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ksmith197

  • 2 years ago

Easy calculus question: if you had something like say a double integral integral[integral (x^2+y^2)/(xy) dy]dx why can't you set x^2 + y^2 = 1?

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  1. ksmith197
    • 2 years ago
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    using the cos^2x + sin^2y identity

  2. ksmith197
    • 2 years ago
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    \[\int\limits_{1}^{4}\int\limits_{1}^{2}(x^2+y^2)/xy)dydx\]

  3. ksmith197
    • 2 years ago
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    why can't the x^2 + y^2 be set = 1

  4. amistre64
    • 2 years ago
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    i think you are confusing a few things

  5. amistre64
    • 2 years ago
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    \[\int\int \frac{x^2+y^2}{xy} dy.dx\] \[\int\int \frac{x^2}{xy}+\frac{y^2}{xy} dy.dx\] \[\int\int \frac{x}{y}+\frac{y}{x} dy.dx\] \[\int\int xln(y)+\frac{y^2}{2x} \ dx\]etc... if that equates to your idea, then try it out

  6. ksmith197
    • 2 years ago
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    ah sorry didn't mean to make you write all of that out that's actually what the original equation was, i just moved evertying around to try something out

  7. ksmith197
    • 2 years ago
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    was just kinda wondering when you can or can't use the x^2+y^2 = cos[x]^2 + sin[y]^2 = 1 identities

  8. amistre64
    • 2 years ago
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    oh, i wrote it for the practice:)

  9. ksmith197
    • 2 years ago
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    just figured that since we are using double integrals, they are equations with x and y, so i was thinking you could use identities to maybe simplify things

  10. ksmith197
    • 2 years ago
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    :) heh

  11. amistre64
    • 2 years ago
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    (..)^2 is not generally equativalent to the function trig^2(...)

  12. No-data
    • 2 years ago
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    You can use that identity to simplify integrals but you have to be carefull.

  13. ksmith197
    • 2 years ago
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    yeah, the integral didn't come out right in mathematica. Or, could you use it but you have to change the limits of integration?

  14. No-data
    • 2 years ago
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    \[x^2+y^2\neq \cos^2x+\sin^2x\]

  15. ksmith197
    • 2 years ago
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    ah right but x^2/r^2 + y^2/r^2 = cos[x]^2 + sin[y]^2 = r^2 right?

  16. TuringTest
    • 2 years ago
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    no

  17. TuringTest
    • 2 years ago
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    \[x=r\cos\theta\]\[y=r\sin\theta\]\[x^2+y^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2\]I'm not sure what you are trying to get out of that

  18. ksmith197
    • 2 years ago
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    er yeah, sorry messed that up. it would still work out though right?? because x^2/1 + y^2/1 = (1)^2 cos^2 + sin^2 = 1^2 blehhh i think i'm mixing this all up

  19. No-data
    • 2 years ago
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    Don't worry.

  20. TuringTest
    • 2 years ago
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    you are thinking something like\[x^2+y^2=r^2\implies{x^2+y^2\over r^2}=1\]perhaps? That would totally screw up your coordinate system though, to put that in the integral

  21. TuringTest
    • 2 years ago
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    thta's some crazy cartesian-polar hybrid

  22. ksmith197
    • 2 years ago
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    hahaha

  23. amistre64
    • 2 years ago
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    but its good to see you thinking outside the mathical box

  24. ksmith197
    • 2 years ago
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    ah i haven't read the chapter on polar stuff yet, so maybe i'll remedy things in the coming week :)

  25. ksmith197
    • 2 years ago
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    thank you all for the help!

  26. TuringTest
    • 2 years ago
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    welcome :)

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