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anonymous

  • 5 years ago

Change the Cartesian Integral into an equivalent polar integral then evaluate: (int(int(dydx)) from -1 to 1 and 0 to sqrt(1-x^2)

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  1. anonymous
    • 5 years ago
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    nvm got it...

  2. anonymous
    • 5 years ago
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    How'd you do it? If I may

  3. anonymous
    • 5 years ago
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    YOu have to graph the boundaries then find the new boundaries: 0 to pi/2 and 0 to 1. Then you use the equation int(int(rdrdtheta) which = pi/2

  4. anonymous
    • 5 years ago
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    limits for r are 0 to 1, theta 0 to pi

  5. anonymous
    • 5 years ago
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    I got 0 to pi /2 because its y...

  6. anonymous
    • 5 years ago
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    \[\int\limits_{}{}\int\limits_{}{}dxdy=\int\limits_{}{}\int\limits_{}{}|\frac{\partial J(x,y)}{\partial (r, \theta)}|dr d \theta=\int\limits_{}{}\int\limits_{}{}r dr d \theta\]

  7. anonymous
    • 5 years ago
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    Integrate out r from 0 to 1, then theta from 0 to pi.

  8. anonymous
    • 5 years ago
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    You get \[\frac{\pi}{2}\]

  9. anonymous
    • 5 years ago
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    ohh i see that thanks

  10. anonymous
    • 5 years ago
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    welcome. i stopped on your question and went for a shower - thought someone was taking it.

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