I need to find the volume of an object set with the following function:
$$x + y + z = 1$$
And all three axis.
So I converted the function into $z = -x - y + 1$, and it gave me kind of a clepsydra, crossing x,y plane in x = y = 1.
So the region seems to be the circle $x^{2}+y^{2}=1$.
However, changing x to r, and y to $\phi$ gives me wrong result - $\pi$, and it should be 1/6.
Is the region I've came up with correct?
First of all, the region of interest here is a right tetrahedron, with a right triangle base of side length $1$ and height $1$, so the volume is
$$\frac13 \cdot \frac 12 (1) (1) = \frac16$$
The volume integral may be done by integrating over, say, $x$ first, then $y$, then $z$:
$$V = \int_0^1 dz \, \int_0^{1-z} dy \, \int_0^{1-y-z} dx $$
Doing the innermost integral:
$$V = \int_0^1 dz \, \int_0^{1-z} dy \, (1-y-z) = \int_0^1 dz \, \left [(1-z)^2 - \frac12 (1-z)^2 \right ] = \frac12 \int_0^1 dz \, (1-z)^2$$
Doing the final integral,
$$V = \frac12 \cdot \frac13 \cdot 1 = \frac16$$
Your region is incorrect (and I'm not sure how you got it).
It seems that you wish to find the volume of the region bounded by the plane $x+y+z=1$ and the planes $x=0,y=0,z=0.$ The upper boundary surface is $z=1-x-y,$ and the lower boundary surface is $z=0$. "Collapsing" everything to a region in the $xy$-plane, we see that the upper boundary curve is $y=1-x$ and the lower boundary curve is $y=0$. "Collapsing" everything to an interval on the $x$-axis, we see that the upper boundary value is $x=1$ and the lower boundary value is $x=0.$ What you're looking at, then, is $$\int_0^1\int_0^{1-x}\int_0^{1-x-y}\,dz\,dy\,dx.$$ Can you evaluate that?
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} {\cal V} & \equiv\bbox[5px,#ffd]{\iiint_{\pars{0,1}^{\,3}}\, \bracks{x + y + z < 1}\dd x\,\dd y\,\dd z} \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \bracks{x < 1 - y - z}\dd x\,\dd y\,\dd z \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1} \bracks{0 < 1 - y - z < 1}\int_{0}^{1 - y - z} \dd x\dd y\,\dd z \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1} \bracks{-z < y < 1 - z}\pars{1 - y - z}\dd y \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1} \bracks{y < 1 - z}\pars{1 - y - z}\dd y\,\dd z \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1 - z}\pars{1 - y - z}\dd y\dd z \\[5mm] = &\ {1 \over 2}\int_{0}^{1}\pars{1 - z}^{2}\dd z = \bbx{1 \over 6} \\ & \end{align}
The object formed by the intersection of a plane with the axes is simply a pyramid, not a circle. The volume is indeed $1/6$, since like any other cone, the volume is equal to: $$V = \frac{1}{3} B\cdot H$$ Where $B$ is the base area (half of a square $=1/2$), and $H$ is the height which is one.
