Surface integrals

A line integral totals a vector field's effect along a curve; a surface integral does the analogous job across a two-dimensional surface, and the version that matters most is flux — how much of a field passes through the surface. If F⃗\vec F is the velocity field of a flowing fluid, the flux integral gives the volume of fluid crossing the surface per unit time; if F⃗\vec F is a current density, it gives the total current through it. In MA10002's vector calculus unit, surface integrals arrive right after line integrals and path independence, and they exist to set up the two big theorems that close the course: Gauss's divergence theorem and Stokes's theorem, both of which are built entirely out of flux integrals like the one developed here. The scope stays practical: every surface you will meet is given as a graph z=g(x,y)z=g(x,y) over a region RR in the xyxy-plane, so the whole machinery reduces to picking a consistent normal and turning a surface integral into an ordinary double integral you already know how to evaluate.

The core insight

Flux through a graph surface z=g(x,y)z=g(x,y) reduces to the double integral ∬R(−F1gx−F2gy+F3) dx dy\iint_R(-F_1g_x-F_2g_y+F_3)\,dx\,dy over its projection RR, once zz has been substituted into F⃗\vec F.

The core idea

For a vector field F⃗\vec F and an oriented surface SS with unit normal n^\hat{\mathbf n}, the surface (flux) integral is

∬SF⃗⋅n^ dS.\iint_S \vec F\cdot \hat{\mathbf n}\,dS.

where: F⃗\vec F is the field being measured, n^\hat{\mathbf n} is the chosen unit normal on SS, and dSdS is the scalar surface-area element; together n^ dS\hat{\mathbf n}\,dS acts as one vector object, not two separate pieces to compute independently. When SS is presented as a graph, z=g(x,y)z=g(x,y), over a region RR in the xyxy-plane, the vector surface element is built directly from the partial derivatives of gg:

n^ dS=(−∂g∂x ı^−∂g∂y ȷ^+k^)dx dy.\hat{\mathbf n}\,dS = \left(-\frac{\partial g}{\partial x}\,\hat\imath - \frac{\partial g}{\partial y}\,\hat\jmath + \hat k\right) dx\,dy.

where: ∂g/∂x\partial g/\partial x and ∂g/∂y\partial g/\partial y are the surface's slopes in the xx and yy directions, and the +k^+\hat k component signals this is the upward-pointing normal. No separate 1+gx2+gy2\sqrt{1+g_x^2+g_y^2} normalizing factor is needed here — it is already built in and cancels against ∣n^∣|\hat{\mathbf n}| once n^ dS\hat{\mathbf n}\,dS is treated as a single vector. Two conventions matter before evaluating anything. First, choosing the downward normal instead of upward simply flips the sign of the whole integral — the surface hasn't changed, only which way you're counting flux as positive. Second, for a closed surface (a sphere, or the full boundary of a solid region), the standing convention is to take n^\hat{\mathbf n} pointing outward, and the integral is written with a circle through the double integral sign, ∯SF⃗⋅n^ dS\displaystyle\oiint_S \vec F\cdot\hat{\mathbf n}\,dS — this is exactly the object Gauss's divergence theorem relates to a volume integral of ∇⋅F⃗\nabla\cdot\vec F.

Figure
A tilted plane surface over a square base region with an upward normal arrow and shading showing the projection onto the xy-plane
Flux through a graph surface z = g(x, y) surface S: z = g(x, y) n̂ projected region R in the xy-plane x y z

Substituting n^ dS\hat{\mathbf n}\,dS into the flux integral converts it to a plain double integral over RR:

∬SF⃗⋅n^ dS=∬R(−F1∂g∂x−F2∂g∂y+F3)dx dy,\iint_S \vec F\cdot\hat{\mathbf n}\,dS = \iint_R \left(-F_1\frac{\partial g}{\partial x} - F_2\frac{\partial g}{\partial y} + F_3\right) dx\,dy,

where: F1,F2,F3F_1,F_2,F_3 are the components of F⃗\vec F evaluated on the surface, i.e. with zz replaced by g(x,y)g(x,y) everywhere it appears. The working procedure is always the same four moves: identify g(x,y)g(x,y) and its projection RR; differentiate to get gxg_x and gyg_y; substitute z=g(x,y)z=g(x,y) into F⃗\vec F; then form the scalar integrand −F1gx−F2gy+F3-F_1g_x-F_2g_y+F_3 and integrate it over RR as an ordinary double integral.

Derivation AReducing a flux integral to a double integral over the base region

The goal is to show, starting only from the definition of the surface integral ∬SF⃗⋅n^ dS\iint_S\vec F\cdot\hat{\mathbf n}\,dS and the parametrization of a graph surface, exactly why the vector element collapses to n^ dS=(−gx ı^−gy ȷ^+k^) dx dy\hat{\mathbf n}\,dS=(-g_x\,\hat\imath-g_y\,\hat\jmath+\hat k)\,dx\,dy with no leftover square-root factor, so that the flux formula used throughout MA10002 can be trusted rather than merely memorized.

