Surface integral

The surface integral or surface integral is a generalization of the concept of integral on flat or curved surfaces . The integration area is therefore not a one-dimensional interval , but a two-dimensional set in three-dimensional space . For a more general representation in with see: Integration on Manifolds . ${\ displaystyle {\ mathcal {F}}}$ ${\ displaystyle \ mathbb {R} ^ {3}}$${\ displaystyle \ mathbb {R} ^ {n}}$${\ displaystyle n \ geq 2}$

A general distinction is made between a scalar and a vector surface integral, depending on the shape of the integrand and the so-called surface element . They are

${\ displaystyle \ iint _ {\ mathcal {F}} f \, \ mathrm {d} \ sigma}$with scalar function and scalar surface element as well as${\ displaystyle f}$${\ displaystyle \ mathrm {d} \ sigma}$
${\ displaystyle \ iint _ {\ mathcal {F}} {\ vec {v}} \ cdot \ mathrm {d} {\ vec {\ sigma}}}$with vector-valued function and vectorial surface element .${\ displaystyle {\ vec {v}}}$${\ displaystyle \ mathrm {d} {\ vec {\ sigma}}}$

terms and definitions

When integrating over surfaces, parameterizations of the surface take the place of the integration variable and surface elements take the place of the infinitesimal (infinitely small) interval width . ${\ displaystyle \ mathrm {d} x}$

Parameterization

As a two-dimensional set, a surface can be represented ( parameterized ) as a function of two variables . If a set, the edge of which does not contain double points, is continuously differentiable , not infinitely long and, furthermore, is a mapping of into , then one says, is the parameterization of the area if is At this point it should be pointed out that a large part of the difficulties in dealing with surface integrals is related to the parameterization. It is not clear a priori that different parameterizations produce the same value for the integral. A change of coordinates for surface integrals is not trivial and is therefore a motivation for the use of differential forms . ${\ displaystyle B \ subset \ mathbb {R} ^ {2}}$${\ displaystyle \ varphi}$${\ displaystyle B}$${\ displaystyle \ mathbb {R} ^ {3}}$${\ displaystyle \ varphi}$${\ displaystyle {\ mathcal {F}}}$${\ displaystyle {\ mathcal {F}} = \ varphi (B)}$

In general, an area can be represented with two parameters and in the following form: ${\ displaystyle \ mathbb {R} ^ {3}}$${\ displaystyle u}$${\ displaystyle v}$

${\ displaystyle \ varphi \ colon B \ to \ mathbb {R} ^ {3}, \ quad \ left (u, v \ right) \ mapsto {\ vec {\ varphi}} \ left (u, v \ right) = \ left ({\ begin {matrix} x \ left (u, v \ right) \\ y \ left (u, v \ right) \\ z \ left (u, v \ right) \\\ end {matrix }} \ right)}$

On the surface of the form of curves or the coordinate lines . These cover the area with a coordinate network, with two coordinate lines running through each point. Thus every point on the surface has unique coordinates . ${\ displaystyle {\ vec {\ varphi}} \ left (u, v \ right)}$ ${\ displaystyle u = \ mathrm {const}}$${\ displaystyle v = \ mathrm {const}}$ ${\ displaystyle \ left (u_ {0}, v_ {0} \ right)}$

Example 1: Representation of parameters

The surface of a sphere with a radius can be parameterized as follows: is the rectangle and ${\ displaystyle R}$${\ displaystyle B}$${\ displaystyle [0, \ pi] \ times [0.2 \ pi]}$

${\ displaystyle {\ vec {\ varphi}} (u, v) = {\ begin {pmatrix} R \ sin (u) \ cos (v) \\ R \ sin (u) \ sin (v) \\ R \ cos (u) \ end {pmatrix}}}$.

