2
$\begingroup$

Given three smooth functions $f,g,h:\mathbb{R}\to\mathbb{R}^3$ how would one find the volume traversed by the triangle $\triangle(f(t),g(t),h(t))$ from $t=a$ to $t=b$?

Assume that all triangles $\triangle(f,g,h)$ are distinct (i.e., there are no intersections between the areas of any two triangles $\triangle(f(s),g(s),h(s))$ and $\triangle(f(t),g(t),h(t))$ where $s\ne t$).

$\endgroup$

1 Answer 1

2
$\begingroup$

For each $t \in [a,b]$, the triangle $\triangle (f(t),g(t),h(t))$ can be parametrized by $$ (s_1,s_2) \mapsto s_1 f(t) + s_2 g(t) + (1-s_1-s_2) h(t), $$ where $(s_1,s_2)$ are taken from the triangle $$ \Delta = \{ (s_1,s_2) \in \mathbb{R}^2 : s_1,s_2 \ge 0, \ s_1+s_2 \le 1 \}. $$ Then the region you describe is exactly $\Phi([a,b] \times \Delta)$, where $$ \Phi \colon [a,b] \times \Delta \to \mathbb{R}^3, \quad \Phi(t,s_1,s_2) = s_1 f(t) + s_2 g(t) + (1-s_1-s_2) h(t). $$ Its volume can be computed by a change of variables: $$ |\Phi([a,b] \times \Delta)| = \int_{\Phi([a,b] \times \Delta)} dx dy dz = \int_{[a,b] \times \Delta} |\det D \Phi(t,s_1,s_2)| \, dt ds_1 ds_2. $$

$\endgroup$
1
  • $\begingroup$ [+1] clear explanation. $\endgroup$
    – Jean Marie
    Apr 18, 2017 at 18:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .