{"id":1454,"date":"2025-10-17T15:29:45","date_gmt":"2025-10-17T15:29:45","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1454"},"modified":"2025-12-15T13:36:26","modified_gmt":"2025-12-15T13:36:26","slug":"surface-integrals-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/surface-integrals-learn-it-5\/","title":{"raw":"Surface Integrals: Learn It 5","rendered":"Surface Integrals: Learn It 5"},"content":{"raw":"

Surface Integral of a Vector Field<\/h2>\r\nWith the idea of orientable surfaces in place, we can now define a surface integral of a vector field. The definition is analogous to the definition of the flux of a vector field along a plane curve.\r\n\r\n
\r\n

Flux of a Vector Field Along a Curve<\/strong><\/p>\r\n

If [latex]\\mathbf{F}[\/latex] is a two-dimensional vector field and [latex]C[\/latex] is a plane curve, then the flux of [latex]\\mathbf{F}[\/latex] along [latex]C[\/latex] involves chopping [latex]C[\/latex] into small pieces, choosing a point inside each piece, and calculating [latex]\\mathbf{F} \\cdot \\mathbf{N}[\/latex] at the point (where [latex]\\mathbf{N}[\/latex] is the unit normal vector at the point).<\/p>\r\n\r\n<\/section>The definition of a surface integral of a vector field proceeds in the same fashion, except now we chop surface [latex]S[\/latex] into small pieces, choose a point in each (two-dimensional) piece, and calculate [latex]\\mathbf{F} \\cdot \\mathbf{N}[\/latex] at the point.\r\n\r\n

\r\n

To place this definition in a real-world setting, let [latex]S[\/latex] be an oriented surface with unit normal vector [latex]\\mathbf{N}[\/latex]. Let [latex]\\mathbf{v}[\/latex] be a velocity field of a fluid flowing through [latex]S[\/latex], and suppose the fluid has density [latex]\\rho(x,y,z)[\/latex].<\/p>\r\n

Imagine the fluid flows through [latex]S[\/latex], but [latex]S[\/latex] is completely permeable so that it does not impede the fluid flow (Figure 4). The mass flux<\/strong> of the fluid is the rate of mass flow per unit area. The mass flux is measured in mass per unit time per unit area.<\/p>\r\n\r\n

[caption id=\"attachment_1496\" align=\"aligncenter\" width=\"231\"]\"A Figure 4. Fluid flows across a completely permeable surface [latex]S[\/latex].[\/caption]<\/div>\r\n<\/section>\r\n
\r\n

How could we calculate the mass flux of the fluid across [latex]S[\/latex]?<\/p>\r\n

The rate of flow, measured in mass per unit time per unit area, is [latex]\\rho \\mathbf{v}[\/latex]. To calculate the mass flux across [latex]S[\/latex], chop [latex]S[\/latex] into small pieces [latex]S_{ij}[\/latex].<\/p>\r\n

If [latex]S_{ij}[\/latex] is small enough, then it can be approximated by a tangent plane at some point [latex]P[\/latex] in [latex]S_{ij}[\/latex]. Therefore, the unit normal vector at [latex]P[\/latex] can be used to approximate [latex]\\mathbf{N}(x,y,z)[\/latex] across the entire piece [latex]S_{ij}[\/latex], because the normal vector to a plane does not change as we move across the plane.<\/p>\r\n

The component of the vector [latex]\\rho \\mathbf{v}[\/latex] at [latex]P[\/latex] in the direction of [latex]\\mathbf{N}[\/latex] is [latex]\\rho \\mathbf{v} \\cdot \\mathbf{N}[\/latex] at [latex]P[\/latex]. Since [latex]S_{ij}[\/latex] is small, the dot product [latex]\\rho \\mathbf{v} \\cdot \\mathbf{N}[\/latex] changes very little as we vary across [latex]S_{ij}[\/latex], and therefore [latex]\\rho \\mathbf{v} \\cdot \\mathbf{N}[\/latex] can be taken as approximately constant across [latex]S_{ij}[\/latex].<\/p>\r\n