Step 1. Parametrize the surface. Since SS is the graph z=g(x,y)z=g(x,y), every point on it is r⃗(x,y)=x ı^+y ȷ^+g(x,y) k^\vec r(x,y) = x\,\hat\imath + y\,\hat\jmath + g(x,y)\,\hat k, with (x,y)(x,y) ranging over RR. This turns a surface described implicitly into one described by two free parameters, exactly as a curve is parametrized by one.

Step 2. Build the two tangent vectors. Differentiating r⃗\vec r with respect to xx and yy gives the tangent directions lying in the surface: r⃗x=ı^+gx k^\vec r_x = \hat\imath + g_x\,\hat k and r⃗y=ȷ^+gy k^\vec r_y = \hat\jmath + g_y\,\hat k. Both vectors lie flat in the tangent plane at that point, so any vector perpendicular to both is normal to the surface there.

Step 3. Cross the tangents to get the normal element. The standard result for a surface element is n^ dS=(r⃗x×r⃗y) dx dy\hat{\mathbf n}\,dS = (\vec r_x\times\vec r_y)\,dx\,dy. Computing the cross product component by component, r⃗x×r⃗y=(ı^+gxk^)×(ȷ^+gyk^)=ı^×ȷ^+gy(ı^×k^)+gx(k^×ȷ^)+gxgy(k^×k^)\vec r_x\times\vec r_y = (\hat\imath+g_x\hat k)\times(\hat\jmath+g_y\hat k) = \hat\imath\times\hat\jmath + g_y(\hat\imath\times\hat k) + g_x(\hat k\times\hat\jmath) + g_xg_y(\hat k\times\hat k), and using ı^×ȷ^=k^\hat\imath\times\hat\jmath=\hat k, ı^×k^=−ȷ^\hat\imath\times\hat k=-\hat\jmath, k^×ȷ^=−ı^\hat k\times\hat\jmath=-\hat\imath, k^×k^=0⃗\hat k\times\hat k=\vec 0, this collapses to −gx ı^−gy ȷ^+k^-g_x\,\hat\imath - g_y\,\hat\jmath + \hat k.

Step 4. Read off the result and confirm the orientation. So n^ dS=(−gx ı^−gy ȷ^+k^) dx dy\hat{\mathbf n}\,dS = (-g_x\,\hat\imath-g_y\,\hat\jmath+\hat k)\,dx\,dy, matching the formula used above, and its k^\hat k-component is +1+1, confirming this is the upward-pointing choice (since r⃗x×r⃗y\vec r_x\times\vec r_y was taken in that order). No normalization step was needed because r⃗x×r⃗y\vec r_x\times\vec r_y already packages both the direction and the correct magnitude of the area element into one vector — the length-scaling factor 1+gx2+gy2\sqrt{1+g_x^2+g_y^2} that would appear if you separated n^\hat{\mathbf n} and dSdS individually is exactly what's absorbed by keeping them together. Reversing the order, r⃗y×r⃗x=−(r⃗x×r⃗y)\vec r_y\times\vec r_x = -(\vec r_x\times\vec r_y), produces the downward normal — consistent with the sign-flip rule stated above.

Worked example

Problem. Evaluate ∬SF⃗⋅n^ dS\displaystyle\iint_S \vec F\cdot\hat{\mathbf n}\,dS with the upward normal, where F⃗(x,y,z)=z k^\vec F(x,y,z) = z\,\hat k (units N) and SS is the part of the plane z=4−x−yz=4-x-y lying above the square R: 0≤x≤1, 0≤y≤1R:\ 0\le x\le 1,\ 0\le y\le 1 (metres).

Step 1. Identify gg and differentiate. Here g(x,y)=4−x−yg(x,y)=4-x-y, so gx=−1g_x=-1 and gy=−1g_y=-1.

Step 2. Substitute the surface into F⃗\vec F. The field has F1=0, F2=0, F3=zF_1=0,\ F_2=0,\ F_3=z; replacing zz with g(x,y)=4−x−yg(x,y)=4-x-y gives F3=4−x−yF_3=4-x-y.

Step 3. Form the integrand for the upward normal: −F1gx−F2gy+F3=−0(−1)−0(−1)+(4−x−y)=4−x−y-F_1g_x-F_2g_y+F_3 = -0(-1)-0(-1)+(4-x-y) = 4-x-y.

