Surface Integrals of Vector Fields

Solved Problems

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.

Solution.

The surface of the cone is given by the vector \(\mathbf{r}:\)

\[\mathbf{r}\left( {x,y} \right) = x \cdot \mathbf{i} + y \cdot \mathbf{j} + \sqrt {{x^2} + {y^2}} \cdot \mathbf{k}.\]

The domain of integration \(D\left( {x,y} \right)\) is the circle defined by the equation \({{x^2} + {y^2}} \le 4.\)

Find the vector area element \(d\mathbf{S}\) normal to the surface and pointing upwards. The partial derivatives are

\[\frac{{\partial \mathbf{r}}}{{\partial x}} = 1 \cdot \mathbf{i} + \frac{x}{{\sqrt {{x^2} + {y^2}} }} \cdot \mathbf{k},\;\;\; \frac{{\partial \mathbf{r}}}{{\partial y}} = 1 \cdot \mathbf{j} + \frac{y}{{\sqrt {{x^2} + {y^2}} }} \cdot \mathbf{k}.\]

Then

\[ \frac{{\partial \mathbf{r}}}{{\partial x}} \times \frac{{\partial \mathbf{r}}}{{\partial y}} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 0 & {\frac{x}{{\sqrt {{x^2} + {y^2}} }}}\\ 0 & 1 & {\frac{y}{{\sqrt {{x^2} + {y^2}} }}} \end{array}} \right| = - \frac{x}{{\sqrt {{x^2} + {y^2}} }} \cdot \mathbf{i} - \frac{y}{{\sqrt {{x^2} + {y^2}} }} \cdot \mathbf{j} + \mathbf{k},\]

so that the vector area element is

\[d\mathbf{S} = \left( { - \frac{x}{{\sqrt {{x^2} + {y^2}} }}, - \frac{y}{{\sqrt {{x^2} + {y^2}} }},1} \right)dxdy.\]

The vector field \(\mathbf{F}\) on the surface of the cone is given by

\[\mathbf{F}\left( {x,y,z} \right) = y \cdot \mathbf{i} - x \cdot \mathbf{j} + \sqrt {{x^2} + {y^2}} \cdot \mathbf{k}.\]

Hence, the flux of the vector field through \(S\) (or, in other words, the surface integral of the vector field) is

\[I = \iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \iint\limits_{D\left( {x,y} \right)} {\Big[ {y \cdot \Big( { - \frac{x}{{\sqrt {{x^2} + {y^2}} }}} \Big) }} + {{ \left( { - x} \right) \cdot \Big( { - \frac{y}{{\sqrt {{x^2} + {y^2}} }}} \Big) }} + {{ \sqrt {{x^2} + {y^2}} \cdot 1} \Big]dxdy} = \iint\limits_{D\left( {x,y} \right)} {\sqrt {{x^2} + {y^2}} dxdy} .\]

By changing to polar coordinates, we have

\[I = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^2 {{r^2}dr} = 2\pi \left[ {\left. {\left( {\frac{{{r^3}}}{3}} \right)} \right|_0^2} \right] = \frac{{16\pi }}{3}.\]

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.

Solution.

The following is the spherical coordinate parametrization of the unit sphere:

\[\mathbf{r}\left( {\psi ,\theta } \right) = \cos \psi \sin \theta \cdot \mathbf{i} + \sin \psi \sin \theta \cdot \mathbf{j} + \cos \theta \cdot \mathbf{k},\]

where \(0 \le \psi \le 2\pi ,\) \(0 \le \theta \le \pi.\) As a result, the vector \(\mathbf{F}\) over the given surface can be written as

\[\mathbf{F}\left( {r,\psi ,\theta } \right) = - \sin \psi \sin \theta \cdot \mathbf{i} + \cos \psi \sin \theta \cdot \mathbf{j} - \cos \theta \cdot \mathbf{k},\]

