Let $T$ be the tetrahedral region given by vertices at $(1, 0, 0)$, $(0, 0, −1)$, $(−2, 0, 0)$, and $(−1, −1, −1)$. Compute the volume of this tetrahedron using triple integrals
My progress: I have trouble on evaluating the plane for which this solid is located. There are some similar problems where origin is included, but this problem is different. It would be very thankful if you could help me on this.
On
If the triple integral is not a must, the volume can be calculated much easily. Note that the base of the tetrahedral can be viewed in the plane $y=0$ with vertexes $(1, 0, 0)$, $(0, 0, −1)$ and $(−2, 0, 0)$, whose area is $A=\frac12\cdot 1\cdot 3= \frac32$ and the corresponding height is $h=1$. Thus, the volume is
$$V=\frac13 Ah=\frac13\cdot \frac32 \cdot 1 = \frac12$$
First, let us denote the vertices of the tetrahedron as $P_0=(1,0,0)$, $P_1=(0,0,-1)$, $P_2=(-2,0,0)$ and $P_3=(-1,-1,-1)$. Note that a tetrahedron is a 3-simplex, so we can define it using barycentric coordinates:
\begin{align} T &= \left\lbrace u_0 P_0 + u_1 P_1 + u_2 P_2 + u_3 P_3 \ \middle|\ \forall_i\ 0 \leq u_i \leq 1, \; {\textstyle \sum}_i\ u_i = 1 \right\rbrace \\ &= \left\lbrace P_0 + u_1(P_1-P_0) + u_2(P_2-P_0) + u_3(P_3-P_0) \ \middle|\ \forall_i\ 0 \leq u_i \leq 1 \right\rbrace \end{align}
Let $U = \left\lbrace (u_1,u_2,u_3) \in \mathbb{R}^3 \ \middle|\ 0 < u_1,u_2,u_3 < 1 \right\rbrace$ be an open subset of $\mathbb{R}^3$ and $f : U \to \mathbb{R}^3$ be a function defined by the following formula:
\begin{align} f(u_1,u_2,u_3) &= P_0 + u_1(P_1-P_0) + u_2(P_2-P_0) + u_3(P_3-P_0) \\ &= (1,0,0) + u_1(-1,0,-1) + u_2(-3,0,0) + u_3(-2,-1,-1) \\ &= (1-u_1-3u_2-2u_3, -u_3, -u_1-u_3) \\ \end{align}
Notice that $f(U) = \text{int}(T)$ and $f$ is a diffeomorphism on its domain (why?). Then just evaluate the volume integral over $T$ by changing variables using the mapping $f$ and its Jacobian determinant and applying Fubini's theorem:
\begin{align} \text{Vol}(T) &= \iiint_T 1\ d\mathbf{x} \\ &= \iiint_U \left| \det Df(\mathbf{u}) \right|\ d\mathbf{u} \\ &= \int_{0}^{1} \;\; \int_{0}^{1-u_1} \!\!\! \int_{0}^{1-u_1-u_2} \left| \det Df(u_1,u_2,u_3) \right| \ du_3\ du_2\ du_1 \end{align}