Let's do this. A general case you may find helpful. Tetrahedron-wise.

The tetrahedron in the first octant bounded by the coordinate planes and TKH's plane: $\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

Of course, $\displaystyle a>0,b>0,c>0$

6 different integrals and they all evaluate to $\displaystyle \frac{1}{6}abc$.

I hope.

$\displaystyle \int_{0}^{a}\int_{0}^{b(1-\frac{x}{a})}\int_{0}^{c(1-\frac{x}{a}-\frac{y}{b})}dzdydx$

$\displaystyle \int_{0}^{c}\int_{0}^{a(1-\frac{z}{c})}\int_{0}^{b(1-\frac{x}{a}-\frac{z}{c})}dydxdz$

$\displaystyle \int_{0}^{c}\int_{0}^{b(1-\frac{z}{c})}\int_{0}^{a(1-\frac{y}{b}-\frac{z}{c})}dxdydz$

$\displaystyle \int_{0}^{b}\int_{0}^{a(1-\frac{y}{b})}\int_{0}^{c(1-\frac{x}{a}-\frac{y}{b})}dzdxdy$

$\displaystyle \int_{0}^{a}\int_{0}^{c(1-\frac{x}{a})}\int_{0}^{b(1-\frac{x}{a}-\frac{z}{c})}dydzdx$

$\displaystyle \int_{0}^{b}\int_{0}^{c(1-\frac{y}{b})}\int_{0}^{a(1-\frac{y}{b}-\frac{z}{c})}dxdzdy$

WHEW!!!

Evaluate. They should all be (abc)/6.

I think your second one is the same only arranged differently in the plane.