Calculate the vector area element \(d\mathbf{S}.\) The partial derivatives are

\[\frac{{\partial \mathbf{r}}}{{\partial \psi }} = - \sin \psi \sin \theta \cdot \mathbf{i} + \cos \psi \sin \theta \cdot \mathbf{j} + 0 \cdot \mathbf{k},\]

\[\frac{{\partial \mathbf{r}}}{{\partial \theta }} = \cos \psi \sin \theta \cdot \mathbf{i} + \sin \psi \cos \theta \cdot \mathbf{j} - \sin \theta \cdot \mathbf{k}.\]

Hence,

\[ \frac{{\partial \mathbf{r}}}{{\partial \psi }} \times \frac{{\partial \mathbf{r}}}{{\partial \theta }} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ { - \sin \psi \sin \theta } & {\cos \psi \sin \theta } & 0\\ {\cos \psi \cos \theta } & {\sin \psi \cos \theta } & { - \sin \theta } \end{array}} \right| = - \cos \psi \,{\sin ^2}\theta \cdot \mathbf{i} - \sin \psi \,{\sin ^2}\theta \cdot \mathbf{j} - \left( {{{\sin }^2}\psi \sin \theta \cos \theta + {{\cos }^2}\psi \sin \theta \cos \theta } \right) \cdot \mathbf{k} = - \cos \psi \,{\sin ^2}\theta \cdot \mathbf{i} - \sin \psi \,{\sin ^2}\theta \cdot \mathbf{j} - \sin \theta \cos \theta \cdot \mathbf{k}.\]

Thus, we have

\[d\mathbf{S} = \left( { - \cos \psi \,{{\sin }^2}\theta , - \sin \psi \,{{\sin }^2}\theta , - \sin\theta \cos \theta } \right)d\psi d\theta .\]

(This vector is oriented downwards.)

Then the surface integral (the flux of the vector field) is

\[ \iint\limits_S {\mathbf{F}\left( {r,\psi ,\theta } \right) \cdot d\mathbf{S}} = {{\iint\limits_{D\left( {\psi ,\theta } \right)} {\left[ {\left( { - \sin \psi \sin \theta } \right) \cdot}\kern0pt{ \left( { - \cos \psi \,{{\sin }^2}\theta } \right)} \right.} } + {\cos \psi \sin \theta \cdot}\kern0pt{ \left( { - \sin \psi \,{{\sin }^2}\theta } \right) }} + {{\left. {\left( { - \cos \theta } \right) \cdot}\kern0pt{ \left( { - \cos \theta \sin \theta } \right)} \right]d\psi d\theta }} = \int\limits_{D\left( {\psi ,\theta } \right)} {\left[ {\sin \psi \cos \psi \,{{\sin }^3}\theta - \sin \psi \cos \psi \,{{\sin }^3}\theta } \right.} \left. { + \sin \theta \,{{\cos }^2}\theta } \right]d\psi d\theta = \int\limits_{D\left( {\psi ,\theta } \right)} {\sin \theta \,{{\cos }^2}\theta d\psi d\theta } = \int\limits_0^{2\pi } {d\psi } \int\limits_0^\pi {\sin \theta \,{{\cos }^2}\theta d\theta } = - 2\pi \int\limits_0^\pi {{{\cos }^2}\theta d\left( {\cos \theta } \right)} = - 2\pi \left[ {\left. {\left( {\frac{{{{\cos }^3}\theta }}{3}} \right)} \right|_0^\pi } \right] = - \frac{{2\pi }}{3}\left( {{{\cos }^3}\pi - {{\cos }^3}0} \right) = \frac{{4\pi }}{3}.\]

Example 5.

Evaluate the surface integral \[\iint\limits_S {{\frac{{dydz}}{x} + \frac{{dzdx}}{y} + \frac{{dxdy}}{z}\normalsize}},\] 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 \(0 \le u \le 1,\) \(0 \le v \le {\frac{\pi }{2}}.\)

Solution.

