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

1. anonymous

2. anonymous

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

3. anonymous

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

4. anonymous

anyway if it is i get an answer of 4437.33

5. anonymous

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

6. anonymous

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. anonymous

after replacing the square root i get an answer of 416

8. anonymous

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

9. anonymous

Thanks champ that'll do it for me