anyone know double integrals?

Question

eyust707 · Answer

yes

eyust707 · Answer

calc 3?

anonymous · Answer

yes

eyust707 · Answer

im in it right now.. whats the integral?

anonymous · Answer

e^(y^3)   0<x<3 and        sqr root(x/3) <y<1

anonymous · Answer

i got 0 for lower bound, by subbing in 0 and 3, however i dont think its correct, got an answer of e+C

eyust707 · Answer

not exactly sure what the initial question was..

eyust707 · Answer

use the equation thing down there and write out the question just as it is in the book

anonymous · Answer

$$e^((y)^3)   where  \omega  is defined by 0<x<3 and \sqrt{(x/3)}<y<1$$

anonymous · Answer

it wouldnt type work right   it is e raised to the y cubed,   not e raised to the y quantity cubed

eyust707 · Answer

$$e^{y^3}$$

anonymous · Answer

yes

eyust707 · Answer

dude honestly weve never done any problems like this.. what are the instructions for the problem?

anonymous · Answer

evaluate the integral

anonymous · Answer

yall havent had double integrals ?or where you have to change the bounds?

eyust707 · Answer

hmm ok are you sure that it doesnt look like this

$$\int\limits_{0}^{3}\int\limits_{\sqrt{x/3}}^{1} e ^{y^3} dy dx$$

anonymous · Answer

i stated it how it was. but i too converted it to what you have there. that is the question. my teacher is not from the u.s

eyust707 · Answer

damn haha the way you wrote it is super wierd.. i would interpret what you wrote as the above. and that i can solve

eyust707 · Answer

so first we integrate the inside with respect to y

anonymous · Answer

to get $$e^(x^3/27)$$

eyust707 · Answer

what did you get as your antiderivative for e^(y^3)

anonymous · Answer

1/   (3y^2)

anonymous · Answer

?

eyust707 · Answer

hmm i dont think that is it

anonymous · Answer

times e^y^3

eyust707 · Answer

hmm i honestly forget how to find that antiderivative... http://www.wolframalpha.com/input/?i=derive+1%2F+%283y^2%29+*+e^%28y^3%29 it doesnt seem to be correct tho

anonymous · Answer

u=y^3
du=3y^2 dx
$$ \int\limits_{0}^{3}  1\div (3y^2) \int\limits_{\sqrt{x \div3}}^{1}   e^y^3   (3y^2 dy) dx$$

anonymous · Answer

what did you get for the antiderivative?

eyust707 · Answer

i wasnt able to find one. wolfram alpha's answer had a gamma function which i have not ever studied. I would make sure youve written the problem down correctly because i dont believe we can integrate that useing only the tools we have from calc 1 2 and 3.

eyust707 · Answer

my only other suggestion would be to draw a picture and try and switch the x and y's

eyust707 · Answer

or possibly break it up into two integrals???

anonymous · Answer

|dw:1322289444987:dw|haha, i hope everyone is as stumped as i. i thought that answer was e+c.