# Finding the volume of the tetrahedron using triple integrals. D is the tetrahedron bounded by the coordinate planes and the plane 3x+3y+z=3, then express the volume of D as a triple integral.

Finding the volume of the tetrahedron using triple integrals
D is the tetrahedron bounded by the coordinate planes and the plane $3x+3y+z=3$, then express the volume of D as a triple integral.
My Try:
The z-limits are $0\le z\le 3-3x-3y$
If y is the limit on the xy- plane then $z=0$ $0\le y\le 1-x$.
Similarly, $0\le x\le 3$.
Finally, the integral will be
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fairymischiefv9
Explanation:
It is correct for y and z bounds but for x we have
- $y=z=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}0\le x\le 1$
and therefore $V={\int }_{0}^{1}dx{\int }_{0}^{1-x}dy{\int }_{0}^{3-3x-3y}dz$