How to write six possible iterated integral?

Question

I want to write out all the possible six iterated integrals for finding the volume of region in the first octant enclosed by the $ x^2+z^2=4 $ and the plane $y=3$.

How can I plot this region of integration in Mathematica and how to write the six permutations to find volume of the given region.

Thanks in advance

Answer 1

Not so fast: mechanodroid's answer doesn't actually answer the question presented, to set up each of the possible iterated integrals for the six possible orders of $x,y,z$. It only answers the question about providing a picture.

As a convention, I will label each in order from inside to outside. In inequality form, our region is the intersection of the regions defined by $0\le x$, $0\le y\le 3$, $0\le z$, and $x^2+z^2\le 4$.
$x,y,z$: With $x$ innermost, the bounds there are $0\le x\le \sqrt{4-z^2}$. The bounds for $y$ are unaffected by $z$, and $z$ can range from $0$ to $2$. $$\int_0^2 \int_0^3 \int_0^{\sqrt{4-z^2}} 1\,dx\,dy\,dz$$ $x,z,y$: With $x$ innermost, the bounds there are $0\le x\le \sqrt{4-z^2}$. The bounds for $z$ are unaffected by $y$, and $y$ can range from $0$ to $3$. $$\int_0^3 \int_0^2 \int_0^{\sqrt{4-z^2}} 1\,dx\,dz\,dy$$ $y,x,z$: With $y$ innermost, its bounds are just $0\le y\le 3$ as always. Then $0\le x\le \sqrt{4-z^2}$ and $0\le z\le 2$. $$\int_0^2 \int_0^{\sqrt{4-z^2}} \int_0^3 1\,dy\,dx\,dz$$ $y,z,x$: With $y$ innermost, its bounds are just $0\le y\le 3$ as always. Then $0\le z\le \sqrt{4-x^2}$ and $0\le x\le 2$. $$\int_0^2 \int_0^{\sqrt{4-x^2}} \int_0^3 1\,dy\,dz\,dx$$ $z,x,y$: With $z$ innermost, its bounds are $0\le z\le \sqrt{4-x^2}$. Then $0\le x\le 2$ and $0\le y\le 3$. $$\int_0^3 \int_0^2 \int_0^{\sqrt{4-x^2}} 1\,dz\,dx\,dy$$ $z,y,x$: With $z$ innermost, its bounds are $0\le z\le \sqrt{4-x^2}$. Then $0\le y\le 3$ and $0\le x\le 2$. $$\int_0^2 \int_0^3 \int_0^{\sqrt{4-x^2}} 1\,dz\,dy\,dx$$

This is a pretty simple one in how things interact; the only thing that changes any of the limits is which one of $x$ and $z$ is farther in. As such, it's rather redundant, and I was using a lot of copy/paste there.
Evaluating the integrals? I'll leave that to you, if you feel like it.

Answer 2

The volume is given by three iterated integrals: $$V = \int_{y=0}^3 \int_{z=0}^2 \int_{x=0}^{\sqrt{4-z^2}}dx\,dz\,dy = 3\int_{z=0}^2 \sqrt{4-z^2}\,dz = 3\pi$$

Indeed, we can plot this in Mathematica with Plot3D[Sqrt[4 - z^2], {y, 0, 3}, {z, 0, 2}]: