Surface Integrals of Scalar Functions

Solved Problems

Example 3.

Find the integral \[\iint\limits_S {xdS},\] where the surface \(S\) is the part of the sphere \[{x^2} + {y^2} + {z^2} = {a^2}\] lying in the first octant.

Solution.

It is convenient to solve this integral in spherical coordinates. The area element for spherical surface is \(dS =\) \({a^2}\sin \theta d\psi d\theta .\) As \(x = a\cos \psi \sin \theta ,\) we can write the integral in the following form:

\[I = \iint\limits_S {xdS} = \iint\limits_{D\left( {\psi ,\theta } \right)} {a\cos \psi \sin \theta {a^2}\sin \theta d\psi d\theta } = {a^3}\iint\limits_{D\left( {\psi ,\theta } \right)} {\cos \psi \,{{\sin }^2}\theta d\psi d\theta } .\]

The domain of integration \(D\left( {\psi ,\theta } \right)\) is defined as

\[D = \left\{ {\left( {\psi ,\theta } \right)|\;0 \le \psi \le \frac{\pi }{2}, 0 \le \theta \le \frac{\pi }{2}} \right\}.\]

Hence, the integral is

\[I = {a^3}\iint\limits_{D\left( {\psi ,\theta } \right)} {\cos \psi \,{{\sin }^2}\theta d\psi d\theta } = {a^3}\int\limits_0^{\frac{\pi }{2}} {\cos \psi d\psi } \int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}\theta d\theta } = {a^3} \cdot \left[ {\left. {\left( {\sin \psi } \right)} \right|_0^{\frac{\pi }{2}}} \right] \int\limits_0^{\frac{\pi }{2}} {\frac{{1 - \cos 2\theta }}{2}d\theta } = {a^3} \cdot 1 \cdot \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {1 - \cos 2\theta } \right)d\theta } = \frac{{{a^3}}}{2}\left[ {\left. {\left( {\theta - \frac{{\sin 2\theta }}{2}} \right)} \right|_0^{\frac{\pi }{2}}} \right] = \frac{{{a^3}}}{2} \cdot \frac{\pi }{2} = \frac{{\pi {a^3}}}{4}.\]

Example 4.

Find the integral \[\iint\limits_S {\frac{{dS}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} ,\] where \(S\) is the part of the cylindrical surface parameterized by

\[\mathbf{r}\left( {u,v} \right) = \left( {a\cos u,a\sin u,v} \right), 0 \le u \le 2\pi , 0 \le v \le H.\]

Solution.

Calculate the partial derivatives

\[\frac{{\partial \mathbf{r}}}{{\partial u}} = \left( {\frac{{\partial x}}{{\partial u}},\frac{{\partial y}}{{\partial u}},\frac{{\partial z}}{{\partial u}}} \right) = \left( { - a\sin u,a\cos u,0} \right),\]

\[\frac{{\partial \mathbf{r}}}{{\partial v}} = \left( {\frac{{\partial x}}{{\partial v}},\frac{{\partial y}}{{\partial v}},\frac{{\partial z}}{{\partial v}}} \right) = \left( {0,0,1} \right)\]

and their cross product:

Then the area element of the given surface is

\[dS = \left| {\frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}}} \right|dudv = \left| {\sqrt {{{\left( {a\cos u} \right)}^2} + {{\left( {a\sin u} \right)}^2}} } \right|dudv = adudv.\]

Now we can calculate the surface integral:

\[\iint\limits_S {\frac{{dS}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} = \iint\limits_{D\left( {u,v} \right)} {\frac{{adudv}}{{\sqrt {{{\left( {a\cos u} \right)}^2} + {{\left( {a\sin u} \right)}^2} + {v^2}} }}} = \iint\limits_{D\left( {u,v} \right)} {\frac{{adudv}}{{\sqrt {{a^2} + {v^2}} }}} = \int\limits_0^{2\pi } {adu} \int\limits_0^H {\frac{{dv}}{{\sqrt {{a^2} + {v^2}} }}} = 2\pi a\int\limits_0^H {\frac{{dv}}{{\sqrt {{a^2} + {v^2}} }}} = 2\pi a\left[ {\left. {\ln \left( {v + \sqrt {{a^2} + {v^2}} } \right)} \right|_{v = 0}^H} \right] = 2\pi a \left[ {\ln \left( {H + \sqrt {{a^2} + {H^2}} } \right) - \ln a} \right] = 2\pi a\ln \frac{{H + \sqrt {{a^2} + {H^2}} }}{a}.\]

Example 5.

Evaluate the integral \[\iint\limits_S {\sqrt {1 + {x^2} + {y^2}} dS}.\] The surface \(S\) is parameterized by the position vector

\[\mathbf{r}\left( {u,v} \right) = u\cos v \cdot \mathbf{i} + u\sin v \cdot \mathbf{j} + v \cdot \mathbf{k}, 0 \le u \le 2, 0 \le v \le \pi.\]

Solution.

We find the partial derivatives and their cross product:

\[\frac{{\partial \mathbf{r}}}{{\partial u}} = \frac{\partial }{{\partial u}}\left( {u\cos v,u\sin v,v} \right) = \left( {\cos v,\sin v,0} \right),\]

\[\frac{{\partial \mathbf{r}}}{{\partial v}} = \frac{\partial }{{\partial v}}\left( {u\cos v,u\sin v,v} \right) = \left( { - u\sin v,u\cos v,1} \right),\]

\[ \frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {\cos v} & {\sin v} & 0\\ { - u\sin v} & {u\cos v} & 1 \end{array}} \right| = \sin v \cdot \mathbf{i} - \cos v \cdot \mathbf{j} + \left( {u\,{{\cos }^2}v + u\,{{\sin }^2}v} \right) \cdot \mathbf{k} = \sin v \cdot \mathbf{i} - \cos v \cdot \mathbf{j} + u \cdot \mathbf{k}.\]

Then the area element is

\[dS = \left| {\frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}}} \right|dudv = \sqrt {1 + {{\left( {u\cos v} \right)}^2} + {{\left( {u\sin v} \right)}^2}} dudv = \sqrt {1 + {u^2}} dudv.\]

Now it is easy to find the given surface integral:

\[\iint\limits_S {\sqrt {1 + {x^2} + {y^2}} dS} = \iint\limits_{D\left( {u,v} \right)} {\sqrt {1 + {{\left( {u\cos v} \right)}^2} + {{\left( {u\sin v} \right)}^2}} \sqrt {1 + {u^2}} dudv} = \iint\limits_{D\left( {u,v} \right)} {\left( {1 + {u^2}} \right)dudv} = \int\limits_0^\pi {dv} \int\limits_0^2 {\left( {1 + {u^2}} \right)du} = \pi \left[ {\left. {\left( {u + \frac{{{u^3}}}{3}} \right)} \right|_0^2} \right] = \frac{{14\pi }}{3}.\]