This parameterization fulfills the spherical equation (see also spherical coordinates ). is the polar angle (usually or ) and the azimuth angle (usually or ). ${\ displaystyle x ^ {2} + y ^ {2} + z ^ {2} = R ^ {2}}$${\ displaystyle u}$${\ displaystyle \ vartheta \,}$${\ displaystyle \ theta \,}$${\ displaystyle v}$${\ displaystyle \ varphi \,}$${\ displaystyle \ phi \,}$

Example 2: Explicit representation

If a function and the area are given in the form , then and are the two parameters; the parameterization of the area looks like this: ${\ displaystyle f \ colon B \ to \ mathbb {R}, \ \ left (x, y \ right) \ mapsto f \ left (x, y \ right)}$${\ displaystyle z = f (x, y)}$${\ displaystyle x}$${\ displaystyle y}$

${\ displaystyle {\ vec {\ varphi}} \ left (x, y \ right) = \ left ({\ begin {matrix} x \\ y \\ f \ left (x, y \ right) \\\ end {matrix}} \ right)}$

Surface element

If in the one-dimensional case this represents the width of an infinitely small interval, then in the two-dimensional case it is obvious to replace it with the area of ​​an infinitely small area . With the parameterization described in the previous section, two tangents can be placed at each point on the surface (see also: Curvilinear coordinates ): One is the tangent that arises if one leaves constant and varies minimally, and one with exchanged variables. That means two tangents to the two coordinate lines in the point under consideration . These tangents can be expressed by two infinitesimal tangent vectors (be the parameterized form of the surface): ${\ displaystyle \ mathrm {d} x}$${\ displaystyle \ mathrm {d} \ sigma}$${\ displaystyle v}$${\ displaystyle u}$${\ displaystyle \ left (u_ {0}, v_ {0} \ right)}$${\ displaystyle {\ vec {\ varphi}} \ left (u, v \ right)}$

${\ displaystyle \ left. {\ frac {\ partial {\ vec {\ varphi}}} {\ partial u}} \ right | _ {u_ {0}, v_ {0}} \ mathrm {d} u}$   and   ${\ displaystyle \ left. {\ frac {\ partial {\ vec {\ varphi}}} {\ partial v}} \ right | _ {u_ {0}, v_ {0}} \ mathrm {d} v}$

In the following, the compact notation is used for the partial derivatives:

${\ displaystyle {\ vec {\ varphi}} _ {u} = {\ frac {\ partial {\ vec {\ varphi}}} {\ partial u}}}$   and   ${\ displaystyle {\ vec {\ varphi}} _ {v} = {\ frac {\ partial {\ vec {\ varphi}}} {\ partial v}}}$

If these tangents are not parallel at any point on the surface, one speaks of a regular parameterization . The cross product of the tangent vectors is then a vector whose length is not equal to zero.

${\ displaystyle \ left | \ left | {\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ right | \ right | \ neq 0}$

The two tangent vectors lie in the tangent plane of the surface at the point under consideration. The area of ​​the parallelogram spanned by the two tangent vectors now corresponds to the amount of their cross product .

If the surface is now parameterized regularly, one defines: ${\ displaystyle {\ vec {\ varphi}} (u, v)}$

• Scalar surface element
${\ displaystyle \ mathrm {d} \ sigma = \ left | \ left | {\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ right | \ right | \ mathrm {d} u \, \ mathrm {d} v}$
• Vectorial surface element
${\ displaystyle \ mathrm {d} {\ vec {\ sigma}} = {\ hat {n}} \ \ mathrm {d} \ sigma = {\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ \ mathrm {d} u \, \ mathrm {d} v}$     with the unit normal vector of the surface element     ${\ displaystyle {\ hat {n}} = {\ frac {{\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v}} {\ left | \ left | { \ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ right | \ right |}}}$

According to the properties of the cross product, the vectorial surface element is perpendicular to the surface; its amount corresponds precisely to the size of the infinitesimal surface piece.

