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psk981 Group Title

change to polar and then integrate

  • 2 years ago
  • 2 years ago

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  1. psk981 Group Title
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    \[\int\limits_{0}^{2}\int\limits_{0}^{\sqrt{4-y ^{2}}} x ^{2}+y^{2} dxdy\]

    • 2 years ago
  2. psk981 Group Title
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    i know it becomes \[r^{2} r dr dTheta\]

    • 2 years ago
  3. TuringTest Group Title
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    yes, which simplifies to\[r^3drd\theta\]where are you stuck?

    • 2 years ago
  4. psk981 Group Title
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    the sqrt term becomes x^2+y^2= 4 right

    • 2 years ago
  5. TuringTest Group Title
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    x^2+y^2=r^2 what is the radius of the circle in question?

    • 2 years ago
  6. psk981 Group Title
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    so i should get r^4/4 evaluated from 0 to 2 then i should get 4 dtheta

    • 2 years ago
  7. psk981 Group Title
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    radius should be 2

    • 2 years ago
  8. TuringTest Group Title
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    yes

    • 2 years ago
  9. TuringTest Group Title
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    and what are the bounds on theta?

    • 2 years ago
  10. psk981 Group Title
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    0 to 2pi

    • 2 years ago
  11. TuringTest Group Title
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    remember that sqrt(4-y^2) is only the *top* of the circle, the bottom would require a negative sign, so the bounds on theta are only...?

    • 2 years ago
  12. psk981 Group Title
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    0 to pi

    • 2 years ago
  13. TuringTest Group Title
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    yes

    • 2 years ago
  14. psk981 Group Title
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    so would i ge 4 pi

    • 2 years ago
  15. TuringTest Group Title
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    yes

    • 2 years ago
  16. psk981 Group Title
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    ur a life saver

    • 2 years ago
  17. TuringTest Group Title
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    no problem, happy to help

    • 2 years ago
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