# Calculus

## Surface Integrals # 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:

${\mathbf{n}\left( {x,y,z} \right)\;\;\text{or}\;\;\;}\kern-0.3pt{- \mathbf{n}\left( {x,y,z} \right).}$

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
${\iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot d\mathbf{S}} } = {\iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot \mathbf{n}dS} \text{ = }}\kern0pt {\iint\limits_{D\left( {u,v} \right)} {\mathbf{F}\left( {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right) \cdot}\kern0pt{ \left[ {\frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}}} \right]dudv} ;}$
• If the surface $$S$$ is oriented inward, then
${\iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot d\mathbf{S}} } = {\iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot \mathbf{n}dS} \text{ = }}\kern0pt {\iint\limits_{D\left( {u,v} \right)} {\mathbf{F}\left( {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right) \cdot}\kern0pt{ \left[ {\frac{{\partial \mathbf{r}}}{{\partial v}} \times \frac{{\partial \mathbf{r}}}{{\partial u}}} \right]dudv}.}$

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
${\iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot d\mathbf{S}} } = {\iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot \mathbf{n}dS} } = {\iint\limits_{D\left( {x,y} \right)} {\mathbf{F}\left( {x,y,z} \right) \cdot}\kern0pt{ \left( { – \frac{{\partial z}}{{\partial x}}\mathbf{i} – \frac{{\partial z}}{{\partial y}}\mathbf{j} + \mathbf{k}} \right)dxdy} ;}$
• If the surface $$S$$ is oriented downward, i.e. the $$k$$th component of the normal vector is negative, then
${\iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot d\mathbf{S}} } = {\iint\limits_S {\mathbf{F}\left( {x,y,z} \right) \cdot \mathbf{n}dS} } = {\iint\limits_{D\left( {x,y} \right)} {\mathbf{F}\left( {x,y,z} \right) \cdot}\kern0pt{ \left( { \frac{{\partial z}}{{\partial x}}\mathbf{i} + \frac{{\partial z}}{{\partial y}}\mathbf{j} – \mathbf{k}} \right)dxdy}.}$

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

${\mathbf{F} \cdot \mathbf{n} } = {\mathbf{F}\left( {P,Q,R} \right) \cdot}\kern0pt{ \mathbf{n}\left( {\cos \alpha ,\cos \beta ,\cos \gamma } \right) } = {P\cos \alpha + Q\cos \beta + R\cos \gamma .}$

Consequently, the surface integral can be written as

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

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:

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

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

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

### Example 3

Evaluate the flux of the vector field $$\mathbf{F}$$ $$= y \cdot \mathbf{i} – x \cdot \mathbf{j}$$ $$+ z \cdot \mathbf{k}$$ through the conic surface $$z$$ $$= \sqrt {{x^2} + {y^2}} ,$$ $$0 \le z \le 2,$$ oriented upwards.

### Example 4

Evaluate the flux of the vector field $$\mathbf{F}\left( {x,y,z} \right)$$ $$= – y \cdot \mathbf{i} + x \cdot \mathbf{j}$$ $$- z \cdot \mathbf{k}$$ through the unit sphere $${x^2} + {y^2} + {z^2}$$ $$= 1$$ that has downward orientation.

### Example 5

Evaluate the surface integral $$\iint\limits_S {{\large\frac{{dydz}}{x}\normalsize} + {\large\frac{{dzdx}}{y}\normalsize} }$$ $$+{ {\large\frac{{dxdy}}{z}\normalsize}} ,$$ where the surface $$S$$ is the part of the ellipsoid parameterized by $$\mathbf{r}\left( {u,v} \right) =$$ $$\big( {a\cos u\cos v,}$$ $${b\sin u\cos v,}$$ $${c\sin v} \big)$$ that has upward orientation. The parameters $$u, v$$ range over $$0 \le u \le 1,$$ $$0 \le v \le {\large\frac{\pi }{2}\normalsize}.$$

### Example 6

Find the surface integral $$\iint\limits_S {2xdydz},$$ where $$S$$ is the surface of the sphere $${x^2} + {y^2} + {z^2}$$ $$= {a^2}$$ oriented downwards.

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

Solution.

We apply the formula

${\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} \text{ = }}\kern0pt {\iint\limits_{D\left( {x,y} \right)} {\mathbf{F} \cdot}\kern0pt{ \left( {\frac{{\partial z}}{{\partial x}}\mathbf{i} + \frac{{\partial z}}{{\partial y}}\mathbf{j} – \mathbf{k}} \right)dxdy} }$

Since

${{\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {x\cos y} \right) }={ \cos y,\;\;\;}}\kern0pt {{\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {x\cos y} \right) }={ – x\sin y,}}$

the flux of the vector field can be written as

$\require{cancel} {\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} \text{ = }}\kern0pt {\iint\limits_{D\left( {x,y} \right)} {\left[ {x \cdot \cos y }\right.}\kern0pt{+ \left.{ \left( { – 1} \right) \cdot \left( { – x\sin y} \right) }\right.}\kern0pt{+ \left.{ z \cdot \left( { – 1} \right)} \right]dxdy} } = {{\iint\limits_{D\left( {x,y} \right)} {\left( {\cancel{x\cos y} }\right.}+{\left.{ x\sin y }\right.}-{\left.{ \cancel{x\cos y}} \right)dxdy} }} = {\iint\limits_{D\left( {x,y} \right)} {x\sin ydxdy} .}$

After simple transformations we find the answer:

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

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