??

  9. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{3}\int\limits_{y/4}^{\sqrt{9-y ^{2}}} 3xy dxdy\]

  10. anonymous
    • 5 years ago
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    if you make the graph you can see that if you set x to go from 0-3 than y will go from 0 till 4x or 0 till sqrt(9-x^2)

  11. anonymous
    • 5 years ago
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    so I guess that way you have to split the integral to two parts

  12. anonymous
    • 5 years ago
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    ok- I'm going to try to solve.

  13. anonymous
    • 5 years ago
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    I am doing it now too, do you have the answer for it?

  14. anonymous
    • 5 years ago
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    no I don't :(

  15. anonymous
    • 5 years ago
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    I have 33.75-(3/32)*ln(3). Does this look right??

  16. anonymous
    • 5 years ago
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    I got different :-)

  17. anonymous
    • 5 years ago
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    ok lets do it step by step

  18. anonymous
    • 5 years ago
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    first you do the x integral to get 3x^2y/2

  19. anonymous
    • 5 years ago
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    I have that!

  20. anonymous
    • 5 years ago
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    now you put back the integral values

  21. anonymous
    • 5 years ago
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    1/2{3(9-y^2)y-3/16y^3)}

  22. anonymous
    • 5 years ago
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    agreed?

  23. anonymous
    • 5 years ago
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    I have 1/2{3(9-y^2)y-3/16y^2)}

  24. anonymous
    • 5 years ago
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    it is y^3 you have 3x^2y and you put 1/4 y for x

  25. anonymous
    • 5 years ago
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    now expanding the brackets gives: 1/2(27-51/16y^3)

  26. anonymous
    • 5 years ago
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    agreed?

  27. anonymous
    • 5 years ago
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    yep!

  28. anonymous
    • 5 years ago
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    now we integrate for y that is: 1/2(27y-51/4*16y^4

  29. anonymous
    • 5 years ago
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    we only need to put back 3 as for 0 everything is 0

  30. anonymous
    • 5 years ago
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    1/2(27*3-51*3^4/(4*16)

  31. anonymous
    • 5 years ago
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    and I dont know what that is :D

  32. anonymous
    • 5 years ago
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    8.2265625

  33. anonymous
    • 5 years ago
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    just to confirm: we have \[(1/2)* ((27*3)-((51*(3^4))/(4*16))\]

  34. anonymous
    • 5 years ago
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    yep!

  35. anonymous
    • 5 years ago
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    charming number

  36. anonymous
    • 5 years ago
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    Thanks for your help! My webGrader still doesn't like it but I'll ask my professor tomorrow!

  37. anonymous
    • 5 years ago
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    I am not hundred percent sure but I guess it is ok, if you have the answer you could write a reply tomorrow

  38. anonymous
    • 5 years ago
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    thanks again!

  39. anonymous
    • 5 years ago
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    In order to do this in x and y, you would have to break the integral up because the y goes from 0 to 4x over interval 0 to 3/4; and y goes from 0 to3 over interval 3/4 to 3. This interval is good for polar coordinates.

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