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Surface Integral — Definition, Formula & Examples

A surface integral extends the idea of a definite integral to functions defined over a curved surface in three-dimensional space, rather than along an interval or flat region. It computes quantities like total flux through a surface or the accumulated value of a scalar field across a curved sheet.

Given a parameterized surface r(u,v)\mathbf{r}(u,v) over a region DD in the uvuv-plane, the surface integral of a scalar function ff is ∬Sf dS=∬Df(r(u,v)) ∥ru×rv∥ du dv\iint_S f\,dS = \iint_D f(\mathbf{r}(u,v))\,\|\mathbf{r}_u \times \mathbf{r}_v\|\,du\,dv. For a vector field F\mathbf{F}, the flux integral is ∬SF⋅dS=∬DF(r(u,v))⋅(ru×rv) du dv\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v)\,du\,dv.

Key Formula

∬SF⋅dS=∬DF(r(u,v))⋅(ru×rv) du dv\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v)\,du\,dv
Where:
  • SS = The surface being integrated over
  • F\mathbf{F} = A vector field defined on the surface
  • r(u,v)\mathbf{r}(u,v) = Parameterization of the surface
  • ru,rv\mathbf{r}_u, \mathbf{r}_v = Partial derivatives of the parameterization with respect to u and v
  • DD = The parameter domain in the uv-plane

How It Works

To evaluate a surface integral, you first parameterize the surface using two parameters uu and vv. Then compute the partial derivatives ru\mathbf{r}_u and rv\mathbf{r}_v and take their cross product. For scalar integrals, you multiply the integrand by the magnitude of that cross product; for flux integrals, you dot the vector field with the cross product itself. Finally, integrate over the parameter domain DD as a standard double integral.

Worked Example

Problem: Compute the flux of F=⟨0,0,z⟩\mathbf{F} = \langle 0, 0, z \rangle upward through the surface z=1−x2−y2z = 1 - x^2 - y^2 for x2+y2≤1x^2 + y^2 \le 1.
Step 1: Write the surface as z=g(x,y)=1−x2−y2z = g(x,y) = 1 - x^2 - y^2. For a graph z=g(x,y)z = g(x,y) with upward orientation, the flux formula is:
∬DF⋅⟨−gx, −gy, 1⟩ dA\iint_D \mathbf{F} \cdot \langle -g_x,\, -g_y,\, 1 \rangle\,dA
Step 2: Compute the partials: gx=−2xg_x = -2x, gy=−2yg_y = -2y. The integrand becomes F⋅⟨2x,2y,1⟩=(0)(2x)+(0)(2y)+z(1)=z=1−x2−y2\mathbf{F} \cdot \langle 2x, 2y, 1 \rangle = (0)(2x) + (0)(2y) + z(1) = z = 1 - x^2 - y^2.
∬D(1−x2−y2) dA\iint_D (1 - x^2 - y^2)\,dA
Step 3: Switch to polar coordinates over the unit disk: x2+y2=r2x^2 + y^2 = r^2, dA=r dr dθdA = r\,dr\,d\theta.
∫02π∫01(1−r2) r dr dθ=2π∫01(r−r3) dr=2π[r22−r44]01=2π⋅14=π2\int_0^{2\pi}\int_0^{1} (1 - r^2)\,r\,dr\,d\theta = 2\pi\int_0^1 (r - r^3)\,dr = 2\pi\left[\frac{r^2}{2} - \frac{r^4}{4}\right]_0^1 = 2\pi \cdot \frac{1}{4} = \frac{\pi}{2}
Answer: The upward flux through the paraboloid is π2\dfrac{\pi}{2}.

Why It Matters

Surface integrals are essential in physics and engineering for computing fluid flux, electric flux through Gaussian surfaces, and heat transfer across boundaries. They also appear as key ingredients in the Divergence Theorem and Stokes' Theorem, which are foundational in electromagnetism and fluid dynamics.

Common Mistakes

Mistake: Forgetting the cross-product magnitude ∥ru×rv∥\|\mathbf{r}_u \times \mathbf{r}_v\| when integrating a scalar function, effectively treating dSdS as du dvdu\,dv.
Correction: The factor ∥ru×rv∥\|\mathbf{r}_u \times \mathbf{r}_v\| accounts for how the surface stretches relative to the parameter plane. Always include it for scalar surface integrals.