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 ${\int}_{0}^{3}{\int}_{0}^{1-x}{\int}_{0}^{3-3x-3y}dz\text{}dy\text{}dx$

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 ${\int}_{0}^{3}{\int}_{0}^{1-x}{\int}_{0}^{3-3x-3y}dz\text{}dy\text{}dx$