To approximate the mass of fluid per unit time flowing across [latex]S_{ij}[\/latex] (and not just locally at point [latex]P[\/latex]), we need to multiply [latex](\\rho \\mathbf{v} \\cdot \\mathbf{N})(P)[\/latex] by the area of [latex]S_{ij}[\/latex]. Therefore, the mass of fluid per unit time flowing across [latex]S_{ij}[\/latex] in the direction of [latex]\\mathbf{N}[\/latex] can be approximated by [latex](\\rho \\mathbf{v} \\cdot \\mathbf{N}) \\Delta S_{ij}[\/latex], where [latex]\\mathbf{N}[\/latex], [latex]\\rho[\/latex], and [latex]\\mathbf{v}[\/latex] are all evaluated at [latex]P[\/latex] (Figure 5).<\/p>\r\n

This is analogous to the flux of two-dimensional vector field [latex]\\mathbf{F}[\/latex] across plane curve [latex]C[\/latex], in which we approximated flux across a small piece of [latex]C[\/latex] with the expression [latex](\\mathbf{F} \\cdot \\mathbf{N}) \\Delta s[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_1497\" align=\"aligncenter\" width=\"497\"]\"A Figure 5. The mass of fluid per unit time flowing across [latex]S_{ij}[\/latex] in the direction of\u00a0[latex]{\\bf{N}}[\/latex] can be approximated by [latex](\\rho{\\bf{v}}\\cdot{\\bf{N}})\\nabla{S_{ij}}[\/latex].[\/caption]\r\n

To approximate the mass flux across [latex]S[\/latex], form the sum [latex]\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n(\\rho{\\bf{v}}\\cdot{\\bf{N}})\\nabla{S_{ij}}[\/latex] As pieces [latex]S_{ij}[\/latex] get smaller, this sum gets arbitrarily close to the mass flux. Therefore, the mass flux is:<\/p>\r\n

[latex]\\displaystyle\\iint_s\\rho{\\bf{v}}\\cdot{\\bf{N}}dS=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n(\\rho{\\bf{v}}\\cdot{\\bf{N}})\\nabla{S_{ij}}[\/latex].<\/p>\r\n

This is a surface integral of a vector field. Letting the vector field [latex]\\rho \\mathbf{v}[\/latex] be an arbitrary vector field [latex]\\mathbf{F}[\/latex] leads to the following definition.<\/p>\r\n\r\n

<\/div>\r\n
\r\n

surface integral of a vector field<\/h3>\r\n

Let [latex]\\mathbf{F}[\/latex] be a continuous vector field with a domain that contains oriented surface [latex]S[\/latex] with unit normal vector [latex]\\mathbf{N}[\/latex]. The surface integral<\/strong> of [latex]\\mathbf{F}[\/latex] over [latex]S[\/latex] is:<\/p>\r\n

[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{d}S=\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S[\/latex].<\/p>\r\n

This integral is also called the flux<\/strong> of [latex]\\mathbf{F}[\/latex] across [latex]S[\/latex].<\/p>\r\n\r\n<\/section>

Notice the parallel between this definition and the definition of vector line integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}{d}s[\/latex]. A surface integral of a vector field is defined in a similar way to a flux line integral across a curve, except the domain of integration is a surface (two-dimensional) rather than a curve (one-dimensional).<\/section>\r\n

Computing Surface Integrals of Vector Fields<\/h3>\r\n

Just as with vector line integrals, surface integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S[\/latex] is easier to compute after surface [latex]S[\/latex] has been parameterized. Let [latex]\\mathbf{r}(u,v)[\/latex] be a parameterization of [latex]S[\/latex] with parameter domain [latex]D[\/latex].<\/p>\r\n

