Surface integrals of vector fields

1 Oriented surfaces

A surface is called oriented if a side has been chosen. Given an oriented surface \(\Sigma \subset \mathbb {R}^3\), it has a “normal” vector field \(N : \Sigma \to \mathbb {R}^3\) such that for each point \(p \in \Sigma \):

  • \(N (p) \) is perpendicular to \(\Sigma \) at \(p\)
  • \(N(p)\) points in the preferred direction
  • \(\vert N (p) \vert = 1\)

The vector field \(N\) is also called an orientation, or the Gauss map of the surface.

Given an oriented surface \(\Sigma \), we say a parametrization \(\varphi : U \to \Sigma \) is compatible with the orientation if at each point, the normal vector

\[ \frac {\partial \varphi }{\partial u} \times \frac {\partial \varphi }{\partial v} (u,v) \]

points in the same direction as \(N (\varphi (u,v)) \). In that case,

\[ N (\varphi (u,v)) = \dfrac { \frac {\partial \varphi }{\partial u} \times \frac {\partial \varphi }{\partial v} (u,v) }{ \vert \frac {\partial \varphi }{\partial u} \times \frac {\partial \varphi }{\partial v} (u,v) \vert } . \]

Not all surfaces are orientable. For example, a Möbius band has only one side, so it is impossible to “pick one side”.

We say a surface is closed if it is compact and does not have boundary. For example, a sphere, an ellipsoid, a torus, and the surface of a polyhedron are closed surfaces. Note that a closed orientable surface in \(\mathbb {R}^3\) has an interior and an exterior. We say that it is positively oriented if the unit normal points “outwards” and negatively oriented if the unit normal points “inwards”.

2 Surface integrals of vector fields

Note that in that case:

\begin{align*} \iint _{\Sigma } F \cdot \, dS & = \iint _U F( \varphi (u,v) ) \cdot \left ( \frac {\partial \varphi }{\partial u} \times \frac {\partial \varphi }{\partial v} \right ) \,\, dudv \\ & = \iint _U F( \varphi (u,v) ) \cdot \left ( \dfrac {\frac {\partial \varphi }{\partial u} \times \frac {\partial \varphi }{\partial v} }{ \vert \frac {\partial \varphi }{\partial u} \times \frac {\partial \varphi }{\partial v} \vert } \right ) \vert \frac {\partial \varphi }{\partial u} \times \frac {\partial \varphi }{\partial v} \vert \,\, dudv \\ & = \iint _U F ( \varphi (u,v) ) \cdot N ( \varphi (u,v) ) \, \vert \frac {\partial \varphi }{\partial u} \times \frac {\partial \varphi }{\partial v} \vert \,\, dudv \\ & = \iint _{\Sigma } ( F \cdot N ) \, dS , \end{align*}

Therefore,

\[ \iint _{\Sigma } F \cdot \, dS = \iint _{\Sigma } ( F \cdot N ) \,\, dS \]

meaning that the integral of the vector field \(F\) is the same as the integral of the scalar function \(F \cdot N\).

Note: if you use a parametrization not compatible with the orientation, the integral changes sign.

These are great to model:

  • Flow of a fluid across an imaginary boundary.
  • Flow of sodium/polasium/glucose/oxygen through the membrane of a cell.
  • Electric flux in electromagnetism.
  • Integral of \(\text {Curl}(F)\) measures amount of rotation of objects over \(\Sigma \).
Let \(\Sigma _1\) be the square with vertices \((0,0,0)\), \((1,0,0)\), \((1,1,0)\), \((0,1,0)\), \(\Sigma _2 \) the square with vertices \((0,0,0)\), \((1,0,0)\), \((1,0,1)\), \((0,0,1)\), and \(\Sigma _3 \) the square with vertices \((0,0,0)\), \((0,1,0)\), \((0,1,1)\), \((0,0,1)\), all oriented towards the first octant. In other words,
  • \(\Sigma _1\) is the unit square in the \(xy\)-plane.
  • \(\Sigma _2\) is the unit square in the \(xz\)-plane.
  • \(\Sigma _3\) is the unit square in the \(yz\)-plane.

Let \(F = \langle 1,0,0 \rangle \). Then

\begin{gather*} \iint _{\Sigma _1} F \cdot dS = \iint _{\Sigma _2} F \cdot dS = 0 \\ \iint _{\Sigma _3} F \cdot dS = 1 \end{gather*}
This is because:
  • The unit normal of \(\Sigma _1 \) is \(\langle 0,0,1 \rangle \).
  • The unit normal of \(\Sigma _2 \) is \(\langle 0,1,0 \rangle \).
  • The unit normal of \(\Sigma _3 \) is \(\langle 1,0,0 \rangle \).

