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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?

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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,

$$V = \frac12 \cdot \frac13 \cdot 1 = \frac16$$

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The region is not a clepsydra, it's a (non-regular) tetrahedron.

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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?

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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.

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