We use the formula

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

Since

\[\frac{{\partial x}}{{\partial u}} = \frac{\partial }{{\partial u}}\left( {a\cos u\cos v} \right) = - a\sin u\cos v,\]

\[\frac{{\partial y}}{{\partial u}} = \frac{\partial }{{\partial u}}\left( {b\sin u\cos v} \right) = b\cos u\cos v,\]

\[\frac{{\partial z}}{{\partial u}} = \frac{\partial }{{\partial u}}\left( {c\sin v} \right) = 0,\]

\[\frac{{\partial x}}{{\partial v}} = \frac{\partial }{{\partial v}}\left( {a\cos u\cos v} \right) = - a\cos u\sin v,\]

\[\frac{{\partial y}}{{\partial v}} = \frac{\partial }{{\partial v}}\left( {b\sin u\cos v} \right) = - b\sin u\sin v,\]

\[\frac{{\partial z}}{{\partial v}} = \frac{\partial }{{\partial v}}\left( {c\sin v} \right) = c\cos v,\]

the determinant can be written as

\[ \left| {\begin{array}{*{20}{c}} {\frac{1}{x}}&{\frac{1}{y}}&{\frac{1}{z}}\\ {\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial z}}{{\partial u}}}\\ {\frac{{\partial x}}{{\partial v}}}&{\frac{{\partial y}}{{\partial v}}}&{\frac{{\partial z}}{{\partial v}}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {\frac{1}{{a\cos u\cos v}}}&{\frac{1}{{b\sin u\cos v}}}&{\frac{1}{{c\sin v}}}\\ {\text{-}{\small{a\sin u\cos v}\normalsize}}&{\small{b\cos u\cos v}\normalsize}&{\small{0}\normalsize}\\ {\text{-}{\small{a\cos u\sin v}\normalsize}}&{\text{-}{\small{b\sin u\sin v}\normalsize}}&{\small{c\cos v}\normalsize} \end{array}} \right| = \frac{1}{{a\cos u\cos v}} \cdot b\cos u\cos v \cdot c\cos v + \frac{1}{{b\sin u\cos v}} \cdot a\sin u\cos v \cdot c\cos v + \frac{1}{{c\sin v}}\left( {a\sin u\cos v \cdot b\sin u\sin v} \right. + \left. {a\cos u\sin v \cdot b\cos u\cos v} \right) = \frac{{bc}}{a}\cos v + \frac{{ac}}{b}\cos v + \frac{{ab}}{c}\left( {{{\sin }^2}u\cos v + {{\cos }^2}u\cos v} \right) = \left( {\frac{{ab}}{c} + \frac{{ac}}{b} + \frac{{bc}}{a}} \right)\cos v.\]

Then the surface integral is

\[I = \iint\limits_{D\left( {u,v} \right)} {\left( {\frac{{ab}}{c} + \frac{{ac}}{b} + \frac{{bc}}{a}} \right)\cos vdudv} = \left( {\frac{{ab}}{c} + \frac{{ac}}{b} + \frac{{bc}}{a}} \right) \int\limits_0^1 {du} \int\limits_0^{\frac{\pi }{2}} {\cos vdv} = \left( {\frac{{ab}}{c} + \frac{{ac}}{b} + \frac{{bc}}{a}} \right) \left[ {\left. {\left( {\sin v} \right)} \right|_0^{\frac{\pi }{2}}} \right] = \frac{{ab}}{c} + \frac{{ac}}{b} + \frac{{bc}}{a}.\]

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.

Solution.

We identify that the components of the vector field \(\mathbf{F}\) are

\[\mathbf{F}\left( {P,Q,R} \right) = \left( {2x,0,0} \right).\]

It is convenient to convert the equation of the sphere to spherical coordinates. Then

\[\mathbf{r}\left( {\psi ,\theta } \right) = a\cos \psi \sin \theta \cdot \mathbf{i} + a\sin \psi \sin \theta \cdot \mathbf{j} + a\cos \theta \cdot \mathbf{k},\]

where \(0 \le \psi \le 2\pi ,\) \(0 \le \theta \le \pi.\) Apply the formula

\[ \iint\limits_S {Pdydz + Qdzdx + Rdxdy} = \iint\limits_{D\left( {\psi ,\theta } \right)} {\left| {\begin{array}{*{20}{c}} P & Q & R\\ {\frac{{\partial x}}{{\partial \psi}}}&{\frac{{\partial y}}{{\partial \psi}}}&{\frac{{\partial z}}{{\partial \psi}}}\\ {\frac{{\partial x}}{{\partial \theta}}}&{\frac{{\partial y}}{{\partial \theta}}}&{\frac{{\partial z}}{{\partial \theta}}} \end{array}} \right|d\psi d\theta} .\]

