Why Mixed Patch Types Require Care
A piecewise surface can contain a curved graph, a flat base, and vertical sides. The flux integral over the whole surface is obtained by computing the flux on each patch with its own outward orientation, then adding the results. The challenge is that the same orientation convention does not always look the same in the coordinates used for each patch.
For a graph, an upward normal is often represented by an oriented area vector whose vertical component is positive. On a vertical face, however, the outward direction may be horizontal. A sign chosen correctly for the graph can therefore be wrong for a side face. The reliable approach is to identify the solid bounded by the surface and determine the outward direction patch by patch.
Flux on Individual Patches
For a vector field \(\mathbf{F}\), the flux through an oriented surface patch \(S\) is the surface integral of the normal component of the field. In a parametrization \(\mathbf{r}(u,v)\), the oriented area vector is \(\mathbf{r}_u\times\mathbf{r}_v\,du\,dv\) if that cross product points in the chosen direction. Reversing its order reverses the orientation and changes the sign of the flux.
A frequently used special case is a graph \(z=f(x,y)\). Its upward oriented area vector is \((-f_x,-f_y,1)\,dx\,dy\). The downward orientation uses the negative of this vector. These vectors already include the surface area scaling, so one does not multiply by a separate scalar area factor.
For a vertical face, it is often convenient to use one horizontal coordinate and \(z\) as parameters. The oriented area vector then points horizontally. For example, on \(x=c\), the outward area vector is either \((1,0,0)\,dy\,dz\) or \((-1,0,0)\,dy\,dz\), depending on which side of the solid the face occupies.
Finding the Outward Direction from the Solid
Suppose a solid is locally described by an inequality \(g(x,y,z)\leq 0\), with \(\nabla g\neq \mathbf{0}\) on the boundary patch. The gradient points in the direction in which \(g\) increases. Since points just inside the solid have \(g\leq0\), the direction of increasing \(g\) points out of the solid. Thus \(\nabla g\) gives an outward normal direction on that patch.
For a solid below the graph \(z=f(x,y)\), the top boundary satisfies \(z-f(x,y)=0\), and the solid has \(z-f(x,y)\leq0\). The gradient of \(z-f(x,y)\) is \((-f_x,-f_y,1)\), precisely the upward graph normal. For a solid above \(z=0\), the bottom boundary has inequality \(-z\leq0\), giving the downward normal \((0,0,-1)\). Similar inequalities settle the direction on vertical faces: \(x\geq0\) gives \(-x\leq0\), so the outward normal at \(x=0\) points in the negative \(x\)-direction.
Separate the boundary into graph, horizontal, and vertical pieces, and record the coordinate bounds on each.
Use the inequalities defining the solid, not a visual guess or a direction copied from another patch.
Use its oriented area vector and the bounds that describe that patch.
Flux can be positive, negative, or zero on individual patches; retain each sign when summing.
Worked Example: A Graph Above a Rectangle
Worked Example: A Graph Above a Rectangle
Let \(E\) be the solid defined by \(0\leq x\leq1\), \(0\leq y\leq1\), and \(0\leq z\leq1+x+y\). Compute the outward flux of \(\mathbf{F}(x,y,z)=(x,y,z)\) across its entire boundary.
The boundary consists of the top graph, the bottom \(z=0\), and four vertical faces. On the top, \(f(x,y)=1+x+y\), so \(f_x=f_y=1\) and the outward area vector is \((-1,-1,1)\,dx\,dy\). The field on the graph is \((x,y,1+x+y)\). Therefore its normal component is
The top contribution is \(\int_0^1\int_0^1 1\,dy\,dx=1\).
The bottom has outward area vector \((0,0,-1)\,dx\,dy\). There, \(z=0\), so \(\mathbf{F}\cdot(0,0,-1)=-z=0\). Its contribution is \(0\).
On \(x=0\), the outward direction is negative \(x\). The field’s \(x\)-component is \(x=0\), so the contribution is \(0\). On \(y=0\), the outward direction is negative \(y\); the \(y\)-component is \(y=0\), so this contribution is also \(0\).
On \(x=1\), the outward direction is positive \(x\). Use \(y,z\) as coordinates: \(0\leq y\leq1\) and \(0\leq z\leq2+y\). The flux is
On \(y=1\), the outward direction is positive \(y\). The bounds are \(0\leq x\leq1\) and \(0\leq z\leq2+x\), so the contribution is
Adding the six contributions gives \(1+0+0+0+\frac52+\frac52=6\). This sum includes the top, bottom, and all four sides, each with its outward orientation.
Check: The top integrand is \(1\) throughout the unit square, and each nonzero side contribution integrates its constant outward field component over the stated face bounds. The six values therefore sum to \(6\).
Worked Example: Nonzero Fluxes That Cancel
Worked Example: Nonzero Fluxes That Cancel
Use the solid \(E\) with \(0\leq x\leq1\), \(0\leq y\leq1\), and \(0\leq z\leq2+x+y\). For \(\mathbf{F}(x,y,z)=(y,z,x)\), compute the outward flux over the entire boundary.
The top graph is \(z=2+x+y\). Its outward area vector is \((-1,-1,1)\,dx\,dy\). Substituting the graph into the field gives \((y,2+x+y,x)\), so the normal component is
Thus the top contribution is \(\int_0^1\int_0^1(-2-2y)\,dy\,dx=-3\). On the bottom, the outward vector is \((0,0,-1)\), and the field has third component \(x\). The contribution is \(\int_0^1\int_0^1(-x)\,dy\,dx=-\frac12\).