In the form presented above, the vectorial surface element is not well-defined , since its direction depends on whether one calculates or . The two possibilities are antiparallel to each other. If you look at closed surfaces, you usually agree that the vectorial surface element pointing outwards is to be used. ${\ displaystyle {\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v}}$${\ displaystyle {\ vec {\ varphi}} _ {v} \ times {\ vec {\ varphi}} _ {u} = - \ left ({\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ right)}$

Example 1: Representation of parameters

As shown above, the surface of the sphere with radius R can be parameterized by the polar angle and the azimuth angle . The area element results from the following calculation: ${\ displaystyle u}$${\ displaystyle v}$

{\ displaystyle {\ begin {aligned} & {\ vec {\ varphi}} = R \ left ({\ begin {matrix} \ sin u \ \ cos v \\\ sin u \ \ sin v \\\ cos u \\\ end {matrix}} \ right), \ quad {\ vec {\ varphi}} _ {u} = R \ left ({\ begin {matrix} \ cos u \ \ cos v \\\ cos u \ \ sin v \\ - \ sin u \\\ end {matrix}} \ right), \ quad {\ vec {\ varphi}} _ {v} = R \ left ({\ begin {matrix} - \ sin u \ \ sin v \\\ sin u \ \ cos v \\ 0 \\\ end {matrix}} \ right), \\ & \ pm \ left ({\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ right) = \ pm R ^ {2} \ sin u \ left ({\ begin {matrix} \ sin u \ \ cos v \\\ sin u \ \ sin v \\\ cos u \\\ end {matrix}} \ right), \ quad \ left | \ left | \ pm \ left ({\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ right) \ right | \ right | = R ^ {2} \ sin u, \\ & {\ hat {n}} = \ pm \ left ({\ begin {matrix} \ sin u \ \ cos v \\\ sin u \ \ sin v \\\ cos u \\\ end {matrix}} \ right), \ quad \ mathrm {d} {\ vec {\ sigma}} = {\ hat {n}} \, \ mathrm {d} \ sigma = {\ hat {n}} \ R ^ {2} \ sin u \ \ mathrm {d} u \, \ mathrm {d} v \\\ end { aligned}}}

Two solutions are possible for the normal vector ( ), depending on the order of and in the cross product. Typically, the positive solution is chosen here, in which points away from the convex spherical surface (so-called "outer normal"). ${\ displaystyle \ pm}$${\ displaystyle {\ vec {\ varphi}} _ {u}}$${\ displaystyle {\ vec {\ varphi}} _ {v}}$${\ displaystyle {\ hat {n}}}$

Example 2: Explicit representation

If the surface is in the form specified, pressing the surface element by the differentials of the coordinates , from. ${\ displaystyle z = f (x, y)}$${\ displaystyle x}$${\ displaystyle y}$

${\ displaystyle {\ vec {\ varphi}} = \ left ({\ begin {matrix} x \\ y \\ f (x, y) \ end {matrix}} \ right), \ quad {\ vec {\ varphi}} _ {x} = \ left ({\ begin {matrix} 1 \\ 0 \\ f_ {x} \ end {matrix}} \ right), \ quad {\ vec {\ varphi}} _ {y } = \ left ({\ begin {matrix} 0 \\ 1 \\ f_ {y} \ end {matrix}} \ right)}$
${\ displaystyle \ pm \ left ({\ vec {\ varphi}} _ {x} \ times {\ vec {\ varphi}} _ {y} \ right) = \ pm \ left ({\ begin {matrix} - f_ {x} \\ - f_ {y} \\ 1 \ end {matrix}} \ right), \ quad \ left | \ left | \ pm \ left ({\ vec {\ varphi}} _ {x} \ times {\ vec {\ varphi}} _ {y} \ right) \ right | \ right | = {\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}} \, \ quad {\ hat {n}} = \ pm {\ frac {1} {\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}}} \ left ({\ begin {matrix} -f_ {x} \\ - f_ {y} \\ 1 \ end {matrix}} \ right)}$

Thus the surface element and the vectorial surface element are the same:

${\ displaystyle \ mathrm {d} \ sigma = {\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}} \ \ mathrm {d} x \, \ mathrm {d} y}$
${\ displaystyle \ mathrm {d} {\ vec {\ sigma}} = {\ hat {n}} \ {\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}} \ \ mathrm {d} x \, \ mathrm {d} y = {\ begin {pmatrix} -f_ {x} \\ - f_ {y} \\ 1 \ end {pmatrix}} \ \ mathrm {d} x \, \ mathrm {d} y}$

