Surface Integrals of Scalar Functions
Consider a scalar function f (x , y , z )S . Let S be given by the position vector
\[\mathbf{r}\left( {u,v} \right) = x\left( {u,v} \right)\mathbf{i} + y\left( {u,v} \right)\mathbf{j} + z\left( {u,v} \right)\mathbf{k},\]
where the coordinates (u , v ) range over some domain D (u , v )uv -plane. Notice that the function f (x , y , z )S , that is
\[f\left[ {\mathbf{r}\left( {u,v} \right)} \right] = f\left[ {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right].\]
The surface integral of scalar function \(f\left( {x,y,z} \right)\) over the surface \(S\) is defined as
\[\iint\limits_S {f\left( {x,y,z} \right)dS} = \iint\limits_{D\left( {u,v} \right)} {f\left( {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right) \left| {\frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}}} \right|dudv} ,\]
where the partial derivatives \({\frac{{\partial \mathbf{r}}}{{\partial u}}}\) and \({\frac{{\partial \mathbf{r}}}{{\partial v}}}\) are given by
\[\frac{{\partial \mathbf{r}}}{{\partial u}} = \frac{{\partial x}}{{\partial u}}\left( {u,v} \right)\mathbf{i} + \frac{{\partial y}}{{\partial u}}\left( {u,v} \right)\mathbf{j} + \frac{{\partial z}}{{\partial u}}\left( {u,v} \right)\mathbf{k},\]
\[\frac{{\partial \mathbf{r}}}{{\partial v}} = \frac{{\partial x}}{{\partial v}}\left( {u,v} \right)\mathbf{i} + \frac{{\partial y}}{{\partial v}}\left( {u,v} \right)\mathbf{j} + \frac{{\partial z}}{{\partial v}}\left( {u,v} \right)\mathbf{k},\]
and \({{\frac{{\partial \mathbf{r}}}{{\partial u}}} \times {\frac{{\partial \mathbf{r}}}{{\partial v}}}}\) is the cross product. The vector \({{\frac{{\partial \mathbf{r}}}{{\partial u}}} \times {\frac{{\partial \mathbf{r}}}{{\partial v}}}}\) is perpendicular to the surface at the point \({\mathbf{r}\left( {u,v} \right)}.\)
The absolute value
\[dS = \left| {{\frac{{\partial \mathbf{r}}}{{\partial u}}} \times {\frac{{\partial \mathbf{r}}}{{\partial v}}}} \right|dudv\]
is called the area element : it represents the area \(dS\) of a small patch of the surface obtained by changing the coordinates \(u\) and \(v\) by small amounts \(du\) and \(dv\) (Figure \(1\)).
Figure 1. The area of the surface \(S\) is given by the integral
\[A = \iint\limits_S {dS} .\]
If the surface \(S\) is defined 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 can be found by the formula
\[\iint\limits_S {f\left( {x,y,z} \right)dS} = \iint\limits_{D\left( {x,y} \right)} {f\left( {x,y,z\left( {x,y} \right)} \right) \sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy} .\]
If a surface \(S\) consists of several "patches" \({S_i},\) then for calculating the surface integral we can apply the additivity property:
\[\iint\limits_S {f\left( {x,y,z} \right)dS} = \sum\limits_{i = 1}^n {\iint\limits_{{S_i}} {f\left( {x,y,z} \right)d{S_i}} } .\]
Solved Problems
Example 1.
Calculate the surface integral \[\iint\limits_S {\left( {x + y + z} \right)dS},\] where \(S\) is the portion of the plane \[x + 2y + 4z = 4\] lying in the first octant \(\left( {x \ge 0}\right.,\) \(y \ge 0,\) \(\left.{z \ge 0} \right).\)
Solution.
