Surface Integrals of Scalar Functions

Consider a scalar function $$f\left( {x,y,z} \right)$$ and a surface $$S.$$ 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 $$\left( {u,v} \right)$$ range over some domain $$D\left( {u,v} \right)$$ of the $$uv$$-plane. Notice that the function $$f\left( {x,y,z} \right)$$ is evaluated only on the points of the surface $$S,$$ 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 $$1$$).

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

Click or tap a problem to see the solution.

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 $$\left( {x \ge 0}\right.,$$ $$y \ge 0,$$ $$\left.{z \ge 0} \right).$$

Example 2

Evaluate the surface integral $\iint\limits_S {{z^2}dS},$ where $$S$$ is the total area of the cone $\sqrt {{x^2} + {y^2}} \le z \le 2.$

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 $$2.$$

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 $$3$$).

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.