Calculus

Surface Integrals

Surface Integrals of Vector Fields

Page 1
Problem 1
Page 2
Problems 2-6

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

Relationship cos(alpha)*dS=dydz

Figure 1.

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

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

Solved Problems

Click on problem description to see 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:

\[
{\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} }
= {\int\limits_0^1 {xdx} \int\limits_{\large\frac{\pi }{4}\normalsize}^{\large\frac{\pi }{3}\normalsize} {\sin ydy} }
= {\left[ {\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^1} \right] \cdot \left[ {\left. {\left( { – \cos y} \right)} \right|_{\large\frac{\pi }{4}\normalsize}^{\large\frac{\pi }{3}\normalsize}} \right] }
= {\frac{1}{2}\left( { – \cos \frac{\pi }{3} + \cos \frac{\pi }{4}} \right) }
= {\frac{1}{2}\left( { – \frac{1}{2} + \frac{{\sqrt 2 }}{2}} \right) }
= {\frac{{\sqrt 2 – 1}}{4}.}
\]
Page 1
Problem 1
Page 2
Problems 2-6