When we take the dot product with \(F\), only the thid one doesn’t cancel. In that case, we get

\begin{align*} \iint _{\Sigma _3 } F \cdot dS & = \iint _{\Sigma _3} ( F \cdot N ) \, dS \\ & = \iint _{\Sigma _3 } 1 \, dS \\ & = Area (\Sigma _3) \\ & = 1 \end{align*}
Let \(\Sigma \) be the portion of the plane \(x - y + z = 4\) inside the cylinder \(x^2 + y ^2 = 9 \), oriented upwards. Compute
\[ \iint _{\Sigma } \langle 3 + z , 2 x^2 , xy - 1 \rangle \cdot dS \]

We use the parametrization \(\varphi : D \to \Sigma \), where \(D \subset \mathbb {R}^2\) is the interior of the circle \( x^2 + y ^2 = 9 \) and \(\varphi \) is given by

\[ \varphi (u,v) = (u,v , 4 - u + v ) \]

Then we compute

\begin{align*} \frac {\partial \varphi }{\partial u } & = \langle 1, 0, -1 \rangle \\ \frac {\partial \varphi }{\partial v } & = \langle 0, 1, 1 \rangle \\ \frac {\partial \varphi }{\partial u } \times \frac {\partial \varphi }{\partial v } & = \langle 1, -1 ,1 \rangle \end{align*}

The cross product points “up” because the third component is positive, so the parametrization is compatible with the orientation. Setting \(F(x,y,z) : = \langle 3 + z , 2x^2, xy - 1 \rangle \), we get

\begin{align*} F (\varphi (u,v)) \cdot \frac {\partial \varphi }{\partial u } \times \frac {\partial \varphi }{\partial v } & = \langle 3 + (4 - u + v) , 2 u^2 , uv - 1 \rangle \cdot \langle 1, -1 ,1 \rangle \\ & = 6 - u + v - 2 u ^2 + uv \end{align*}

Then we compute, using polar coordinates,

\begin{align*} \iint _{\Sigma } F \cdot dS & = \iint _{D} [ 6 - u + v - 2 u ^2 + uv] \, dudv \\ & = \int _0 ^3 \int _0 ^{2\pi } r [ 6 - r \cos \theta + r \sin \theta -2 r^2 \cos ^2 \theta + r^2 \sin \theta \cos \theta ] \, d\theta dr\\ & = \int _0 ^3 [ 12 \pi r - 2 \pi r^3 ] \, dr\\ & = [ 6 \cdot 3^2 - \frac {1}{2} \cdot 3 ^4 ] \pi \\ & = \frac {27}{2} \pi \end{align*}
Let \(\Sigma \) be the triangle with vertices \((3,0,0)\), \((0,6,0)\), \((0,0,5)\), oriented downwards. Compute
\[ \iint _{\Sigma } \langle 3z , y, 2 \rangle \cdot dS \]

Note that \(\Sigma \) is part of the plane \(10x + 5y + 6z = 30\). We use the parametrization \(\varphi : U \to \Sigma \) with \( U \subset \mathbb {R}^2 \) the triangle with vertices \((0,0)\), \((3,0)\), \((0,6)\), and

\[ \varphi ( u ,v ) : = \left ( u,v, 5 - \frac {5}{3} x - \frac {5}{6} y \right ) . \]

Using this change of variables,

\[ \langle 3z , y , 2 \rangle = \langle 15 - 5u - \frac {5}{2} v , v, 2 \rangle \]

Also we compute the Jacobian

\begin{align*} \frac {\partial \varphi }{\partial u } & = \langle 1,0, - \frac {5}{3} \rangle \\ \frac {\partial \varphi }{\partial v } & = \langle 0,1, - \frac {5}{6} \rangle \\ \frac {\partial \varphi }{\partial u } \times \frac {\partial \varphi }{\partial v } & = \langle \frac {5}{3}, \frac {5}{6} , 1 \rangle \end{align*}

Since \(\frac {\partial \varphi }{\partial u } \times \frac {\partial \varphi }{\partial v }\) points upwards, it is NOT compatible with the orientation, so we put a minus sign! Then

\begin{align*} \iint _{\Sigma } \langle 3z, y, 2 \rangle \cdot dS & = -\iint _D \langle 15 - 5u - \frac {5}{2} v , v, 2 \rangle \cdot \langle \frac {5}{3}, \frac {5}{6} , 1 \rangle \, dvdu \\ & =- \int _0^3 \int _0 ^{6 - 2u} [ 25 - \frac {25}{3}u - \frac {25}{6} v + \frac {5}{6} v + 2 ] \, dudv \\ & = - \int _0 ^3 [ 27 (6 - 2u ) - \frac {25}{3} u (6 - 2u) - \frac {5}{3} (6 - 2u)^2 ] \, du \\ & = - \int _0 ^3 [ 162 -54 u - 50 u + \frac {50}{3}u^2 - 60 + 40u - \frac {20}{3} u ^2 ] \, du \\ & = -\int _0^3 [ 102 -64 u + 10 u ^2 ] \, du \\ & = -[ 306 - 32 \cdot 3^2 + 10 \cdot 3^2 ]\\ & = -108 \end{align*}