Since

\[\frac{{\partial x}}{{\partial \psi }} = \frac{\partial }{{\partial \psi }}\left( {a\cos \psi \sin \theta } \right) = - a\sin \psi \sin \theta ,\]

\[\frac{{\partial y}}{{\partial \psi }} = \frac{\partial }{{\partial \psi }}\left( {a\sin \psi \sin \theta } \right) = a\cos \psi \sin \theta ,\]

\[\frac{{\partial z}}{{\partial \psi }} = \frac{\partial }{{\partial \psi }}\left( {a\cos \theta } \right) = 0,\]

\[\frac{{\partial x}}{{\partial \theta }} = \frac{\partial }{{\partial \theta }}\left( {a\cos \psi \sin \theta } \right) = a\cos \psi \cos \theta ,\]

\[\frac{{\partial y}}{{\partial \theta }} = \frac{\partial }{{\partial \theta }}\left( {a\sin \psi \sin \theta } \right) = a\cos \psi \cos \theta ,\]

\[\frac{{\partial z}}{{\partial \theta }} = \frac{\partial }{{\partial \theta }}\left( {a\cos \theta } \right) = - a\sin \theta ,\]

the determinant in the double integral is

\[ \left| {\begin{array}{*{20}{c}} P&Q&R\\ {\frac{{\partial x}}{{\partial \psi }}}&{\frac{{\partial y}}{{\partial \psi }}}&{\frac{{\partial z}}{{\partial \psi }}}\\ {\frac{{\partial x}}{{\partial \theta }}}&{\frac{{\partial y}}{{\partial \theta }}}&{\frac{{\partial z}}{{\partial \theta }}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {2a\cos \psi \sin \theta }&0&0\\ { - a\sin \psi \sin \theta }&{a\cos \psi \sin \theta }&0\\ {a\cos \psi \cos \theta }&{a\sin \psi \cos \theta }&{ - a\sin \theta } \end{array}} \right| = 2a\cos \psi \sin \theta \cdot a\cos \psi \sin \theta \cdot \left( { - a\sin \theta } \right) = - 2{a^3}{\cos ^2}\psi \,{\sin ^3}\theta .\]

This value corresponds to the downward orientation of the surface.

The initial integral becomes

\[I = \iint\limits_S {2xdydz} = - 2a^3\iint\limits_S {{{\cos }^2}\psi \,{{\sin }^3}\theta d\psi d\theta } = - 2{a^3}\int\limits_0^{2\pi } {{{\cos }^2}\psi d\psi } \int\limits_0^\pi {{{\sin }^3}\theta d\theta } .\]

Calculate the last two integrals separately:

\[\int\limits_0^{2\pi } {{{\cos }^2}\psi d\psi } = \frac{1}{2}\int\limits_0^{2\pi } {\left( {1 + \cos 2\psi } \right)d\psi } = \frac{1}{2}\left[ {\left. {\left( {\psi + \frac{{\sin 2\psi }}{2}} \right)} \right|_0^{2\pi }} \right] = \pi ,\]

\[ \int\limits_0^\pi {{{\sin }^3}\theta d\theta } = \int\limits_0^\pi {{{\sin }^2}\theta \sin \theta d\theta } = \int\limits_0^\pi {\left( {{\cos^2}\theta - 1} \right)d\left( {\cos \theta } \right)} = \left. {\left( {\frac{{{\cos^3}\theta }}{3} - \cos \theta } \right)} \right|_0^\pi = \left( {\frac{{{\cos^3}\pi }}{3} - \cos \pi } \right) - \left( {\frac{{{\cos^3}0}}{3} - \cos 0} \right) = \left( { - \frac{1}{3} + 1} \right) - \left( {\frac{1}{3} - 1} \right) = \frac{4}{3}.\]

Hence, the value of the surface integral is

\[I = - 2{a^3} \cdot \pi \cdot \frac{4}{3} = - \frac{{8{a^3}\pi }}{3}.\]