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

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dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0need more info what are we integrating? are those coordinate points

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y between 0 and 4, x between 0 and (16y^2)^(1/2), lastly z between 0 and y

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{4} \int\limits_{0}^{16^{y2}} \int\limits_{0}^{y} 13x dzdxdy\] is this correct

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0anyway if it is i get an answer of 4437.33

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The upper limit of x: 16y^2 needs a sqr root over it and I think standard order is dzdydx

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0oh 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?

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0after replacing the square root i get an answer of 416

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0integrating over dzdydx i get 52(16y^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks champ that'll do it for me
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