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Problem of the Week #48 - February 25th, 2013

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Chris L T521

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Jan 26, 2012
995
Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: A tetrahedron $T$ with vertices $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ and $\mathbf{d}$ can be parameterized by $\mathbf{x}=\mathbf{g}(s,t,u)$, where
\[\mathbf{g}(s,t,u)=\mathbf{d}+u(\mathbf{c}-\mathbf{d})+tu(\mathbf{a}-\mathbf{c})+stu(\mathbf{b}-\mathbf{a})\]
and $0\leq s\leq 1$, $0\leq t\leq 1$ and $0\leq u\leq 1$. Use this to show that
\[\iiint\limits_T f(x,y,z)\,dV= \int_0^1\int_0^1\int_0^1 f(\mathbf{g}(s,t,u))tu^2 \left|\det\left(\begin{bmatrix}\mathbf{b}-\mathbf{a}\\ \mathbf{a}-\mathbf{c}\\ \mathbf{c}-\mathbf{d}\end{bmatrix}\right)\right|\,ds\,dt\,du.\]

Then use this formula to evaluate $\displaystyle\iiint\limits_T (x+y+z)\,dV$, where $T$ is the tetrahedron with vertices $(0,0,0)$, $(1,1,0)$, $(-1,2,1)$, and $(-1,-1,4)$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
This week's problem was mostly answered correctly by jakncoke (you got the derivation correct, but the the computation of $\iiint x+y+z \,dV$ incorrect.)

Here's a hybrid of our two solutions (with some minor changes to jakncoke's answer).

We do the justification part first. If $\mathbb{g}(s,t,u)=\mathbf{d}+u(\mathbf{c}-\mathbf{d})+tu(\mathbf{a}-\mathbf{c})+stu(\mathbf{b}-\mathbf{a})$ with $0\leq s,t,u\leq 1$, then we have that\[\begin{aligned}\mathbf{g}_s(s,t,u) &= tu(\mathbf{b}-\mathbf{a})\\ \mathbf{g}_t(s,t,u) &= u(\mathbf{a}-\mathbf{c}) + su(\mathbf{b}-\mathbf{a})\\ \mathbf{g}_u(s,t,u) &= \mathbf{c}-\mathbf{d}+t(\mathbf{a}-\mathbf{c}) + st(\mathbf{b}-\mathbf{a}).\end{aligned}\]
We now see that our Jacobian matrix is
\[\begin{bmatrix}tu(\mathbf{b}-\mathbf{a}) \\ u(\mathbf{a}-\mathbf{c}) + su(\mathbf{b}-\mathbf{a})\\ \mathbf{c}-\mathbf{d} + t(\mathbf{a}-\mathbf{c})+st(\mathbf{b}-\mathbf{a})\end{bmatrix}=\begin{bmatrix}tu & 0 & 0\\ su & u & 0\\ st & t & 1\end{bmatrix}\begin{bmatrix}\mathbf{b}-\mathbf{a}\\ \mathbf{a}-\mathbf{c}\\ \mathbf{c}-\mathbf{d}\end{bmatrix}\]
and thus
\[\begin{aligned} \left|\det\begin{bmatrix}tu(\mathbf{b}-\mathbf{a}) \\ u(\mathbf{a}-\mathbf{c}) + su(\mathbf{b}-\mathbf{a})\\ \mathbf{c}-\mathbf{d} + t(\mathbf{a}-\mathbf{c})+st (\mathbf{b}-\mathbf{a})\end{bmatrix}\right| &= \left|\det\left( \begin{bmatrix}tu & 0 & 0\\ su & u & 0\\ st & t & 1\end{bmatrix} \begin{bmatrix}\mathbf{b}-\mathbf{a}\\ \mathbf{a}-\mathbf{c}\\ \mathbf{c}-\mathbf{d}\end{bmatrix}\right)\right|\\ &= \left|\det\begin{bmatrix} tu & 0 & 0\\ su & u & 0\\ st & t & 1\end{bmatrix}\right| \cdot \left|\det\begin{bmatrix} \mathbf{b}-\mathbf{a}\\ \mathbf{a}-\mathbf{c}\\ \mathbf{c}-\mathbf{d} \end{bmatrix}\right|\\ &= tu^2\left|\det\begin{bmatrix}\mathbf{b}-\mathbf{a}\\ \mathbf{a}-\mathbf{c}\\ \mathbf{c}-\mathbf{d}\end{bmatrix}\right| \end{aligned} \]

