Calculus

Surface Integrals

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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:

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

    Relationship cos(alpha)*dS=dydz
    Figure 1.

    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 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:

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

    Page 1
    Problem 1
    Page 2
    Problems 2-6