Step 4. Integrate over RR. The inner integral over xx (with yy fixed) is ∫01(4−x−y) dx=[4x−x22−yx]01=72−y\displaystyle\int_0^1(4-x-y)\,dx = \left[4x-\frac{x^2}2-yx\right]_0^1 = \frac72-y. The outer integral is ∫01(72−y)dy=[72y−y22]01=72−12=3\displaystyle\int_0^1\left(\frac72-y\right)dy = \left[\frac72y-\frac{y^2}2\right]_0^1 = \frac72-\frac12 = 3.

So ∬SF⃗⋅n^ dS=3\displaystyle\iint_S\vec F\cdot\hat{\mathbf n}\,dS = 3 (N·m² of flux), a clean, positive number confirming the field flows upward through the tilted plane.

Competitive stretch

Extending to a curved surface: for F⃗=x ı^+y ȷ^+2z k^\vec F=x\,\hat\imath+y\,\hat\jmath+2z\,\hat k flowing upward through the paraboloid z=4−x2−y2z=4-x^2-y^2 (for z≥0z\ge0), the same four steps give gx=−2x, gy=−2yg_x=-2x,\ g_y=-2y, projection R:x2+y2≤4R:x^2+y^2\le4 (a disc of radius 22), and an integrand −x(−2x)−y(−2y)+2(4−x2−y2)=2x2+2y2+8−2x2−2y2=8-x(-2x)-y(-2y)+2(4-x^2-y^2) = 2x^2+2y^2+8-2x^2-2y^2=8 that collapses to the constant 88. Integrating 88 over the disc of area 4π4\pi gives 8×4π=32π≈100.58\times4\pi=32\pi\approx100.5. The lesson: even a curved surface reduces to the same flat double-integral machinery, and sometimes the algebra simplifies dramatically once zz is substituted in.

Common traps

  • Forgetting the sign convention: using gx,gyg_x,g_y with the wrong sign in −F1gx−F2gy+F3-F_1g_x-F_2g_y+F_3 silently flips the answer — always confirm whether the upward or downward normal is wanted before writing the integrand.
  • Substituting z=g(x,y)z=g(x,y) into F⃗\vec F before differentiating gg for gx,gyg_x,g_y — differentiate g(x,y)g(x,y) first as a function of x,yx,y alone, and only then substitute into the field components.
  • Mixing up the region of integration: RR is the projection of SS onto the xyxy-plane, not SS itself, and its boundary must be read off carefully (e.g. from a z≥0z\ge0 constraint, as in the paraboloid example).
  • Treating dSdS (the scalar area element, used for ∬Sf dS\iint_S f\,dS) and n^ dS\hat{\mathbf n}\,dS (the vector flux element, used for ∬SF⃗⋅n^ dS\iint_S\vec F\cdot\hat{\mathbf n}\,dS) as interchangeable — they belong to different kinds of surface integral and combine with the integrand in different ways.

Ask the tutor: Want to practice setting up the flux integral for a surface oriented downward, or one projecting onto a non-rectangular region — ask the AI tutor to walk through the setup step by step.

Why does hat n dS not need an extra square-root factor for magnitude

Because n^ dS\hat{\mathbf n}\,dS is built as one object, the cross product r⃗x×r⃗y\vec r_x\times\vec r_y, not as a unit vector multiplied separately by a scalar area element. That cross product already has the correct length built into it: if you instead computed the true unit normal n^\hat{\mathbf n} and the scalar dSdS on their own, each would carry a factor of 1+gx2+gy2\sqrt{1+g_x^2+g_y^2}, but those factors are inverses of each other and cancel exactly when you multiply n^\hat{\mathbf n} by dSdS again. Keeping them combined from the start just skips a step that would otherwise cancel anyway.

How do I know which region R to integrate over

RR is always the shadow the surface SS casts on the xyxy-plane, so the fastest way to find it is to ask which (x,y)(x,y) values the surface actually covers. For a surface given directly as z=g(x,y)z=g(x,y) over a stated rectangle or disc, RR is just that stated set. For a surface defined by a condition like z≥0z\ge0 on a paraboloid z=4−x2−y2z=4-x^2-y^2, set the condition on g(x,y)g(x,y) itself — here 4−x2−y2≥04-x^2-y^2\ge0 — and solve; that inequality is exactly the boundary of RR, in this case the disc x2+y2≤4x^2+y^2\le4.

Does this flux formula work for any surface, or only ones written as z equals g of x comma y

This exact formula, with gxg_x and gyg_y, is specific to surfaces presented as a graph over the xyxy-plane, which is the case MA10002 focuses on. A surface parametrized more generally, or one you'd naturally solve for xx in terms of y,zy,z or for yy in terms of x,zx,z, uses the same underlying idea — tangent vectors crossed to give a normal element — but the roles of the variables and the resulting formula change accordingly. The graph-surface technique here is the one version you need to carry into Gauss's and Stokes's theorems at this stage.

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Where this leads