Double Integral of e^(x^4) HELP

sarahlynnuoit
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Homework Statement



Integrate Double Integral of e^(x^4), First Bound- a = 3*(sqroot(y)) and b= 2 and 2nd Bound - C = 0 and D = 8?

Homework Equations





The Attempt at a Solution


We have been working on this all day and we have tried changing bounds but we cannot find an solution that can be integrated and we cannot get rid of the e^(x^4)
Please help! :)
 
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Do you by any chance mean:

\int_{0}^{8}\int_{\sqrt[3]{y}}^{2} e^{x^{4}} \ dx\ dy

If so, then know that the lower bound is cube root of y, not 3 square root of y. That may be why you're not getting something that you can integrate when you reorder the variables. If you integrate first with respect to y, and then x, you will get something you can integrate.
 
So given that same problem how do you reset the limits of integration so you can integrate with respect to y first?
 
Sketch the region of integration 2 \leq x \leq y^{1/3} and 0 \leq y \leq 8. Now draw an arrow through your sketched region pointed in the same direction of the y-axis. The boundary the region that your arrow intersects first will be your new lower limit. The boundary of the region that your arrow intersects second will be your new upper limit.
 
y goes from 0 to 2, obviously. Now, for each y what bounds does x have?

That's basically what Samuelb88 is saying.
 
I see thanks for the info guys.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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