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anonymous

  • 5 years ago

Evaluate the triple integral E=13xdV where E is the solid: 0y4, 0x16−y2 , 0zy.

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  1. dumbcow
    • 5 years ago
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    need more info what are we integrating? are those coordinate points

  2. anonymous
    • 5 years ago
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    y between 0 and 4, x between 0 and (16-y^2)^(1/2), lastly z between 0 and y

  3. dumbcow
    • 5 years ago
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    \[\int\limits_{0}^{4} \int\limits_{0}^{16-^{y2}} \int\limits_{0}^{y} 13x dzdxdy\] is this correct

  4. dumbcow
    • 5 years ago
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    anyway if it is i get an answer of 4437.33

  5. anonymous
    • 5 years ago
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    The upper limit of x: 16-y^2 needs a sqr root over it and I think standard order is dzdydx

  6. dumbcow
    • 5 years ago
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    oh oops..um the reason i switched dy and dx is otherwise you wont get a number at the end in other words you will be integrating over x with y variables leaving the answer as an expression of y. is that what they want or do you need this to evaluate to a number?

  7. dumbcow
    • 5 years ago
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    after replacing the square root i get an answer of 416

  8. dumbcow
    • 5 years ago
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    integrating over dzdydx i get 52(16-y^2)

  9. anonymous
    • 5 years ago
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    Thanks champ that'll do it for me

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