Find volume of tetrahedron bounded by $x=0$,$y=0$,$z=0$ and the plane $z=1+x-y$.

Question

I want to find the Volume of a tetrahedron which is bounded by $x=0$,$y=0$,$z=0$ and the plane $z=1+x-y$.

I know I have to find the region formed by the bounds above and then calculate the integral $\int_RzdA$. My question is how can I work to find the region components which I will use in the integral?

Answer 1

HINT.....Do you need to use calculus? The volume of a tetrahedron with neighbouring edge vectors $\underline{a}, \underline{b}, \underline{c}$ is $$|\frac 16\underline{a}\cdot(\underline{b}\times\underline{c})|$$

Answer 2

$\text{Geometrical approach}$:

Notice, the normal distance of the plane $z=1+x-y\iff x-y-z+1=0$ from the origin $O(0, 0, 0)$ $$=\frac{(0)-(0)-(0)+1}{\sqrt{(1)^2+(-1)^2+(-1)^2}}=\frac{1}{\sqrt 3}$$ The given plane plane: $z=1+x-y$ intersects the coordinate axes at $A(-1, 0, 0)$, $B(0, 1, 0)$ & $C(0, 0, 1)$ respectively.

Thus, the tetrahedron $OABC$ has equilateral triangular base $ABC$ whose each side is
$$=AB=\sqrt{(-1-0)^2+(0-1)^2+(0-0)^2}=\sqrt 2=BC=AC$$
Area of $\triangle ABC$ $$=\frac{\sqrt 3}{4}(\sqrt 2)^2=\frac{\sqrt 3}{2}$$

Hence, the volume of the tetrahedron $OABC$ bounded by $x=0$, $y=0$, $z=0$ & the plane $z=1+x-y $ is $$=\frac{1}{3}\times (\text{area of equilateral triangular base}\ )\times (\text{normal height})$$ $$=\frac{1}{3}\times \left(\frac{\sqrt 3}{2}\right)\times \left(\frac{1}{\sqrt 3}\right)=\color{blue}{\frac{1}{6}}$$