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I am trying understand the fundamentals of evaluating multiple integrals over general regions.

Suppose we want to evaluate $$ \iiint_E 1 \cdot\mathrm{d}w\mathrm{d}v\mathrm{d}u, $$

where $E$ is the region bounded by the inequality $u^2 + v^2 + w^2 \le y $. Assuming $u,v,w \in [0,1]$, would it be correct to find the integral as follows? $$ \iiint_E 1 \mathrm{d}w\mathrm{d}v\mathrm{d}u = \int_{0}^{y}\int_{0}^{(y-u^2)^{1/2}}\int_{0}^{(y-(u^2+v^2))^{1/2}} 1 \mathrm{d}w\mathrm{d}v\mathrm{d}u$$

How would the integration limits change if instead we assume $u,v,w \in [-1,1]$? Answers that don't require change of variables would be appreciated.

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This is one of those cases where having a clear picture in mind really helps. It's important to see that $E$ is a sphere, with centre $(0,0,0)$ and radius $\sqrt y$, and you are asked to find its volume.

Then, suddenly, you impose this additional restriction - that $u,v,w\in[0,1]$. So now you're finding the volume of the intersection of a sphere with a cube. If the sphere is small enough, this will be $1/8$ of the original sphere. If it's big enough, this will be the whole cube. If it's somewhere in between, everything is more complicated - you'll need to deal with each integration in two separate sections.

Lastly, you impose a different additional restriction - that $u,v,w\in[-1,1]$. You're finding the volume of the intersection of the same sphere with a different cube. Again, there are three cases - it's either the whole sphere, or the whole cube, or something more complicated for which you'll need to deal with each integration in two separate sections.

So it seems to me that you've actually asked six different questions here; some are easy and some are hard. But the complexity all comes down to the additional restrictions you've imposed on $u,v,w$. Without these extra conditions, the integral is just $$ \int\int\int_E 1 \thinspace dw\thinspace dv\thinspace du = \int_{-\sqrt y}^{\sqrt y} \int_{-\sqrt {y-u^2}}^{\sqrt {y-u^2}} \int_{-\sqrt {y-u^2-v^2}}^{\sqrt {y-u^2-v^2}} 1\thinspace dw\thinspace dv\thinspace du $$

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+1 and thanks, I wasn't sure if the lack of responses was because this question is too basic for math.stackexchange, but instead it looks like I've asked too many questions. I looked at your profile and was just wondering: XSLT? Aren't you bored? – T. Webster Sep 24 '12 at 19:16

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