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anonymous

  • 5 years ago

Anyone willing to verify my answer to a double integral? Thanks

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  1. anonymous
    • 5 years ago
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    what is the question?

  2. anonymous
    • 5 years ago
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    \[\int\limits_{-3}^{2}\int\limits_{0}^{1}y^{2} x\]dy dx

  3. anonymous
    • 5 years ago
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    And what was your answer?

  4. anonymous
    • 5 years ago
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    Should I write it? Might it influence others' solution?

  5. anonymous
    • 5 years ago
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    -5/6

  6. anonymous
    • 5 years ago
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    You want us to verify your answer you should give your answer ;p

  7. anonymous
    • 5 years ago
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    Lol, I meant see what answer you get and then I'll check if it's the same.

  8. anonymous
    • 5 years ago
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    Oh. Yes I also got -5/6

  9. anonymous
    • 5 years ago
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    Thanks!

  10. anonymous
    • 5 years ago
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    No it is not -5/6

  11. anonymous
    • 5 years ago
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    opps..ya it is, sorry

  12. anonymous
    • 5 years ago
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    Yeah I got something different at first, but I thought it was dxdy, not dydx

  13. anonymous
    • 5 years ago
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    :)

  14. anonymous
    • 5 years ago
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    lol! Thanks! Just making sure I know what I'm doing... I've got another similar question which I haven't answered yet as I need help answering it. Want to help out?

  15. anonymous
    • 5 years ago
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    If you ask, we may answer. If you don't ask we certainly will not help ;p

  16. anonymous
    • 5 years ago
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    Haha, Ok here goes:

  17. anonymous
    • 5 years ago
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    \[\int\limits_{-1}^{1}\int\limits_{0}^{\sqrt{1-x ^{2}}}(x ^{2}+y ^{2})^{3/2}\]dydx

  18. anonymous
    • 5 years ago
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    Convert to polar.

  19. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{\pi}\int\limits_{0}^{1}r^3 r\ drd \Theta\]

  20. anonymous
    • 5 years ago
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    Does that make sense? I'm getting Pi/5 for the answer.

  21. anonymous
    • 5 years ago
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    Thanks dude, sorry about the late reply, my internet went and has finally JUST come back!

  22. anonymous
    • 5 years ago
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    Can you help me a bit step by step... Thanks. How did you convert to polar?

  23. anonymous
    • 5 years ago
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    is it the x=rcostheta and y=rsintheta?

  24. anonymous
    • 5 years ago
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    \[r^2 = x^2 + y^2 \rightarrow r = (x^2 + y^2)^{1/2} \rightarrow (x^2 + y^2)^{3/2} = r^3\] \[dydx = r\ drd \Theta\]

  25. anonymous
    • 5 years ago
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    Then you have to convert your limits. Since x goes from -1 to 1 and y goes from 0 to sqrt(1+x^2) you can see that you're integrating over the top half of a circle with radius 1.

  26. anonymous
    • 5 years ago
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    So your radius goes from 0 to 1, and your angle goes from 0 to Pi.

  27. anonymous
    • 5 years ago
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    Can you explain how you realised that it is the top half of a circle?

  28. anonymous
    • 5 years ago
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    Because the y values go from 0 to sqrt(1+x^2) Those are all positive y values inside the circle x^2 + y^2 = 1 If it had said -sqrt(1+x^2) then it would have been the bottom half of the circle.

  29. anonymous
    • 5 years ago
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    If you graph the line y = sqrt(1+x^2) you'll see it forms the top half of a circle. Integrating from y = 0 to y=sqrt(1+x^2) will be integrating over that same half-circle.

  30. anonymous
    • 5 years ago
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    I understand the 0 to 1 limit of y. How did the 0 to Pi limit come about? Thanks

  31. anonymous
    • 5 years ago
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    You have to look at both pieces together. If y goes from 0 to the top of the half circle and x goes from -1 to 1 then the area you're covering is the top half of the circle. That means your r will go from 0 to 1 and your Theta will go from 0 to pi. cos(0) = 1 = x, cos(pi) = -1 = x.

  32. anonymous
    • 5 years ago
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    Oh! Sorry about that! Theta goes from 0 to Pi, I was thinking about x... Understand that part now! Thanks a lot for your help!

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