Projection onto a surface with a known surface element

In the following, we assume that a surface with its surface element and associated normal vector is known. E.g. ${\ displaystyle A}$${\ displaystyle \ mathrm {d} A}$${\ displaystyle {\ hat {n}} _ {A}}$

${\ displaystyle \ mathrm {d} A = \ mathrm {d} x \ mathrm {d} y}$   and   ${\ displaystyle {\ hat {n}} _ {A} = {\ hat {e}} _ {z} = \ left ({\ begin {matrix} 0 \\ 0 \\ 1 \\\ end {matrix} } \ right)}$
• Lateral surface of a circular cylinder with radius :${\ displaystyle \ rho}$
${\ displaystyle \ mathrm {d} A = \ rho \ mathrm {d} \ varphi \ mathrm {d} z}$   and   ${\ displaystyle {\ hat {n}} _ {A} = {\ hat {e}} _ {\ rho} = \ left ({\ begin {matrix} \ cos \ varphi \\\ sin \ varphi \\ 0 \\\ end {matrix}} \ right)}$
• Spherical surface with radius :${\ displaystyle r}$
${\ displaystyle \ mathrm {d} A = r ^ {2} \ sin \ vartheta \ mathrm {d} \ vartheta \ mathrm {d} \ varphi}$   and   ${\ displaystyle {\ hat {n}} _ {A} = {\ hat {e}} _ {r} = \ left ({\ begin {matrix} \ sin \ vartheta \ \ cos \ varphi \\\ sin \ vartheta \ \ sin \ varphi \\\ cos \ vartheta \\\ end {matrix}} \ right)}$

The surface element is to be determined for another surface with a normal vector . The area is roughly given by and thus the normal vector is equal . ${\ displaystyle {\ mathcal {F}}}$${\ displaystyle {\ hat {n}} _ {\ mathcal {F}}}$${\ displaystyle \ mathrm {d} \ sigma}$${\ displaystyle g (x, y, z) = 0}$${\ displaystyle {\ hat {n}} _ {\ mathcal {F}} = \ nabla g / || \ nabla g ||}$

We now project along to . Then the surface elements can be linked using for : ${\ displaystyle {\ mathcal {F}}}$${\ displaystyle {\ hat {n}} _ {A}}$${\ displaystyle A}$${\ displaystyle \ mathrm {d} A = \ mathrm {d} {\ vec {A}} \ cdot {\ hat {n}} _ {A} = | \ mathrm {d} {\ vec {\ sigma}} \ cdot {\ hat {n}} _ {A} | = \ mathrm {d} \ sigma \, | {\ hat {n}} _ {\ mathcal {F}} \ cdot {\ hat {n}} _ {A} |}$${\ displaystyle {\ hat {n}} _ {\ mathcal {F}} \ cdot {\ hat {n}} _ {A} \ neq 0}$

${\ displaystyle \ mathrm {d} \ sigma = {\ frac {\ mathrm {d} A} {| {\ hat {n}} _ {\ mathcal {F}} \ cdot {\ hat {n}} _ { A} |}} = {\ frac {|| \ nabla g || \, \ mathrm {d} A} {| \ nabla g \ cdot {\ hat {n}} _ {A} |}}}$

Each line along the normal vectors may only intersect the surface once. Otherwise you have to divide the area into smaller areas , the projection of which is then clear, or choose a different base area . ${\ displaystyle {\ hat {n}} _ {A}}$${\ displaystyle {\ mathcal {F}}}$${\ displaystyle {\ mathcal {F}}}$${\ displaystyle {\ mathcal {F}} _ {1}, \, {\ mathcal {F}} _ {2}, \, \ dotsc}$${\ displaystyle A}$

The vectorial surface element is:

${\ displaystyle \ mathrm {d} {\ vec {\ sigma}} = {\ hat {n}} _ {\ mathcal {F}} {\ frac {\ mathrm {d} A} {| {\ hat {n }} _ {\ mathcal {F}} \ cdot {\ hat {n}} _ {A} |}} = {\ frac {\ nabla g} {|| \ nabla g ||}} {\ frac {| | \ nabla g || \, \ mathrm {d} A} {| \ nabla g \ cdot {\ hat {n}} _ {A} |}} = {\ frac {\ nabla g \, \ mathrm {d } A} {| \ nabla g \ cdot {\ hat {n}} _ {A} |}}}$

example 1

Given a surface of the form , then applies and thus: ${\ displaystyle {\ mathcal {F}}}$${\ displaystyle z = f (x, y)}$${\ displaystyle g (x, y, z) = zf (x, y)}$

${\ displaystyle \ nabla g = {\ begin {pmatrix} -f_ {x} \\ - f_ {y} \\ 1 \ end {pmatrix}} \, \ quad || \ nabla g || = {\ sqrt { f_ {x} ^ {2} + f_ {y} ^ {2} +1}} \, \ quad {\ hat {n}} _ {\ mathcal {F}} = {\ frac {\ nabla g} { || \ nabla g ||}} = {\ frac {1} {\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}}} {\ begin {pmatrix} -f_ {x} \\ - f_ {y} \\ 1 \ end {pmatrix}}}$

This area is now projected into the plane with and ; is there ${\ displaystyle xy}$${\ displaystyle \ mathrm {d} A = \ mathrm {d} x \ mathrm {d} y}$${\ displaystyle {\ hat {n}} _ {A} = {\ hat {e}} _ {z}}$

${\ displaystyle \ mathrm {d} \ sigma = {\ frac {|| \ nabla g || \, \ mathrm {d} x \ mathrm {d} y} {| \ nabla g \ cdot {\ hat {e} } _ {z} |}} = {\ frac {{\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}} \, \ mathrm {d} x \ mathrm {i.e. } y} {| {\ hat {e}} _ {z} \ cdot {\ hat {e}} _ {z} |}} = {\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}} \, \ mathrm {d} x \ mathrm {d} y}$
${\ displaystyle \ mathrm {d} {\ vec {\ sigma}} = {\ frac {\ nabla g \, \ mathrm {d} x \ mathrm {d} y} {| \ nabla g \ cdot {\ hat { e}} _ {z} |}} = {\ begin {pmatrix} -f_ {x} \\ - f_ {y} \\ 1 \ end {pmatrix}} \ mathrm {d} x \ mathrm {d} y }$

Example 2

We are looking for the surface element of a solid of revolution around the axis with , that is . ${\ displaystyle z}$${\ displaystyle \ rho = f (z)}$${\ displaystyle g (\ rho, \ varphi, z) = \ rho -f (z)}$

${\ displaystyle \ nabla g = {\ hat {e}} _ {\ rho} -f_ {z} {\ hat {e}} _ {z} \, \ quad || \ nabla g || = {\ sqrt {1 + f_ {z} ^ {2}}} \, \ quad {\ hat {n}} _ {\ mathcal {F}} = {\ frac {\ nabla g} {|| \ nabla g ||} } = {\ frac {{\ hat {e}} _ {\ rho} -f_ {z} {\ hat {e}} _ {z}} {\ sqrt {1 + f_ {z} ^ {2}} }}}$

By projecting onto the lateral surface of a circular cylinder with a radius , the surface element is obtained: ${\ displaystyle \ rho = f (z)}$

${\ displaystyle \ mathrm {d} \ sigma = {\ frac {|| \ nabla g || \, \ rho \, \ mathrm {d} \ varphi \ mathrm {d} z} {| \ nabla g \ cdot { \ hat {e}} _ {z} |}} = {\ frac {{\ sqrt {1 + f_ {z} ^ {2}}} \, f (z) \, \ mathrm {d} \ varphi \ mathrm {d} z} {| ({\ hat {e}} _ {\ rho} -f_ {z} {\ hat {e}} _ {z}) \ cdot {\ hat {e}} _ {\ rho} |}} = {\ sqrt {1 + f_ {z} ^ {2}}} \, f (z) \, \ mathrm {d} \ varphi \ mathrm {d} z}$
${\ displaystyle \ mathrm {d} {\ vec {\ sigma}} = ({\ hat {e}} _ {\ rho} -f_ {z} {\ hat {e}} _ {z}) \, f (z) \, \ mathrm {d} \ varphi \ mathrm {d} z}$

