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

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

Mathematics
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need more info what are we integrating? are those coordinate points
y between 0 and 4, x between 0 and (16-y^2)^(1/2), lastly z between 0 and y
\[\int\limits_{0}^{4} \int\limits_{0}^{16-^{y2}} \int\limits_{0}^{y} 13x dzdxdy\] is this correct

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anyway if it is i get an answer of 4437.33
The upper limit of x: 16-y^2 needs a sqr root over it and I think standard order is dzdydx
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?
after replacing the square root i get an answer of 416
integrating over dzdydx i get 52(16-y^2)
Thanks champ that'll do it for me

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