# 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

${f\left[ {\mathbf{r}\left( {u,v} \right)} \right] \text{ = }}\kern0pt{ f\left[ {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right].}$

The surface integral of scalar function $$f\left( {x,y,z} \right)$$ over the surface $$S$$ is defined as

${\iint\limits_S {f\left( {x,y,z} \right)dS} \text{ = }}\kern0pt {{\iint\limits_{D\left( {u,v} \right)} {f\left( {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right) \cdot}\kern0pt{ \left| {\frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}}} \right|dudv} ,}}$

where the partial derivatives $${\large\frac{{\partial \mathbf{r}}}{{\partial u}}\normalsize}$$ and $${\large\frac{{\partial \mathbf{r}}}{{\partial v}}\normalsize}$$ 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 $${{\large\frac{{\partial \mathbf{r}}}{{\partial u}}\normalsize} \times {\large\frac{{\partial \mathbf{r}}}{{\partial v}}\normalsize}}$$ is the cross product. The vector $${{\large\frac{{\partial \mathbf{r}}}{{\partial u}}\normalsize} \times {\large\frac{{\partial \mathbf{r}}}{{\partial v}}\normalsize}}$$ is perpendicular to the surface at the point $${\mathbf{r}\left( {u,v} \right)}.$$

The absolute value $$dS =$$ $$\left| {{\large\frac{{\partial \mathbf{r}}}{{\partial u}}\normalsize} \times {\large\frac{{\partial \mathbf{r}}}{{\partial v}}\normalsize}} \right|dudv$$ 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

${\iint\limits_S {f\left( {x,y,z} \right)dS} } = {\iint\limits_{D\left( {x,y} \right)} {f\left( {x,y,z\left( {x,y} \right)} \right) \cdot}\kern0pt{ \sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy} .}$

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 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.

### Example 4

Find the integral $$\iint\limits_S {\large\frac{{dS}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\normalsize} ,$$ where $$S$$ is the part of the cylindrical surface parameterized by $$\mathbf{r}\left( {u,v} \right) =$$ $$\left( {a\cos u,a\sin u,v} \right),$$ $$0 \le u \le 2\pi ,$$ $$0 \le v \le H.$$

### Example 5

Evaluate the integral $$\iint\limits_S {\sqrt {1 + {x^2} + {y^2}} dS}.$$ The surface $$S$$ is parameterized by the position vector $$\mathbf{r}\left( {u,v} \right)$$ $$= u\cos v \cdot \mathbf{i}$$ $$+ u\sin v \cdot \mathbf{j}$$ $$+ v \cdot \mathbf{k},$$ $$0 \le u \le 2,$$ $$0 \le v \le \pi.$$

### 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:

${\frac{{\partial z}}{{\partial x}} = – \frac{1}{4},\;\;\;}\kern0pt{\frac{{\partial z}}{{\partial y}} = – \frac{1}{2}.}$

Applying the formula

${\iint\limits_S {f\left( {x,y,z} \right)dS} } = {\iint\limits_{D} {f\left( {x,y,z\left( {x,y} \right)} \right) \cdot}\kern0pt{ \sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy},}$

we can express the surface integral in terms of the double integral:

${I = \iint\limits_S {\left( {x + y + z} \right)dS} } = {{\iint\limits_D {\left( {x + y + 1 – \frac{x}{4} – \frac{y}{2}} \right) \cdot}\kern0pt{ \sqrt {1 + {{\left( { – \frac{1}{4}} \right)}^2} + {{\left( { – \frac{1}{2}} \right)}^2}} dxdy} }} = {\iint\limits_D {\left( {\frac{{3x}}{4} + \frac{y}{2} + 1} \right)\frac{{\sqrt {21} }}{4}dxdy}}$

The region of integration $$D$$ is the triangle shown in Figure $$2.$$

Calculate the given integral:

${I \text{ = }}\kern0pt{ \frac{{\sqrt {21} }}{4}\int\limits_0^2 {\Big[ {\int\limits_0^{4 – 2y} {\Big( {\frac{{3x}}{4} }}}+{{{ \frac{y}{2} + 1} \Big)dx} } \Big]dy} } = {\frac{{\sqrt {21} }}{{16}}\int\limits_0^2 {\Big[ {\int\limits_0^{4 – 2y} {\big( {3x + 2y }}}}+{{{{ 4} \big)dx} } \Big]dy} } = {\frac{{\sqrt {21} }}{{16}}\int\limits_0^2 {\Big[ {\left. {\Big( {\frac{{3{x^2}}}{2} + 2yx }}\right.}}}+{{{\left.{{ 4x} \Big)} \right|_{x = 0}^{4 – 2y}} \Big]dy} } = {{\frac{{\sqrt {21} }}{{16}}\int\limits_0^2 {\Big[ {\frac{3}{2}{{\left( {4 – 2y} \right)}^2} }}}+{{{ 2\left( {4 – 2y} \right)y }}}+{{{ 4\left( {4 – 2y} \right)} \Big]dy} }} = {{\frac{{\sqrt {21} }}{{32}}\int\limits_0^2 {\Big[ {3{{\left( {16 – 16y + 4{y^2}} \right)}^2} }}}+{{{ 16y – 8{y^2} + 32 – 16y} \Big]dy} }} = {\frac{{\sqrt {21} }}{{32}}\int\limits_0^2 {\left( {80 – 48y + 4{y^2}} \right)dy} } = {\frac{{\sqrt {21} }}{4}\int\limits_0^2 {\left( {20 – 12y + {y^2}} \right)dy} } = {\frac{{\sqrt {21} }}{4}\left[ {\left. {\left( {20y – 6{y^2} + \frac{{{y^3}}}{3}} \right)} \right|_0^2} \right] } = {\frac{{\sqrt {21} }}{4}\left( {40 – 24 + \frac{8}{3}} \right) } = {\frac{{7\sqrt {21} }}{3}.}$

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Problem 1
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Problems 2-5