How to parameterize the faces of a cube for a surface integral

Key takeaway: When a problem gives a box like $0\le x\le 1$, $0\le y\le 1$, $0\le z\le 1$, the safest move is to treat the boundary as six separate square faces. Each face gets its own two-parameter map, and the outward normal comes from the orientation of that face.

For the cube $0\le x\le 1,\quad 0\le y\le 1,\quad 0\le z\le 1,$ its boundary is made of six faces. A clean parametrization is to write each face as a surface with two free variables.

1) The six face parametrizations

• Bottom face $z=0$: $\mathbf r(x,y)=(x,y,0),\quad 0\le x\le 1,\ 0\le y\le 1.$
• Top face $z=1$: $\mathbf r(x,y)=(x,y,1),\quad 0\le x\le 1,\ 0\le y\le 1.$
• Front face $y=0$: $\mathbf r(x,z)=(x,0,z),\quad 0\le x\le 1,\ 0\le z\le 1.$
• Back face $y=1$: $\mathbf r(x,z)=(x,1,z),\quad 0\le x\le 1,\ 0\le z\le 1.$
• Left face $x=0$: $\mathbf r(y,z)=(0,y,z),\quad 0\le y\le 1,\ 0\le z\le 1.$
• Right face $x=1$: $\mathbf r(y,z)=(1,y,z),\quad 0\le y\le 1,\ 0\le z\le 1.$

2) For the vector field

$\mathbf F(x,y,z)=\mathbf j+x^2 z\,\mathbf k,$ we read this as $\mathbf F(x,y,z)=(0,1,x^2z).$

If the task is a flux integral over the cube, then on each face you compute $\iint_S \mathbf F\cdot \mathbf n\,dS$ using the outward unit normal for that face.

3) A quick shortcut

Because the field has only a $\mathbf j$-component and a $\mathbf k$-component:

• faces with normals in the $\pm \mathbf i$ direction often give zero flux from the $\mathbf j$ term,
• faces with normals in the $\pm \mathbf j$ or $\pm \mathbf k$ direction need direct checking.

If your course allows it, the Divergence Theorem is usually much faster for the whole closed cube.

For this field, $\nabla\cdot \mathbf F=\frac{\partial}{\partial x}(0)+\frac{\partial}{\partial y}(1)+\frac{\partial}{\partial z}(x^2z)=x^2.$ Then the outward flux through the closed cube is $\iiint_V x^2\,dV=\int_0^1\int_0^1\int_0^1 x^2\,dz\,dy\,dx=\int_0^1 x^2\,dx=\frac13.$

So if you only needed the parametrization, list the six faces. If you needed the total flux, the divergence theorem gives the answer quickly.

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Pitfalls the pros know 👇 A common mistake is to parametrize the whole cube with one map. That does not work for a closed surface like this; you need one parametrization per face. Another easy slip is mixing up orientation: the outward normal on $x=0$ points in the negative $x$-direction, while on $x=1$ it points in the positive $x$-direction.

What if the problem changes? If the problem asks for just one face, keep the other two coordinates fixed and let the remaining two vary in $[0,1]$. If the box were $a\le x\le b$, $c\le y\le d$, $e\le z\le f$, the same pattern works with intervals adjusted accordingly. For a non-cubic rectangular solid, nothing conceptual changes; only the bounds do.

Tags: surface parametrization, outward normal, divergence theorem