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    can any of you help me with this

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    (Original post by xyazk)
    can any of you help me with this

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    This is now in its own thread since it isn't a maths challenge question
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    well thankyou for this. I'm new to this site so I don't really know how to ask for help
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    (Original post by xyazk)
    can any of you help me with this

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    By the time you reach inputting sqrt limit you would have had x^4 term wouldn't you which would mean taking the square of the expression inside the square root and integrating with respect to z...
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    (Original post by xyazk)
    can any of you help me with this
    What's confusing about it? This can be done in the order it's already written without the need to change orders of integration.
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    (Original post by RDKGames)
    What's confusing about it? This can be done in the order it's already written without the need to change orders of integration.
    i ask this because i didnt get the answer. sorry if this question seem too easy for you
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    (Original post by xyazk)
    i ask this because i didnt get the answer. sorry if this question seem too easy for you
    What did you get?
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    (Original post by Merdan)
    By the time you reach inputting sqrt limit you would have had x^4 term wouldn't you which would mean taking the square of the expression inside the square root and integrating with respect to z...
    Name:  image-4c45c243-a454-4b9e-9366-70f2243276e6963109092-compressed.jpg.jpeg
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    (Original post by xyazk)
    i think my last ingration is wrong. how do you solve it?
    Second to last line, note that (\sqrt{16-z^2})^4=(16-z^2)^2=z^4-32z^2+256
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    thank you for helping
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