At \(x=0\), the outward normal points in the negative \(x\)-direction. The first component of the field is \(y\), and \(0\leq z\leq2+y\). The flux is
At \(x=1\), the outward direction is positive \(x\); the bounds are \(0\leq y\leq1\) and \(0\leq z\leq3+y\). The contribution is
At \(y=0\), the outward direction is negative \(y\). The second component of the field is \(z\), and \(0\leq z\leq2+x\), giving
At \(y=1\), the outward direction is positive \(y\), with \(0\leq z\leq3+x\). Its contribution is
The total is \(-3-\frac12-\frac43+\frac{11}{6}-\frac{19}{6}+\frac{37}{6}=0\). In sixths, the numerator is \(-18-3-8+11-19+37=0\).
Check: Each side integral uses the corresponding field component dotted with that face’s outward direction. In particular, the \(y=1\) upper bound is \(3+x\), giving \(\frac12(9+3+\frac13)=\frac{37}{6}\). The displayed contributions sum to zero.
Worked Example: A Triangular Base and a Sloping Side
Worked Example: A Triangular Base and a Sloping Side
Let \(T\) be the solid defined by \(x\geq0\), \(y\geq0\), \(x+y\leq1\), and \(0\leq z\leq1+x\). Find the outward flux of the constant field \(\mathbf{F}=(1,2,3)\) across its boundary.
The top is \(z=1+x\) over the triangle \(x\geq0\), \(y\geq0\), \(x+y\leq1\). Its outward area vector is \((-1,0,1)\,dx\,dy\), and \(\mathbf{F}\cdot(-1,0,1)=2\). The triangle has area \(\frac12\), so the top flux is \(2(\frac12)=1\). The bottom has outward vector \((0,0,-1)\); its normal component is \(-3\), so its flux is \(-3(\frac12)=-\frac32\).
At \(x=0\), the outward direction is negative \(x\). Here \(0\leq y\leq1\) and \(0\leq z\leq1\), so the flux is \(-1\cdot1\cdot1=-1\). At \(y=0\), the outward direction is negative \(y\). The bounds are \(0\leq x\leq1\) and \(0\leq z\leq1+x\). Thus the flux is
The remaining vertical face lies on \(x+y=1\). Parametrize it by \(\mathbf{r}(x,z)=(x,1-x,z)\), where \(0\leq x\leq1\) and \(0\leq z\leq1+x\). Since \(\mathbf{r}_x=(1,-1,0)\) and \(\mathbf{r}_z=(0,0,1)\), the cross product \(\mathbf{r}_z\times\mathbf{r}_x=(1,1,0)\) points outward from \(x+y\leq1\). The flux on this face is
The total is \(1-\frac32-1-3+\frac92=0\). This calculation checks that the sloping vertical face has been included with its outward orientation, rather than treated as another horizontal or graph patch.
Check: The diagonal face’s parametrization has oriented area vector \((1,1,0)\,dx\,dz\), whose dot product with the field is \(3\). Its height is \(1+x\), so its contribution is indeed \(3\int_0^1(1+x)\,dx=\frac92\).
A Sign Error to Avoid
In the second example, the top is the upper boundary of the solid, so its outward direction is upward. Suppose instead that the downward graph vector \((1,1,-1)\) is used there. It reverses the correct top contribution from \(-3\) to \(+3\). The other five contributions sum to \(3\), so this incorrect choice would produce a total of \(3+3=6\), not the computed total of \(0\).
Why the Patchwise Method Works
The method follows from the definition of a surface integral and the additivity of integration. If a surface is divided into patches whose interiors do not overlap, integrating over the whole surface is the same as summing the integrals over those patches. Each patch uses its own parametrization and oriented area vector, but all of those vectors must represent the same prescribed orientation—in these examples, outward from the solid.
The seams do not create an extra flux term. A seam is a curve, not a surface patch, and a curve has zero area in a surface integral. Consequently, the total flux is found by adding the patch integrals only. This reasoning requires a finite, piecewise smooth surface and a vector field defined and integrable on the patches.
The outward direction test can also be justified directly. On a regular boundary described by \(g=0\), with the solid on the side \(g\leq0\), the directional derivative of \(g\) in the direction \(\nabla g\) is \(\nabla g\cdot\nabla g=\lVert\nabla g\rVert^2>0\). Moving a small distance in that direction increases \(g\), hence moves from the solid toward its exterior. Therefore \(\nabla g\) is an outward normal direction. This supplies a consistent sign test even when a face is sloped or its orientation is not visually obvious.
Check Your Understanding
For each question, identify the outward orientation before setting up any integral.
- A solid lies below \(z=f(x,y)\). What oriented area vector represents the outward direction on its top graph?
- A solid is restricted by \(x\geq0\). Which direction is outward on its face \(x=0\), and what inequality gives that direction?
- Why is it valid to add the fluxes over a finite collection of patches meeting along seams?
- For the solid \(0\leq z\leq2+x+y\) over the unit square, what is the outward oriented area vector on \(y=0\)?
- If a patch’s chosen normal is reversed, how does its flux contribution change?