The unit normal vector is given by [latex]\\mathbf{N} = \\frac{\\mathbf{t}_u \\times \\mathbf{t}_v}{||\\mathbf{t}_u \\times \\mathbf{t}_v||}[\/latex]. Using the surface integral formula for scalar-valued functions, we have:<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S&=\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S \\\\\r\n&=\\displaystyle\\iint_S{\\bf{F}}\\cdot\\frac{{\\bf{t}}_u\\times{\\bf{t}}_v}{||{\\bf{t}}_u\\times{\\bf{t}}_v||}dS \\\\\r\n&=\\displaystyle\\iint_D\\left({\\bf{F}}({\\bf{r}}(u,v))\\cdot\\frac{{\\bf{t}}_u\\times{\\bf{t}}_v}{||{\\bf{t}}_u\\times{\\bf{t}}_v||}\\right)||{\\bf{t}}_u\\times{\\bf{t}}_v||dA \\\\\r\n&=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA\r\n\\end{aligned}[\/latex].<\/p>\r\n

Notice that the magnitude terms cancel, simplifying the computation considerably.<\/p>\r\n\r\n

\r\n

computing surface integrals<\/h3>\r\n

For a vector field [latex]\\mathbf{F}[\/latex] and oriented surface [latex]S[\/latex] with parameterization [latex]\\mathbf{r}(u,v)[\/latex] over domain [latex]D[\/latex]:<\/p>\r\n

[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA[\/latex].<\/p>\r\n\r\n<\/section>

Calculate the surface integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S[\/latex], where [latex]{\\bf{F}}=\\langle-y,x,0\\rangle[\/latex] and [latex]S[\/latex] is the surface with parameterization [latex]{\\bf{r}}(u,v)=\\langle{u},v^2-u,u+v\\rangle[\/latex], [latex]0\\leq{u}\\leq3, \\ 0\\leq{v}\\leq4[\/latex].[reveal-answer q=\"812467498\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"812467498\"]\r\n

The tangent vectors are [latex]{\\bf{t}}_u=\\langle1,-1,1\\rangle[\/latex] and [latex]{\\bf{t}}_v=\\langle0,2v,1\\rangle[\/latex]. Therefore,<\/p>\r\n

[latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\langle-1-2v,-1,2v\\rangle[\/latex].<\/p>\r\n

By\u00a0[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA[\/latex],<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\iint_S{\\bf{F}}\\cdot{d}S&=\\displaystyle\\int_0^4\\displaystyle\\int_0^3{\\bf{F}}({\\bf{r}}(u,v))\\cdot{\\bf{t}}_u\\times{\\bf{t}}_v)dudv \\\\\r\n&=\\displaystyle\\int_0^4\\displaystyle\\int_0^3\\langle{u}-v^2,u,0\\rangle\\cdot\\langle-1-2v,-1,2v\\rangle{d}udv \\\\\r\n&=\\displaystyle\\int_0^4\\displaystyle\\int_0^3[(u-v^2)(-1-2v)-u]dudv \\\\\r\n&=\\displaystyle\\int_0^4\\displaystyle\\int_0^3(2v^3+v^2-2uv-2u]dudv \\\\\r\n&=\\displaystyle\\int_0^4[2v^3u+v^2u-vu^2-u^2]_0^3dv \\\\\r\n&=\\displaystyle\\int_0^4(6v^3+3v^2-9v-9)dv \\\\\r\n&=\\left[\\frac{3v^2}2+v^3-\\frac{9v^2}2-9v\\right]_0^4 \\\\\r\n&=340\r\n\\end{aligned}[\/latex].<\/p>\r\nTherefore, the flux of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0across [latex]S[\/latex] is [latex]340[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Let [latex]{\\bf{v}}(x,y,z)=\\langle2x,2y,z\\rangle[\/latex] represent a velocity field (with units of meters per second) of a fluid with constant density [latex]80[\/latex] kg\/m3<\/sup>. Let [latex]S[\/latex] be hemisphere [latex]x^{2}+y^{2}+z^{2}=9[\/latex] with [latex]z\\geq 0[\/latex] such that [latex]S[\/latex] is oriented outward. Find the mass flow rate of the fluid across [latex]S[\/latex].[reveal-answer q=\"682315602\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"682315602\"]\r\n

A parameterization of the surface is:<\/p>\r\n

[latex]{\\bf{r}}(\\phi,\\theta)=\\langle3\\cos\\theta\\sin\\phi,3\\sin\\theta\\sin\\phi,3\\cos\\phi\\rangle, \\ 0\\leq\\theta\\leq2\\pi, \\ 0\\leq\\phi\\leq\\pi\/2[\/latex].<\/p>\r\n

