Change the order of triple integration

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?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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