We rewrite the equation of the plane in the form
\[z = z\left( {x,y} \right) = 1 - \frac{x}{4} - \frac{y}{2}.\]
Find the partial derivatives:
\[\frac{{\partial z}}{{\partial x}} = - \frac{1}{4},\;\; \frac{{\partial z}}{{\partial y}} = - \frac{1}{2}.\]
Applying the formula
\[\iint\limits_S {f\left( {x,y,z} \right)dS} = \iint\limits_{D} {f\left( {x,y,z\left( {x,y} \right)} \right) \sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy},\]
we can express the surface integral in terms of the double integral:
\[I = \iint\limits_S {\left( {x + y + z} \right)dS} = \iint\limits_D {\left( {x + y + 1 - \frac{x}{4} - \frac{y}{2}} \right) \sqrt {1 + {{\left( { - \frac{1}{4}} \right)}^2} + {{\left( { - \frac{1}{2}} \right)}^2}} dxdy} = \iint\limits_D {\left( {\frac{{3x}}{4} + \frac{y}{2} + 1} \right)\frac{{\sqrt {21} }}{4}dxdy}.\]
The region of integration \(D\) is the triangle shown in Figure \(2.\)
Figure 2. Calculate the given integral:
\[I = \frac{{\sqrt {21} }}{4}\int\limits_0^2 {\Big[ {\int\limits_0^{4 - 2y} {\Big( {\frac{{3x}}{4} }}}+{{{ \frac{y}{2} + 1} \Big)dx} } \Big]dy} = \frac{{\sqrt {21} }}{{16}}\int\limits_0^2 {\Big[ {\int\limits_0^{4 - 2y} {\big( {3x + 2y }}}}+{{{{ 4} \big)dx} } \Big]dy} = \frac{{\sqrt {21} }}{{16}}\int\limits_0^2 {\Big[ {\left. {\Big( {\frac{{3{x^2}}}{2} + 2yx }}\right.}}+{{\left.{{ 4x} \Big)} \right|_{x = 0}^{4 - 2y}} \Big]dy} = \frac{{\sqrt {21} }}{{16}}\int\limits_0^2 {\Big[ {\frac{3}{2}{{\left( {4 - 2y} \right)}^2} }}+{{ 2\left( {4 - 2y} \right)y }}+{{ 4\left( {4 - 2y} \right)} \Big]dy} = \frac{{\sqrt {21} }}{{32}}\int\limits_0^2 {\Big[ {3{{\left( {16 - 16y + 4{y^2}} \right)}^2} }}+{{ 16y - 8{y^2} + 32 - 16y} \Big]dy} = \frac{{\sqrt {21} }}{{32}}\int\limits_0^2 {\left( {80 - 48y + 4{y^2}} \right)dy} = \frac{{\sqrt {21} }}{4}\int\limits_0^2 {\left( {20 - 12y + {y^2}} \right)dy} = \frac{{\sqrt {21} }}{4}\left[ {\left. {\left( {20y - 6{y^2} + \frac{{{y^3}}}{3}} \right)} \right|_0^2} \right] = \frac{{\sqrt {21} }}{4}\left( {40 - 24 + \frac{8}{3}} \right) = \frac{{7\sqrt {21} }}{3}.\]
Example 2.
Evaluate the surface integral \[\iint\limits_S {{z^2}dS},\] where \(S\) is the total area of the cone \[\sqrt {{x^2} + {y^2}} \le z \le 2.\]
Solution.
Let \({S_1}\) be side surface of the cone, and \({S_2}\) be its base. We can write the given integral as the sum of two integrals:
\[I = {I_1} + {I_2} = \iint\limits_{{S_1}} {{z^2}d{S_1}} + \iint\limits_{{S_2}} {{z^2}d{S_2}} .\]
Find the first integral \({I_1},\) using the formula
\[{I_1} = \iint\limits_{{S_1}} {{z^2}d{S_1}} = \iint\limits_D {f\left( {x,y,z\left( {x,y} \right)} \right) \sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy} .\]
Here the partial derivatives are
\[\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\sqrt {{x^2} + {y^2}} = \frac{x}{{\sqrt {{x^2} + {y^2}} }},\]
\[\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\sqrt {{x^2} + {y^2}} = \frac{y}{{\sqrt {{x^2} + {y^2}} }}.\]
Then
\[\sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} = \sqrt {1 + \frac{{{x^2}}}{{{x^2} + {y^2}}} + \frac{{{y^2}}}{{{x^2} + {y^2}}}} = \sqrt 2 .\]
Since \(z = 2\) on the base of the cone, then the domain \(D\left( {x,y} \right)\) is defined by the inequality \({z^2} + {y^2} \le 4\) (Figure \(3\)).
Figure 3. Hence, the integral \({I_1}\) is written as
\[{I_1} = \iint\limits_{{S_1}} {{z^2}d{S_1}} = \sqrt 2 \iint\limits_{D\left( {x,y} \right)} {\left( {{x^2} + {y^2}} \right)dxdy} .\]
In polar coordinates we have
\[{I_1} = \sqrt 2 \iint\limits_{D\left( {x,y} \right)} {\left( {{x^2} + {y^2}} \right)dxdy} = \sqrt 2 \int\limits_0^{2\pi } {\int\limits_0^2 {{r^2} \cdot rdrd\varphi } } = \sqrt 2 \int\limits_0^{2\pi } d \varphi \int\limits_0^2 {{r^3}dr} = \sqrt 2 \cdot 2\pi \cdot \left[ {\left. {\left( {\frac{{{r^4}}}{4}} \right)} \right|_0^2} \right] = \frac{{\sqrt 2 \pi }}{2}\left( {{2^4} - 0} \right) = 8\sqrt 2 \pi .\]
Consider now the second integral \({I_2}.\) The base of the cone is described by the equation \(z = 2.\) Therefore,
\[{I_2} = \iint\limits_{{S_2}} {{2^2}d{S_2}} = 4\iint\limits_{{S_2}} {d{S_2}} ,\]
where \(\iint\limits_{{S_2}} {d{S_2}} \) represents the area of the base, which is \(\pi \cdot {2^2} \) \(= 4\pi .\) Then
\[{I_2} = 4\iint\limits_{{S_2}} {d{S_2}} = 4 \cdot 4\pi = 16\pi .\]
Thus, the full value of the initial surface integral is
\[I = {I_1} + {I_2} = 8\sqrt 2 \pi + 16\pi = 8\pi \left( {\sqrt 2 + 2} \right).\]
See more problems on Page 2.