The tangent vectors are:<\/p>\r\n

[latex]{\\bf{t}}_\\theta\\langle-3\\sin\\theta\\sin\\phi,3\\cos\\theta\\sin\\phi,0\\rangle\\text{ and }{\\bf{t}}_\\phi\\langle3\\cos\\theta\\cos\\phi,3\\sin\\theta\\cos\\phi,-3\\sin\\phi\\rangle[\/latex],<\/p>\r\n

and their cross product is:<\/p>\r\n

[latex]{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta=\\langle9\\cos\\theta\\sin^2\\phi,9\\sin\\theta\\sin^2\\phi,9\\sin\\phi\\cos\\phi\\rangle d\\phid\\theta[\/latex].<\/p>\r\n

Notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. Therefore we use the orientation [latex]{\\bf{N}}=\\langle9\\cos\\theta\\sin^2\\phi,9\\sin\\theta\\sin^2\\phi,9\\sin\\phi\\cos\\phi\\rangle[\/latex] for the sphere.<\/p>\r\n

By the surface integral equation,<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\iint_S\\rho{\\bf{v}}\\cdot{d}{\\bf{S}}&=80\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^{\\pi\/2}{\\bf{v}}({\\bf{r}}(\\phi,\\theta))\\cdot({\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta)d\\phi{d}\\theta \\\\\r\n&=80\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^{\\pi\/2} \\\\\r\n&=80\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^{\\pi\/2}\\langle6\\cos\\theta\\sin\\phi,6\\sin\\theta\\sin\\phi,3\\cos\\phi\\rangle\\cdot\\langle9\\cos\\theta\\sin^2\\phi,9\\sin\\theta\\sin^2\\phi,9\\sin\\phi\\cos\\phi\\rangle \\\\\r\n&=80\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^{\\pi\/2}54\\sin^2\\phi+27\\cos^2\\phi\\sin\\phi{d}\\phi{d}\\theta \\\\\r\n&=80\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^{\\pi\/2}54(1-\\cos^2\\phi)\\sin\\phi+27\\cos^2\\phi\\sin\\phi{d}\\phi{d}\\theta \\\\\r\n&=80\\displaystyle\\int_0^{2\\pi}\\displaystyle\\int_0^{\\pi\/2}54\\sin\\phi-27\\cos^2\\phi\\sin\\phi{d}\\phi{d}\\theta \\\\\r\n&=80\\displaystyle\\int_0^{2\\pi}[-54\\cos\\phi+9\\cos^3\\phi]_{\\phi=0}^{\\phi=2\\pi}d\\theta \\\\\r\n&=80\\displaystyle\\int_0^{2\\pi}45d\\theta=7,200\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n

Therefore, the mass flow rate is [latex]7,200\\pi\\text{ kg\/sec\/m^2}[\/latex].\u00a0<\/span><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>","rendered":"

Surface Integral of a Vector Field<\/h2>\n

With the idea of orientable surfaces in place, we can now define a surface integral of a vector field. The definition is analogous to the definition of the flux of a vector field along a plane curve.<\/p>\n

\n

Flux of a Vector Field Along a Curve<\/strong><\/p>\n

If [latex]\\mathbf{F}[\/latex] is a two-dimensional vector field and [latex]C[\/latex] is a plane curve, then the flux of [latex]\\mathbf{F}[\/latex] along [latex]C[\/latex] involves chopping [latex]C[\/latex] into small pieces, choosing a point inside each piece, and calculating [latex]\\mathbf{F} \\cdot \\mathbf{N}[\/latex] at the point (where [latex]\\mathbf{N}[\/latex] is the unit normal vector at the point).<\/p>\n<\/section>\n

The definition of a surface integral of a vector field proceeds in the same fashion, except now we chop surface [latex]S[\/latex] into small pieces, choose a point in each (two-dimensional) piece, and calculate [latex]\\mathbf{F} \\cdot \\mathbf{N}[\/latex] at the point.<\/p>\n