The integrals

With the parameterizations and the surface elements you can now define the surface integrals. These multi-dimensional integrals are Lebesgue integrals , but can be calculated as multiple Riemann integrals in most applications .

The scalar surface integral

The scalar surface integral of a scalar function over a surface with regular parameterization with is defined as ${\ displaystyle f \ colon \ mathbb {R} ^ {3} \ rightarrow \ mathbb {R}}$${\ displaystyle {\ mathcal {F}}}$${\ displaystyle \ varphi \ colon B \ rightarrow \ mathbb {R} ^ {3}}$${\ displaystyle B \ subset \ mathbb {R} ^ {2}}$

${\ displaystyle \ iint _ {\ mathcal {F}} f ({\ vec {x}}) \, \ mathrm {d} \ sigma = \ iint _ {B} f \ left ({\ vec {\ varphi} } (u, v) \ right) \, || {\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} || \, \, \ mathrm {d} (u, v)}$

For example , if one sets , the scalar surface integral is simply the area of the surface. ${\ displaystyle f ({\ vec {x}}) = 1}$

The vectorial surface integral

The vector surface integral of a vector-valued function over a surface with regular parameterization with is defined as ${\ displaystyle f \ colon \ mathbb {R} ^ {3} \ rightarrow \ mathbb {R} ^ {3}}$${\ displaystyle {\ mathcal {F}}}$${\ displaystyle \ varphi \ colon B \ rightarrow \ mathbb {R} ^ {3}}$${\ displaystyle B \ subset \ mathbb {R} ^ {2}}$

${\ displaystyle \ iint _ {\ mathcal {F}} {\ vec {f}} ({\ vec {x}}) \ cdot \ mathrm {d} {\ vec {\ sigma}} = \ iint _ {B. } {\ vec {f}} \ left ({\ vec {\ varphi}} (u, v) \ right) \ cdot ({\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi }} _ {v}) \, \, \ mathrm {d} (u, v) =: \ Phi _ {\ mathcal {F}} ({\ vec {f}})}$.

A clear presentation of this integral takes place via the flow of a vector field through the surface : The size indicates the contribution to the total flow made by the infinitesimally small surface vector ; namely, how much of flows through the patch . The flux is maximum when the vector field is parallel to the surface normal , and zero when it is perpendicular to , i.e. tangential to the surface - then it "flows" along the surface, but not through it. ${\ displaystyle \ Phi}$ ${\ displaystyle {\ vec {f}}}$${\ displaystyle {\ mathcal {F}}}$${\ displaystyle {\ vec {f}} \ cdot \ mathrm {d} {\ vec {\ sigma}}}$${\ displaystyle \ Phi _ {\ mathcal {F}} ({\ vec {f}})}$${\ displaystyle \ mathrm {d} {\ vec {\ sigma}} = {\ hat {n}} \, \ mathrm {d} \ sigma}$${\ displaystyle {\ vec {f}}}$${\ displaystyle \ mathrm {d} {\ sigma}}$${\ displaystyle {\ vec {f}}}$${\ displaystyle {\ hat {n}}}$${\ displaystyle {\ vec {f}}}$${\ displaystyle {\ hat {n}}}$${\ displaystyle {\ vec {f}}}$

literature

• G. Bärwolff: Higher mathematics for natural scientists and engineers . 2nd Edition. Spectrum Academic Publishing House, 2006, ISBN 978-3-8274-1688-9
• KF Riley, MP Hobson: Mathematical Methods for Physics and Engineering . 3. Edition. Cambridge University Press, 2006, ISBN 978-0-521-67971-8