Introduction
A closed surface assembled from different kinds of patches requires more than one surface formula. A curved graph may need a graph area element, a cylindrical wall a parametrization, and a planar face a constant normal. The central task is to orient every patch outward and compute its contribution consistently.
This tutorial works through three direct calculations. Each finds the flux by integrating over the surface patches themselves and then adding the results. No Divergence Theorem is used.
Flux on a Patch and Outward Orientation
Let \(\mathbf F\) be a continuous vector field defined on a surface \(S\). If the surface is oriented by a unit normal vector \(\mathbf n\), its flux is the integral of the normal component of the field over the surface. In symbols, the flux is \(\iint_S \mathbf F\cdot\mathbf n\,dS\). Positive contributions point along the chosen normal; negative contributions point against it.
It is often convenient to combine the normal and the area element into a vector area element. For a parametrized patch \(\mathbf r(u,v)\), the oriented vector area element is \((\mathbf r_u\times\mathbf r_v)\,du\,dv\), or its negative if the cross product points the wrong way. Thus the patch flux is \(\iint \mathbf F(\mathbf r(u,v))\cdot(\mathbf r_u\times\mathbf r_v)\,du\,dv\), with the cross product ordered or reversed to match the required orientation.
For an upward-oriented graph \(z=g(x,y)\), this vector area element is \((-g_x,-g_y,1)\,dx\,dy\). For a downward-oriented graph, its negative is used. A planar face can be handled with the same principle: use a constant normal and integrate over the face’s two-dimensional region.
Why the Patch Contributions Add
Suppose a closed surface is divided into finitely many smooth patches, each with its outward orientation. The flux over the whole surface is the sum of the fluxes over the patches. The seams where patches meet do not affect the surface integral: they are curves, which have zero surface area.
This reasoning requires a well-defined continuous field on the surface, smooth parametrizations on the patches, and a consistent outward orientation. It does not require the patches to use the same parametrization or formula.
Worked Example: A Paraboloid Cap, a Cylinder, and a Bottom
Worked Example: A Paraboloid Cap, a Cylinder, and a Bottom
Let \(S\) bound the solid \(0\leq z\leq 2-x^2-y^2\), \(x^2+y^2\leq1\). Its boundary consists of the paraboloid cap, the bottom disk \(z=0\), and the cylindrical wall \(x^2+y^2=1\), \(0\leq z\leq1\). Find the outward flux of \(\mathbf F(x,y,z)=(x,y,z)\).
Paraboloid cap. Write the cap as \(z=g(x,y)=2-x^2-y^2\). The solid lies below this graph, so its outward normal points upward. Since \(g_x=-2x\) and \(g_y=-2y\), its upward vector area element is \((2x,2y,1)\,dx\,dy\). On the graph, \(\mathbf F=(x,y,2-x^2-y^2)\). Therefore the dot product is
The projection is the unit disk. In polar coordinates, \(x^2+y^2=r^2\) and \(dx\,dy=r\,dr\,d\theta\), so the cap flux is
Bottom disk. The solid is above \(z=0\), so the outward normal is \((0,0,-1)\). On this disk, the third component of \(\mathbf F\) is \(z=0\), and hence \(\mathbf F\cdot(0,0,-1)=0\). The bottom contributes \(0\).
Cylindrical wall. Parameterize it by \(\mathbf r(\theta,z)=(\cos\theta,\sin\theta,z)\), with \(0\leq\theta\leq2\pi\) and \(0\leq z\leq1\). The tangent vectors are \(\mathbf r_\theta=(-\sin\theta,\cos\theta,0)\) and \(\mathbf r_z=(0,0,1)\). Their cross product is \((\cos\theta,\sin\theta,0)\), which points outward. On the wall, \(\mathbf F=(\cos\theta,\sin\theta,z)\), so its dot product with this vector is \(1\). The wall flux is
Adding the three contributions gives \(\frac{5\pi}{2}+0+2\pi=\frac{9\pi}{2}\). As a check, the cap integrand \(2+r^2\) is positive, the cylinder’s outward dot product is \(1\), and the bottom dot product vanishes, matching the signs and values used in the sum.
Worked Example: A Curved Roof Above a Square
Worked Example: A Curved Roof Above a Square
Consider the solid \(0\leq x\leq1\), \(0\leq y\leq1\), \(0\leq z\leq1+x^2\). Its boundary has a curved roof, a bottom, and four planar sides. Find the outward flux of \(\mathbf F(x,y,z)=(x,y,z)\).
Roof. The roof is \(z=g(x,y)=1+x^2\), and the solid lies beneath it, so use the upward vector area element. Since \(g_x=2x\) and \(g_y=0\), it is \((-2x,0,1)\,dx\,dy\). The field on the roof is \((x,y,1+x^2)\), giving
Thus the roof flux is \(\int_0^1\int_0^1(1-x^2)\,dy\,dx=[x-x^3/3]_0^1=2/3\).
Bottom. Its outward normal is \((0,0,-1)\), while \(z=0\) on the bottom. The dot product is zero, so this face contributes \(0\).
Side \(x=0\). The outward normal is \((-1,0,0)\). The field’s \(x\)-component is zero there, so this side contributes \(0\).
