Change the order of triple integration

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SUMMARY

The discussion centers on changing the order of integration for the triple integral \(\int_{0}^{2}\int_{0}^{y^3}\int_{0}^{y^2}dzdxdy\) to the order dydzdx. The user Mark44 suggests the new integral as \(\int_{0}^{8}\int_{0}^{\sqrt[3]{x}}\int_{0}^{\sqrt{z}}dydzdx\), asserting that both integrals represent the same volume. The key insight is that both integrals yield the same value when evaluated, as they describe the same geometric region in three-dimensional space.

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coco87
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Homework Statement


Rewrite \int_{0}^{2}\int_{0}^{y^3}\int_{0}^{y^2}dzdxdy as an integral with order dydzdx.

Homework Equations


N/A


The Attempt at a Solution


Honestly, I got as far as sketching it:
dzdxdy-graph.jpg


and after sketching it, I'm lost...

I can't figure out how to set z or y, but I'm fairly sure that x is [0,8].

could anyone possibly offer some help?

Thanks
 
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Is this all you're given in the problem? Just change the order of integration? I'm concerned that you might be omitting some information in the problem.

BTW, I've always found it to be easier to include the variable in one of the limits of integration, when I'm dealing with iterated integrals.
\int_{y = 0}^{2}\int_{x = 0}^{y^3}\int_{z = 0}^{y^2}dz~dx~dy
 
Mark44:

Thank you for your tip :)

I'm actually not leaving anything out at all. Many of these questions are vague (and rather annoying). I think I might have it, but am not sure:

\int_{0}^{8}\int_{0}^{\sqrt[3]{x}}\int_{0}^{\sqrt{z}}dydzdx. However, this seems way to easy to be true...
 
Well, you can check. both integrals represent the volume of the region, so both integrals should produce the same value.
 
Mark44:

Hmm, how would I check without a function to integrate?
 

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