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Can some fast calculate the double integral using polar coordinates and cartesian coordinates? D={(x,z,y)|-1≤x≤0,-√(1-x^2 )≤y≤√(1-x^2 )} f(x,y,z)=x^2

  • one year ago
  • one year ago

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    i got the answer to π/8 and 1, but that can't be true?

    • one year ago
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    |dw:1352299213766:dw|

    • one year ago
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    if expression is like above then just evaluate it with usual process ... doesn't seem that difficult.

    • one year ago
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    not quite sure i understand. i mean on polar coordinates the set would be: Dp={(x,z,y)┤|-1≤r≤0 ,π/2≤θ≤3π/2} right? and then the following (see file)

    • one year ago
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    how did you change the limits?

    • one year ago
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    |dw:1352299738473:dw|

    • one year ago
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    don't know if you can see it but there is a square root around 1-x^2

    • one year ago
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    well the reason why i set my limit for theta to that is becuase the set should look like something like this: |dw:1352300003766:dw|

    • one year ago
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    Unless i have misunderstod something?

    • one year ago
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    it's the left half of a circle, I agree

    • one year ago
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    sorry .. i completely missed sqrt() ..

    • one year ago
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    well kinda my falt i should have written sqrt() .. insted of √(1-x^2

    • one year ago
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    what bounds did you use?

    • one year ago
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    |dw:1352300314417:dw|

    • one year ago
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    yeah, that's what I got

    • one year ago
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    oh no, r^3 though...

    • one year ago
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    is it not r^2?

    • one year ago
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    you have r^2 from the x^2 and r from dA

    • one year ago
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    yeah .. it's r^3

    • one year ago
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    can evne find that in my book that extra r, but seems to be right yea

    • one year ago
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    can't* even*

    • one year ago
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    But is this polar set wrong then? Dp={(x,z,y)┤|-1≤r≤0 ,π/2≤θ≤3π/2}

    • one year ago
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    \[ dx \;dy = r \; dr \; d\theta \]

    • one year ago
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    no wait can see it my self.. it is just 0<=r<=1

    • one year ago
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    right, the angle covers the negative part

    • one year ago
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    I still get pi/8 by the way...

    • one year ago
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    Just think i need to take a few more looks becuase i keep geting a wrong result.

    • one year ago
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    So we can all agree to this: (sorry for have pulling both you through all thís)

    • one year ago
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    That's what I got

    • one year ago
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    Alright thanks alot to both of you.

    • one year ago
  31. TuringTest Group Title
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    Welcome.

    • one year ago
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