Evaluate the triple integral

E=13xdV

where E is the solid: 0y4, 0x16−y2 , 0zy.

Question

Evaluate the triple integral

E=13xdV

where E is the solid: 0y4, 0x16−y2 , 0zy.

dumbcow · Answer

need more info what are we integrating?
are those coordinate points

anonymous · Answer

y between 0 and 4, x between 0 and (16-y^2)^(1/2), lastly z between 0 and y

dumbcow · Answer

$$\int\limits_{0}^{4} \int\limits_{0}^{16-^{y2}} \int\limits_{0}^{y} 13x   dzdxdy$$
is this correct

dumbcow · Answer

anyway if it is
i get an answer of 4437.33

anonymous · Answer

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

dumbcow · Answer

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?

dumbcow · Answer

after replacing the square root i get an answer of 416

dumbcow · Answer

integrating over dzdydx i get 52(16-y^2)

anonymous · Answer

Thanks champ that'll do it for me