Question

Find the volume of the tetrahedron in the first octant bounded by the coordinate planes and the plane passing through ​, ​, and .

Answer 1

Let's assume a plane

$ax+by+cz=1$

with each of a, b, and c greater than 0. Then the plane has intercepts with coordinates

$y=z=0 \implies ax = 1 \implies x = \dfrac1a$

$x=z=0 \implies by = 1 \implies y = \dfrac1b$

$x=y=0 \implies cz = 1 \implies z = \dfrac1c$

The volume of the tetrahedron is then

$\displaystyle \int_0^{\frac1a} \int_0^{\frac{1-ax}b} \int_0^{\frac{1-ax-by}c} dz \, dy \, dx = \boxed{\frac1{6abc}}$

i.e. 1/6 times the product of the intercepts' non-zero coordinates