Missing term from manual verification of divergence theorem exercise

Question

Let $F(x,y,z)=(R(x,y,z),0,0)$ with $R\in C^1$. Let $\Omega $ be the volume bounded by $x=0,y=0,z=0,x+y+z=1$. Prove by direction computation that $$\int_{\partial \Omega}F\cdot ds=\int _\Omega \nabla \cdot Fdxdydz.$$

For the LHS, the boundary consists of four triangular faces, and three of those have normals perpendicular to the $z$ axis hence contribute nothing. For the fourth we parametrize via $u(x,y,z)=1-z-y,v(x,y,z)=y,w(x,y,z)=z$ and obtain $$\int_{z=0}^1\int _{y=0}^{1-z}R(1-z-y,y,z)dydz.$$

However, on the RHS we have $$\int_{z=0}^1\int_{y=0}^{1-z}\int _{x=0}^{1-z-y}\frac{\partial R}{\partial x}dxdydz$$which gives me $$\int_{z=0}^1\int _{y=0}^{1-z}[R(1-z-y,y,z)-R(0,y,z)]dydz.$$

What do I do with the $R(0,y,z)$ term? I'm clueless. I want to say something like $x=0$ means $y=z-1$ and then maybe do something with that, but I don't see how to read the equality $x+y+z=1$ from the limits of the integral...

Answer 1

Your computation of the LHS isn't quite right. The surface integral of $\mathbf{F}$ along $\partial\Omega$ is \begin{equation} \int_{\partial\Omega}\mathbf{F}\cdot d\mathbf{S} = \int_{\partial\Omega}(\mathbf{F}\cdot\mathbf{n})dS, \end{equation} where $\mathbf{n}$ is an outward-pointing normal vector to $\partial\Omega$. Since only the first component of $\mathbf{F}$ is nonzero, $\mathbf{F}\cdot\mathbf{n}$ will be nonzero only when $\mathbf{n}$ has a nonzero $x$-component. Along the $xy$- and $xz$-planes (corresponding to $z=0$ and $y=0$), $\mathbf{n}$ is perpendicular to $\mathbf{e}_1$, the standard unit vector in the $x$-direction. So these faces contribute nothing to the LHS. But along the $yz$-plane (corresponding to $x=0$), $\mathbf{n}=\mathbf{e}_1$, so $\mathbf{F}\cdot\mathbf{n}=R(x,y,z)$. Since this plane corresponds to $x=0$, this will give you the $R(0,y,z)$ term that shows up on the RHS.