\n

To place this definition in a real-world setting, let [latex]S[\/latex] be an oriented surface with unit normal vector [latex]\\mathbf{N}[\/latex]. Let [latex]\\mathbf{v}[\/latex] be a velocity field of a fluid flowing through [latex]S[\/latex], and suppose the fluid has density [latex]\\rho(x,y,z)[\/latex].<\/p>\n

Imagine the fluid flows through [latex]S[\/latex], but [latex]S[\/latex] is completely permeable so that it does not impede the fluid flow (Figure 4). The mass flux<\/strong> of the fluid is the rate of mass flow per unit area. The mass flux is measured in mass per unit time per unit area.<\/p>\n

\n
\"A
Figure 4. Fluid flows across a completely permeable surface [latex]S[\/latex].<\/figcaption><\/figure>\n<\/div>\n<\/section>\n
\n

How could we calculate the mass flux of the fluid across [latex]S[\/latex]?<\/p>\n

The rate of flow, measured in mass per unit time per unit area, is [latex]\\rho \\mathbf{v}[\/latex]. To calculate the mass flux across [latex]S[\/latex], chop [latex]S[\/latex] into small pieces [latex]S_{ij}[\/latex].<\/p>\n

If [latex]S_{ij}[\/latex] is small enough, then it can be approximated by a tangent plane at some point [latex]P[\/latex] in [latex]S_{ij}[\/latex]. Therefore, the unit normal vector at [latex]P[\/latex] can be used to approximate [latex]\\mathbf{N}(x,y,z)[\/latex] across the entire piece [latex]S_{ij}[\/latex], because the normal vector to a plane does not change as we move across the plane.<\/p>\n

The component of the vector [latex]\\rho \\mathbf{v}[\/latex] at [latex]P[\/latex] in the direction of [latex]\\mathbf{N}[\/latex] is [latex]\\rho \\mathbf{v} \\cdot \\mathbf{N}[\/latex] at [latex]P[\/latex]. Since [latex]S_{ij}[\/latex] is small, the dot product [latex]\\rho \\mathbf{v} \\cdot \\mathbf{N}[\/latex] changes very little as we vary across [latex]S_{ij}[\/latex], and therefore [latex]\\rho \\mathbf{v} \\cdot \\mathbf{N}[\/latex] can be taken as approximately constant across [latex]S_{ij}[\/latex].<\/p>\n

To approximate the mass of fluid per unit time flowing across [latex]S_{ij}[\/latex] (and not just locally at point [latex]P[\/latex]), we need to multiply [latex](\\rho \\mathbf{v} \\cdot \\mathbf{N})(P)[\/latex] by the area of [latex]S_{ij}[\/latex]. Therefore, the mass of fluid per unit time flowing across [latex]S_{ij}[\/latex] in the direction of [latex]\\mathbf{N}[\/latex] can be approximated by [latex](\\rho \\mathbf{v} \\cdot \\mathbf{N}) \\Delta S_{ij}[\/latex], where [latex]\\mathbf{N}[\/latex], [latex]\\rho[\/latex], and [latex]\\mathbf{v}[\/latex] are all evaluated at [latex]P[\/latex] (Figure 5).<\/p>\n

This is analogous to the flux of two-dimensional vector field [latex]\\mathbf{F}[\/latex] across plane curve [latex]C[\/latex], in which we approximated flux across a small piece of [latex]C[\/latex] with the expression [latex](\\mathbf{F} \\cdot \\mathbf{N}) \\Delta s[\/latex].<\/p>\n

\"A
Figure 5. The mass of fluid per unit time flowing across [latex]S_{ij}[\/latex] in the direction of\u00a0[latex]{\\bf{N}}[\/latex] can be approximated by [latex](\\rho{\\bf{v}}\\cdot{\\bf{N}})\\nabla{S_{ij}}[\/latex].<\/figcaption><\/figure>\n

To approximate the mass flux across [latex]S[\/latex], form the sum [latex]\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n(\\rho{\\bf{v}}\\cdot{\\bf{N}})\\nabla{S_{ij}}[\/latex] As pieces [latex]S_{ij}[\/latex] get smaller, this sum gets arbitrarily close to the mass flux. Therefore, the mass flux is:<\/p>\n

