Calculus

Surface Integrals

Surface Integrals of Scalar Functions

Page 1
Problem 1
Page 2
Problems 2-5

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 \(S,\) i.e.

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

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

\[A = \iint\limits_S {dS} .\]
The area element dS

Figure 1.

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 on problem description to see 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

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

Figure 2.

we can express the surface integral through 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:

\[
{I \text{ = }}\kern0pt{ \frac{{\sqrt {21} }}{4}\int\limits_0^2 {\Big[ {\int\limits_0^{4 – 2y} {\Big( {\frac{{3x}}{4} }}}+{{{ \frac{y}{2} + 1} \Big)dx} } \Big]dy} }
= {\frac{{\sqrt {21} }}{{16}}\int\limits_0^2 {\Big[ {\int\limits_0^{4 – 2y} {\big( {3x + 2y }}}}+{{{{ 4} \big)dx} } \Big]dy} }
= {\frac{{\sqrt {21} }}{{16}}\int\limits_0^2 {\Big[ {\left. {\Big( {\frac{{3{x^2}}}{2} + 2yx }}\right.}}}+{{{\left.{{ 4x} \Big)} \right|_{x = 0}^{4 – 2y}} \Big]dy} }
= {{\frac{{\sqrt {21} }}{{16}}\int\limits_0^2 {\Big[ {\frac{3}{2}{{\left( {4 – 2y} \right)}^2} }}}+{{{ 2\left( {4 – 2y} \right)y }}}+{{{ 4\left( {4 – 2y} \right)} \Big]dy} }}
= {{\frac{{\sqrt {21} }}{{32}}\int\limits_0^2 {\Big[ {3{{\left( {16 – 16y + 4{y^2}} \right)}^2} }}}+{{{ 16y – 8{y^2} + 32 – 16y} \Big]dy} }}
= {\frac{{\sqrt {21} }}{{32}}\int\limits_0^2 {\left( {80 – 48y + 4{y^2}} \right)dy} }
= {\frac{{\sqrt {21} }}{4}\int\limits_0^2 {\left( {20 – 12y + {y^2}} \right)dy} }
= {\frac{{\sqrt {21} }}{4}\left[ {\left. {\left( {20y – 6{y^2} + \frac{{{y^3}}}{3}} \right)} \right|_0^2} \right] }
= {\frac{{\sqrt {21} }}{4}\left( {40 – 24 + \frac{8}{3}} \right) }
= {\frac{{7\sqrt {21} }}{3}.}
\]
Page 1
Problem 1
Page 2
Problems 2-5