# Surface Integrals of Vector Fields

We consider a vector field $$\mathbf{F}\left( {x,y,z} \right)$$ and a surface $$S,$$ which is defined by the position vector

$\mathbf{r}\left( {u,v} \right) = x\left( {u,v} \right) \cdot \mathbf{i} + y\left( {u,v} \right) \cdot \mathbf{j} + z\left( {u,v} \right) \cdot \mathbf{k}.$

Suppose that the functions $$x\left( {u,v} \right),$$ $$y\left( {u,v} \right),$$ $$z\left( {u,v} \right)$$ are continuously differentiable in some domain $$D\left( {u,v} \right)$$ and the rank of the matrix

$\left[ {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial z}}{{\partial u}}}\\ {\frac{{\partial x}}{{\partial v}}}&{\frac{{\partial y}}{{\partial v}}}&{\frac{{\partial z}}{{\partial v}}} \end{array}} \right]$

is equal to $$2.$$

We denote by $$\mathbf{n}\left( {x,y,z} \right)$$ a unit normal vector to the surface $$S$$ at the point $$\left( {x,y,z} \right).$$ If the surface $$S$$ is smooth and the vector function $$\mathbf{n}\left( {x,y,z} \right)$$ is continuous, there are only two possible choices for the unit normal vector:

$\mathbf{n}\left( {x,y,z} \right)\;\;\text{or}\;\; - \mathbf{n}\left( {x,y,z} \right).$

If the choice of the vector is done, the surface $$S$$ is called oriented.

If $$S$$ is a closed surface, by convention, we choose the normal vector to point outward from the surface.

The surface integral of the vector field $$\mathbf{F}$$ over the oriented surface $$S$$ (or the flux of the vector field $$\mathbf{F}$$ across the surface $$S$$) can be written in one of the following forms:

• If the surface $$S$$ is oriented outward, then
$\iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot d\mathbf{S}} = \iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot \mathbf{n}dS} = \iint\limits_{D\left( {u,v} \right)} {\mathbf{F}\left( {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right) \left[ {\frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}}} \right]dudv} ;$
• If the surface $$S$$ is oriented inward, then
$\iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot d\mathbf{S}} = \iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot \mathbf{n}dS} = \iint\limits_{D\left( {u,v} \right)} {\mathbf{F}\left( {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right) \left[ {\frac{{\partial \mathbf{r}}}{{\partial v}} \times \frac{{\partial \mathbf{r}}}{{\partial u}}} \right]dudv}.$

Here $$d\mathbf{S} = \mathbf{n}dS$$ is called the vector element of the surface. Dot means the scalar product of the appropriate vectors. The partial derivatives in the formulas are calculated in the following way:

$\frac{{\partial \mathbf{r}}}{{\partial u}} = \frac{{\partial x}}{{\partial u}}\left( {u,v} \right) \cdot \mathbf{i} + \frac{{\partial y}}{{\partial u}}\left( {u,v} \right) \cdot \mathbf{j} + \frac{{\partial z}}{{\partial u}}\left( {u,v} \right) \cdot \mathbf{k},$
$\frac{{\partial \mathbf{r}}}{{\partial v}} = \frac{{\partial x}}{{\partial v}}\left( {u,v} \right) \cdot \mathbf{i} + \frac{{\partial y}}{{\partial v}}\left( {u,v} \right) \cdot \mathbf{j} + \frac{{\partial z}}{{\partial v}}\left( {u,v} \right) \cdot \mathbf{k}.$

If the surface $$S$$ is given explicitly by the equation $$z = z\left( {x,y} \right),$$ where $$z\left( {x,y} \right)$$ is a differentiable function in the domain $$D\left( {x,y} \right),$$ then the surface integral of the vector field $$\mathbf{F}$$ over the surface $$S$$ is defined in one of the following forms:

• If the surface $$S$$ is oriented upward, i.e. the $$k$$th component of the normal vector is positive, then
$\iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot d\mathbf{S}} = \iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot \mathbf{n}dS} = \iint\limits_{D\left( {x,y} \right)} {\mathbf{F}\left( {x,y,z} \right) \left( { - \frac{{\partial z}}{{\partial x}}\mathbf{i} - \frac{{\partial z}}{{\partial y}}\mathbf{j} + \mathbf{k}} \right)dxdy} ;$
• If the surface $$S$$ is oriented downward, i.e. the $$k$$th component of the normal vector is negative, then
$\iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot d\mathbf{S}} = \iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot \mathbf{n}dS} = \iint\limits_{D\left( {x,y} \right)} {\mathbf{F}\left( {x,y,z} \right) \left( { \frac{{\partial z}}{{\partial x}}\mathbf{i} + \frac{{\partial z}}{{\partial y}}\mathbf{j} - \mathbf{k}} \right)dxdy}.$

We can also write the surface integral of vector fields in the coordinate form.

Let $$P\left( {x,y,z} \right),$$ $$Q\left( {x,y,z} \right),$$ $$R\left( {x,y,z} \right)$$ be the components of the vector field $$\mathbf{F}.$$ Suppose that $$\cos \alpha,$$ $$\cos \beta,$$ $$\cos \gamma$$ are the angles between the outer unit normal vector $$\mathbf{n}$$ and the $$x$$-axis, $$y$$-axis, and $$z$$-axis, respectively. Then the scalar product $$\mathbf{F} \cdot \mathbf{n}$$ is

$\mathbf{F} \cdot \mathbf{n} = \mathbf{F}\left( {P,Q,R} \right) \mathbf{n}\left( {\cos \alpha ,\cos \beta ,\cos \gamma } \right) = P\cos \alpha + Q\cos \beta + R\cos \gamma .$

Consequently, the surface integral can be written as

$\iint\limits_S {\left( {\mathbf{F} \cdot \mathbf{n}} \right)dS} = \iint\limits_S {\left( {P\cos \alpha + Q\cos \beta + R\cos \gamma } \right)dS} .$

As $$\cos \alpha \cdot dS = dydz$$ (Figure $$1$$), and, similarly, $$\cos \beta \cdot dS = dzdx,$$ $$\cos \gamma \cdot dS = dxdy,$$ we obtain the following formula for calculating the surface integral:

$\iint\limits_S {\left( {\mathbf{F} \cdot \mathbf{n}} \right)dS} = \iint\limits_S {\left( {P\cos \alpha + Q\cos \beta + R\cos \gamma } \right)dS} = \iint\limits_S {Pdydz + Qdzdx + Rdxdy}.$

If the surface $$S$$ is given in parametric form by the vector $$\mathbf{r}\big( {x\left( {u,v} \right),y\left( {u,v} \right),}$$ $${z\left( {u,v} \right)} \big),$$ the latter formula can be written as

$\iint\limits_S {\left( {\mathbf{F} \cdot \mathbf{n}} \right)dS} = \iint\limits_S {Pdydz + Qdzdx + Rdxdy} = \iint\limits_{D\left( {u,v} \right)} {\left| {\begin{array}{*{20}{c}} P&Q&R\\ {\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial z}}{{\partial u}}}\\ {\frac{{\partial x}}{{\partial v}}}&{\frac{{\partial y}}{{\partial v}}}&{\frac{{\partial z}}{{\partial v}}} \end{array}} \right|dudv} ,$

where the coordinates $$\left( {u,v} \right)$$ range over some domain $$D\left( {u,v} \right).$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Evaluate the flux of the vector field $\mathbf{F}\left( {x,y,z} \right) = \left( {x, - 1,z} \right)$ across the surface $$S$$ that has downward orientation and is given by the equation

$z = x\cos y, 0 \le x \le 1, \frac{\pi }{4} \le y \le \frac{\pi }{3}.$

### Example 2

Find the flux of the vector field $\mathbf{F}\left( {x,y,z} \right) = \left( {y,x,z} \right)$ through the surface $$S,$$ parameterized by the vector

$\mathbf{r}\left( {u,v} \right) = \left( {\cos v,\sin v,u} \right), 0 \le u \le 2, \frac{\pi }{2} \le v \le \pi .$

### Example 1.

Evaluate the flux of the vector field $\mathbf{F}\left( {x,y,z} \right) = \left( {x, - 1,z} \right)$ across the surface $$S$$ that has downward orientation and is given by the equation

$z = x\cos y, 0 \le x \le 1, \frac{\pi }{4} \le y \le \frac{\pi }{3}.$

Solution.

