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psk981

  • 3 years ago

change to polar and then integrate

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  1. psk981
    • 3 years ago
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    \[\int\limits_{0}^{2}\int\limits_{0}^{\sqrt{4-y ^{2}}} x ^{2}+y^{2} dxdy\]

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

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

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

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

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

  7. psk981
    • 3 years ago
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    radius should be 2

  8. TuringTest
    • 3 years ago
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    yes

  9. TuringTest
    • 3 years ago
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    and what are the bounds on theta?

  10. psk981
    • 3 years ago
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    0 to 2pi

  11. TuringTest
    • 3 years ago
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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...?

  12. psk981
    • 3 years ago
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    0 to pi

  13. TuringTest
    • 3 years ago
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    yes

  14. psk981
    • 3 years ago
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    so would i ge 4 pi

  15. TuringTest
    • 3 years ago
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    yes

  16. psk981
    • 3 years ago
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    ur a life saver

  17. TuringTest
    • 3 years ago
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    no problem, happy to help

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