Find the volume of the tetrahedron using triple integrals

Question

Find the volume of the tetrahedron with vertices $(0,0,0), (0,0,1), (1,0,1), (0,1,1)$, equation $$x+y=z$$ bounded by $x=0$, $y=0$ and $z=1$.

Then $x+y=1$. My integral: is this correct? $$\int_0^1 \int_0^{z} \int_0^{z-y} \,dx\,dy\,dz$$

Answer 1

Your limits are correct, considering the YZ axis, if you want another example to understand better I have given the limits considering the XZ plane.

The tetrahedron is having its base on the XZ plane and YZ plane so it will be a good idea to consider the limits from the XZ plane to the $x+y=z$ plane.

considering the limits from the YZ plane to the $x+y=z$

$y=0$ to $y=z-x$

Now as the limits for the volume has been taken care of, the area formed on the XZ plane is considered. If the strip is considered parallel to the z axis then the limits

$z=x$ to $z=1$ and the limits of the x are $x=0$ to $x=1$ . This limit has been chosen to make it easier to solve the last integral using the Beta function.

The equation will be $$\int_0^1\int_x^{1}\int_0^{z-x}dy dz dx$$