[latex]\\displaystyle\\iint_s\\rho{\\bf{v}}\\cdot{\\bf{N}}dS=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n(\\rho{\\bf{v}}\\cdot{\\bf{N}})\\nabla{S_{ij}}[\/latex].<\/p>\n

This is a surface integral of a vector field. Letting the vector field [latex]\\rho \\mathbf{v}[\/latex] be an arbitrary vector field [latex]\\mathbf{F}[\/latex] leads to the following definition.<\/p>\n

<\/div>\n
\n

surface integral of a vector field<\/h3>\n

Let [latex]\\mathbf{F}[\/latex] be a continuous vector field with a domain that contains oriented surface [latex]S[\/latex] with unit normal vector [latex]\\mathbf{N}[\/latex]. The surface integral<\/strong> of [latex]\\mathbf{F}[\/latex] over [latex]S[\/latex] is:<\/p>\n

[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{d}S=\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S[\/latex].<\/p>\n

This integral is also called the flux<\/strong> of [latex]\\mathbf{F}[\/latex] across [latex]S[\/latex].<\/p>\n<\/section>\n

Notice the parallel between this definition and the definition of vector line integral [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}{d}s[\/latex]. A surface integral of a vector field is defined in a similar way to a flux line integral across a curve, except the domain of integration is a surface (two-dimensional) rather than a curve (one-dimensional).<\/section>\n

Computing Surface Integrals of Vector Fields<\/h3>\n

Just as with vector line integrals, surface integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S[\/latex] is easier to compute after surface [latex]S[\/latex] has been parameterized. Let [latex]\\mathbf{r}(u,v)[\/latex] be a parameterization of [latex]S[\/latex] with parameter domain [latex]D[\/latex].<\/p>\n

The unit normal vector is given by [latex]\\mathbf{N} = \\frac{\\mathbf{t}_u \\times \\mathbf{t}_v}{||\\mathbf{t}_u \\times \\mathbf{t}_v||}[\/latex]. Using the surface integral formula for scalar-valued functions, we have:<\/p>\n

[latex]\\begin{aligned} \\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S&=\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S \\\\ &=\\displaystyle\\iint_S{\\bf{F}}\\cdot\\frac{{\\bf{t}}_u\\times{\\bf{t}}_v}{||{\\bf{t}}_u\\times{\\bf{t}}_v||}dS \\\\ &=\\displaystyle\\iint_D\\left({\\bf{F}}({\\bf{r}}(u,v))\\cdot\\frac{{\\bf{t}}_u\\times{\\bf{t}}_v}{||{\\bf{t}}_u\\times{\\bf{t}}_v||}\\right)||{\\bf{t}}_u\\times{\\bf{t}}_v||dA \\\\ &=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA \\end{aligned}[\/latex].<\/p>\n

Notice that the magnitude terms cancel, simplifying the computation considerably.<\/p>\n

\n

computing surface integrals<\/h3>\n

For a vector field [latex]\\mathbf{F}[\/latex] and oriented surface [latex]S[\/latex] with parameterization [latex]\\mathbf{r}(u,v)[\/latex] over domain [latex]D[\/latex]:<\/p>\n

[latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S=\\displaystyle\\iint_D({\\bf{F}}({\\bf{r}}(u,v))\\cdot({\\bf{t}}_u\\times{\\bf{t}}_v))dA[\/latex].<\/p>\n<\/section>\n

Calculate the surface integral [latex]\\displaystyle\\iint_S{\\bf{F}}\\cdot{\\bf{N}}{d}S[\/latex], where [latex]{\\bf{F}}=\\langle-y,x,0\\rangle[\/latex] and [latex]S[\/latex] is the surface with parameterization [latex]{\\bf{r}}(u,v)=\\langle{u},v^2-u,u+v\\rangle[\/latex], [latex]0\\leq{u}\\leq3, \\ 0\\leq{v}\\leq4[\/latex].<\/p>\n