Side \(x=1\). The outward normal is \((1,0,0)\). The field’s normal component is \(x=1\). The face is described by \(0\leq y\leq1\) and \(0\leq z\leq2\), giving flux \(\int_0^1\int_0^2 1\,dz\,dy=2\).
Side \(y=0\). Its outward normal is \((0,-1,0)\), and the field’s \(y\)-component is zero on this side. Its contribution is \(0\).
Side \(y=1\). Its outward normal is \((0,1,0)\), and the field’s normal component is \(y=1\). The height is \(0\leq z\leq1+x^2\), so this contribution is
The total outward flux is \(\frac23+0+2+0+\frac43=4\). Checking the arithmetic, the roof and \(y=1\) side contribute \(\frac23+\frac43=2\), and the \(x=1\) side contributes another \(2\). All other faces have zero normal component, so the patch sum is \(4\).
Worked Example: A Quarter-Cylinder with Flat Ends
Worked Example: A Quarter-Cylinder with Flat Ends
Let \(S\) bound the solid \(x^2+y^2\leq1\), \(x\geq0\), \(y\geq0\), \(0\leq z\leq1\). Its patches are a quarter of a cylindrical wall, top and bottom quarter-disks, and the planar faces \(x=0\) and \(y=0\). Find the outward flux of \(\mathbf F(x,y,z)=(x,y,z)\).
Cylindrical wall. Use \(\mathbf r(\theta,z)=(\cos\theta,\sin\theta,z)\), \(0\leq\theta\leq\pi/2\), \(0\leq z\leq1\). As in the preceding parametrization, \(\mathbf r_\theta\times\mathbf r_z=(\cos\theta,\sin\theta,0)\), which points outward. The field’s dot product with it is \(\cos^2\theta+\sin^2\theta=1\). Hence the wall flux is \(\int_0^{\pi/2}\int_0^1 1\,dz\,d\theta=\pi/2\).
Top quarter-disk. At \(z=1\), the outward normal is \((0,0,1)\). The field’s normal component is \(1\), so the flux equals the disk’s area. In polar coordinates, this is \(\int_0^{\pi/2}\int_0^1 r\,dr\,d\theta=\pi/4\).
Bottom quarter-disk. The outward normal is \((0,0,-1)\), but \(z=0\) there. Its flux is \(0\).
Face \(x=0\). The outward normal is \((-1,0,0)\), and \(F_x=x=0\) on that face, so its flux is \(0\).
Face \(y=0\). The outward normal is \((0,-1,0)\), and \(F_y=y=0\) on that face, so its flux is \(0\).
The total is \(\pi/2+\pi/4+0+0+0=3\pi/4\). As a check, the curved wall’s normal component is identically \(1\) over angular width \(\pi/2\) and height \(1\), while the top contributes the quarter-disk area \(\pi/4\); together these give \(3\pi/4\).
A Reliable Patch-by-Patch Workflow
For a new composite-surface problem, it helps to list the patches before integrating. Then determine each patch’s outward orientation from the solid it bounds. Only after the normal has been fixed should the field and area element be substituted.
Record each planar or curved piece and the parameter region or coordinate bounds that cover it.
For every patch, decide which side points away from the enclosed solid. Use that orientation in the normal or cross product.
Substitute the field into the oriented surface integral and evaluate it over that patch’s bounds.
Sum signed contributions, then check that the orientation and zero contributions are consistent with the geometry.
Common Pitfall: Reversing One Patch
A closed surface uses outward normals everywhere, not merely a mixture of normals that look plausible in isolation. For example, the bottom of a solid above \(z=0\) points downward. If the bottom normal in the first example were incorrectly taken as \((0,0,1)\), the orientation would be inward on that face. In that example the erroneous choice happens to give zero as well, because \(F_z=0\) on the bottom. The numerical answer would not reveal the mistake.
That is why an orientation check must be geometric, not just numerical. Ask whether the normal points away from the solid. On a graph roof with the solid below, outward is upward; on the bottom with the solid above, outward is downward. A zero flux contribution does not excuse an incorrectly oriented normal.
Why Direct Decomposition Matters
Directly integrating over patches makes the geometry visible in the calculation. It shows which parts of the field cross each face, which contributions cancel or vanish, and where a sign comes from. It is also a useful independent method when a closed surface has a complicated boundary description: each patch can be treated with the surface formula appropriate to its shape, then combined by additivity.
The main discipline is to preserve the outward orientation throughout. Once each patch’s vector area element and bounds are correct, the total flux is the ordinary sum of the resulting integrals.
Check Your Understanding
Use the patch orientations and direct surface-integral formulas to answer these questions.
- For a graph \(z=g(x,y)\) with the solid below it, which graph vector area element gives the outward orientation?
- Why may the flux across a piecewise smooth closed surface be calculated by adding the fluxes of its patches?
- On the bottom of a solid lying above \(z=0\), what is the outward normal direction?
- For the quarter-cylinder example, why is the cylindrical wall’s outward vector area element \((\cos\theta,\sin\theta,0)\,d\theta\,dz\)?
- Why does a zero patch contribution not establish that the normal used on that patch was oriented correctly?