We apply the formula

$\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \iint\limits_{D\left( {x,y} \right)} {\mathbf{F} \left( {\frac{{\partial z}}{{\partial x}}\mathbf{i} + \frac{{\partial z}}{{\partial y}}\mathbf{j} - \mathbf{k}} \right)dxdy}.$

Since

$\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {x\cos y} \right) = \cos y,\;\;\; \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {x\cos y} \right) = - x\sin y,$

the flux of the vector field can be written as

$\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \iint\limits_{D\left( {x,y} \right)} {\left[ {x \cdot \cos y + \left( { - 1} \right) \cdot \left( { - x\sin y} \right) + z \cdot \left( { - 1} \right)} \right]dxdy} = \iint\limits_{D\left( {x,y} \right)} {\left( {\cancel{x\cos y} + x\sin y - \cancel{x\cos y}} \right)dxdy} = \iint\limits_{D\left( {x,y} \right)} {x\sin ydxdy} .$

After some algebra we find the answer:

$\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \int\limits_0^1 {xdx} \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\sin ydy} = \left[ {\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^1} \right] \cdot \left[ {\left. {\left( { - \cos y} \right)} \right|_{\frac{\pi }{4}}^{\frac{\pi }{3}}} \right] = \frac{1}{2}\left( { - \cos \frac{\pi }{3} + \cos \frac{\pi }{4}} \right) = \frac{1}{2}\left( { - \frac{1}{2} + \frac{{\sqrt 2 }}{2}} \right) = \frac{{\sqrt 2 - 1}}{4}.$

### Example 2.

Find the flux of the vector field $\mathbf{F}\left( {x,y,z} \right) = \left( {y,x,z} \right)$ through the surface $$S,$$ parameterized by the vector

$\mathbf{r}\left( {u,v} \right) = \left( {\cos v,\sin v,u} \right), 0 \le u \le 2, \frac{\pi }{2} \le v \le \pi .$

Solution.

First we calculate the partial derivatives:

$\frac{{\partial \mathbf{r}}}{{\partial u}} = \left( {0,0,1} \right),\;\; \frac{{\partial \mathbf{r}}}{{\partial v}} = \left( { - \sin v,\cos v,0} \right).$

It follows that

$\frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0&0&1\\ { - \sin v}&{\cos v}&0 \end{array}} \right| = - \cos v \cdot \mathbf{i} - \sin v \cdot \mathbf{j}.$

Hence, the vector area element is

$d\mathbf{S} = \left[ {\frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}}} \right]dudv = \left( { - \cos v, - \sin v,0} \right)dudv.$

As $$x = \cos v,$$ $$y = \sin v$$ and $$z = v,$$ the vector field $$\mathbf{F}$$ can be represented in the following form:

$\mathbf{F}\left( {r,u,v} \right) = \left( {\sin v,\cos v,u} \right).$

Then the original surface integral across $$S$$ is

$\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \iint\limits_{D\left( {u,v} \right)} {\left[ {\sin v \cdot \left( { - \cos v} \right) + \cos v \cdot \left( { - \sin v} \right) + 0} \right]dudv} = \iint\limits_{D\left( {u,v} \right)} {\left( { - 2\sin v\cos v} \right)dudv} = - \int\limits_0^2 {du} \int\limits_{\frac{\pi }{2}}^\pi {\sin 2vdv} = - 2 \cdot \left[ {\left. {\left( { - \frac{{\cos 2v}}{2}} \right)} \right|_{\frac{\pi }{2}}^\pi } \right] = \cos 2\pi - \cos \pi = 2.$