Therefore, by change of variables, we have that
\[\begin{aligned} \iiint\limits_T f(x,y,z)\,dV &= \int_0^1\int_0^1\int_0^1 f(\mathbf{g}(s,t,u)) \left|\det\begin{bmatrix}tu(\mathbf{b}-\mathbf{a}) \\ u(\mathbf{a}-\mathbf{c}) + su(\mathbf{b}-\mathbf{a})\\ \mathbf{c}-\mathbf{d} + t(\mathbf{a}-\mathbf{c})+st (\mathbf{b}-\mathbf{a})\end{bmatrix}\right|\,ds\,dt\,du\\ &= \int_0^1\int_0^1\int_0^1 f(\mathbf{g}(s,t,u)) tu^2\left|\det\begin{bmatrix}\mathbf{b}-\mathbf{a}\\ \mathbf{a}-\mathbf{c}\\ \mathbf{c}-\mathbf{d}\end{bmatrix}\right| \,ds\,dt\,du. \end{aligned}\]
which is what was to be shown.

For the second part of the problem, let us define the corners of the tetrahedron $T$ as follows:
\[\begin{aligned}\mathbf{a} &= (0,0,0)\\ \mathbf{b} &= (1,1,0) \\ \mathbf{c} &= (-1,2,1)\\ \mathbf{d} &= (-1,-1,4).\end{aligned}\]

It now follows that
\[\begin{aligned} \mathbf{b}-\mathbf{a} &= (1,1,0)\\ \mathbf{a}-\mathbf{c} &= (1,-2,-1)\\ \mathbf{c}-\mathbf{d} &= (0,3,-3) \end{aligned}\]
and thus we have that
\[\begin{aligned}\mathbf{g}(s,t,u) &= (-1,-1,4) + u(0,3,-3) +tu(1,-2,-1)+ stu(1,1,0)\\ &= (-1+tu+stu, -1+3u-2tu+stu, 4-3u-tu). \end{aligned}\]
and
\[\left|\det\begin{bmatrix}\mathbf{b}-\mathbf{a}\\ \mathbf{a}-\mathbf{c}\\ \mathbf{c}-\mathbf{d}\end{bmatrix}\right| =\left|\det\begin{bmatrix} 1 & 1 & 0\\ 1 & -2 & -1 \\ 0 & 3 & -3\end{bmatrix}\right| = \left|\det\begin{bmatrix} -2 & -1\\ 3 & -3\end{bmatrix} - \det\begin{bmatrix}1 & 0\\ 0 & -3\end{bmatrix}\right|=|9+3| = 12\]

Furthermore, if $f(x,y,z)=x+y+z$, we have that
\[\begin{aligned}f(\mathbf{g}(s,t,u)) &= -1+tu+stu+(-1+3u-2tu+stu)+(4-3u-tu)\\ &= 2-2tu+2stu\\ &= 2(1-tu+stu). \end{aligned}\]

We now plug all of this into our formula to see that

\[\begin{aligned} \iiint\limits_T x+y+z\,dV &= \int_0^1 \int_0^1 \int_0^1 2(1-tu+stu)tu^2(12)\,ds\,dt\,du\\ &= 24\int_0^1 \int_0^1 \int_0^1 tu^2-t^2u^3+st^2u^3\,ds\,dt\,du\\ &= 24\int_0^1 \int_0^1 \left[s(tu^2-t^2u^3)+\frac{1}{2}s^2t^2u^3 \right]_0^1\,dt\,du\\ &= 24\int_0^1\int_0^1 tu^2-\frac{1}{2}t^2u^3\,dt\,du\\ &= 24 \int_0^1\left[ \frac{1}{2}t^2u^2-\frac{1}{6}t^3u^3\right]_0^1\,du\\ &= 24\int_0^1 \frac{u^2}{2}-\frac{u^3}{6}\,du\\ &= 24\left[\frac{u^3}{6} - \frac{u^4}{24}\right]_0^1\\ &= 24\left[\frac{1}{6}-\frac{1}{24}\right]\\ &= 24\cdot\frac{3}{24}\\ &= 3\end{aligned}\]

This now completes the solution to the problem. $\hspace{4in}\clubsuit$
 
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