# Clarifications needed in Divergence Theorem

Consider this problem: $F=[4x,x^2y,-x^2z]$ and $S$ is the tetrahedron described by $(0,0,0);(1,0,0);(0,1,0);(0,0,1)$

we have to find the surface integral.

Question 1: Applying divergence theorem $\nabla\cdot F =4$ which implies the surface integral turns into $\int \int \int 4 dV$ or otherwise 4 times the volume of tetrahedron. We know volume of tetrahedron is $\sqrt{2}a^3/12$ which means the answer should be $\sqrt{2}/3$. But the answer is not that.

Question 2: If we split $dV$ into $dxdydz$ and then integrate, the limit my book says should be $0 to 1-y-z$,$0 to 1$,$0 to 1$ respectively. Can you talk a bit more about it. Intuitively I got an understanding, but how is it different the limits had it been from x+y+z=1?

Question 3: Just for the kicks I tried integrating this on the surface of the tetrahedron. I get the normal vector for only one surface, i.e the plane x+y+z =1 the others, turn into an equation where one of the variable (x,y,z) is zero [as the tetrahedron is bounded by that plane] which implies when turning into parametric form of u-v plane, the r(u,v) will be dependent on only 1 variable. For instance, one of the planes will be $x+z\leq 1$ => r(u,v)=[u,0,1-u] which implies $r_v=0$, hence cross product of $r_u$ and $r_v$ will be zero, hence the integral will be zero on that surface. This was my modus operandi. So I surface integrated over only $x+y+z \leq 1$ and I am getting a pretty way off the answer. Can someone lend me a hand here.

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The volume of a simplex in $\mathbb{R}^n$ (tetrahedron in $\mathbb{R}^3$), one of whose vertices is the origin, is $\frac{1}{n!}$ times the determinant of the matrix with the other vertices as columns. Thus, the volume of your tetrahedron is $$\frac{1}{3!}\det\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\frac16$$ and $4\cdot\frac16=\frac23$, which is hopefully the correct answer.

The volume integral should be $$\int_0^1\int_0^{1-z}\int_0^{1-z-y}4\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$ That is, $z$ can take on any value in $[0,1]$. The integral in $y$ is taken over each particular value of $z$, so $y$ can take any value in $[0,1-z]$. The integral in $x$ is taken over each particular value of $z$ and $y$, so $x$ can take any value in $[0,1-z-y]$.

On the surface $x+y+z=1$, the surface normal is $\nabla(x+y+z-1)=(1,1,1)$. Normalizing this gives the unit normal: $n=\frac{1}{\sqrt{3}}(1,1,1)$.

To integrate over this surface, the integral would be \begin{align} &\int_0^1\int_0^{1-y}F\cdot n\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\;\mathrm{d}x\,\mathrm{d}y\\ &=\int_0^1\int_0^{1-y}F\cdot n\sqrt{3}\;\mathrm{d}x\,\mathrm{d}y\\ &=\int_0^1\int_0^{1-y}(4x,x^2y,-x^2z)\cdot(1,1,1)\frac{1}{\sqrt{3}}\sqrt{3}\;\mathrm{d}x\,\mathrm{d}y\\ &=\int_0^1\int_0^{1-y}(4x+x^2y-x^2z)\;\mathrm{d}x\,\mathrm{d}y\\ &=\int_0^1\int_0^{1-y}(4x+x^2y-x^2(1-x-y))\;\mathrm{d}x\,\mathrm{d}y\\ &=\int_0^1\int_0^{1-y}(x^3-x^2+4x+2x^2y)\;\mathrm{d}x\,\mathrm{d}y \end{align} The other surfaces should be much simpler.

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Q1: The volume of a regular tetrahedron of side length $a$ may be $\sqrt{2}a^3/12$; but the tetrahedron $T$ at stake here is not regular. It can be viewed as a pyramid with base area ${1\over2}$ and height $1$, so its volume is ${1\over6}$. This implies that the flux $\Phi$ computed as a volume integral amounts to ${2\over3}$.

Q2: The triple integral in question is $$\Phi=\int_0^1\Bigl(\int_0^{1-x}\bigl(\int_0^{1-x-y}4\ dz\bigr)\ dy\Bigr)\ dx\ .$$ To compute it you have to begin with the innermost integral: For given $(x,y)\in\Delta_1$ (the base triangle) the allowable $z$ have to satisfy $z\geq0$ and $x+y+z\leq1$. This leads to the integration interval $[0,1-x-y]$ for $z$, and the value of the innermost integral is therefore $=4(1-x-y)=:g(x,y)$. In the second step we have to integrate the function $g$ over $\Delta_1$. For given $x\in[0,1]$ (the base of $\Delta_1$) the allowable $y$ have to satisfy $y\geq0$ and $x+y\leq1$. This leads to the integration interval $[0,1-x]$ for $y$. The value of the integral $\int_0^{1-x} g(x,y)\ dy$ is a function $x\mapsto h(x)$ which in the third step has to be integrated from $0$ to $1$.

Q3: The surface $\partial T$ of our tetrahedron consists of four triangles. In order to compute the total flux $\Phi:=\int_{\partial T}{\bf F}\cdot d{\bf \vec\omega}$ we therefore have to compute four separate integrals over four parameter triangles.

For the base triangle $\Delta_1$ we use the trivial parametrization ${\bf f}:(x,y)\mapsto(x,y,0)$ with $d\vec\omega=(0,0,-1)\ {\rm d}(x,y)$ (we need the outward normal). Therefore we get $$\int_{\Delta_1}{\bf F}\cdot d\vec\omega=\int_0^1\int_0^{1-x}\bigl(4x,x^2y,0)\cdot(0,0,-1)\ dy\ dx=0\ .$$

I hope you can do the other three yourself.

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Regarding Q#2, I was not actually asking about the way to integrate but rather the limits. How did they come decide upon the limits – Soham Jun 28 '12 at 17:36
@Soham: See my edit. – Christian Blatter Jun 28 '12 at 18:02