{"id":1452,"date":"2025-10-17T15:29:38","date_gmt":"2025-10-17T15:29:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1452"},"modified":"2025-11-06T18:51:27","modified_gmt":"2025-11-06T18:51:27","slug":"surface-integrals-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/surface-integrals-learn-it-3\/","title":{"raw":"Surface Integrals: Learn It 3","rendered":"Surface Integrals: Learn It 3"},"content":{"raw":"

Surface Integral of a Scalar-Valued Function<\/h2>\r\n

Now that we can parameterize surfaces and calculate their surface areas, we can define surface integrals. Let's start with the surface integral of a scalar-valued function.<\/p>\r\n

A surface integral of a scalar-valued function is similar to a scalar line integral, but in one higher dimension. For a line integral, the domain of integration is a parameterized curve (one-dimensional). For a surface integral, the domain of integration is a parameterized surface (two-dimensional).<\/p>\r\n

The definition follows the same logic as line integrals. For scalar line integrals, we chop the domain curve into tiny pieces, choose a point in each piece, evaluate the function at that point, and take a limit of the corresponding Riemann sum. For scalar surface integrals, we chop the domain surface into tiny pieces and proceed in the same fashion.<\/p>\r\n\r\n

Defining the Surface Integral<\/h3>\r\n

Let [latex]S[\/latex] be a piecewise smooth surface with parameterization [latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex] with parameter domain [latex]D[\/latex]. Let [latex]f(x,y,z)[\/latex] be a function with a domain that contains [latex]S[\/latex].<\/p>\r\n

For simplicity, assume the parameter domain [latex]D[\/latex] is a rectangle (we can extend this to any parameter domain). Divide rectangle [latex]D[\/latex] into subrectangles [latex]D_{ij}[\/latex] with horizontal width [latex]\\Delta u[\/latex] and vertical length [latex]\\Delta v[\/latex].<\/p>\r\n

Suppose [latex]i[\/latex] ranges from [latex]1[\/latex] to [latex]m[\/latex] and [latex]j[\/latex] ranges from [latex]1[\/latex] to [latex]n[\/latex], so [latex]D[\/latex] is subdivided into [latex]mn[\/latex] rectangles. This division of [latex]D[\/latex] into subrectangles gives a corresponding division of [latex]S[\/latex] into pieces [latex]S_{ij}[\/latex].<\/p>\r\n

Choose point [latex]P_{ij}[\/latex] in each piece [latex]S_{ij}[\/latex], evaluate [latex]f[\/latex] at [latex]P_{ij}[\/latex], and multiply by the area of [latex]S_{ij}[\/latex] to form the Riemann sum:<\/p>\r\n

[latex]\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n f(P_{ij}) \\Delta S_{ij}[\/latex]<\/p>\r\n\r\n

\r\n

surface integral of a scalar-valued function<\/h3>\r\n

The\u00a0surface integral of a scalar-valued function<\/span><\/strong> of [latex]f[\/latex] over a piecewise smooth surface [latex]S[\/latex] is:<\/p>\r\n

[latex]\\large{\\displaystyle\\iint_Sf(x,y,z)dS=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(P_{ij})\\Delta{S_{ij}}}[\/latex].<\/p>\r\n\r\n<\/section>

\r\n

Notice the similarities between this definition and the definition of a scalar line integral. In the line integral, we chop a curve into pieces, evaluate a function at a point in each piece, and let the length of the pieces shrink to zero. In the surface integral, we chop a surface into pieces, evaluate a function at a point in each piece, and let the area of the pieces shrink to zero.<\/p>\r\n\r\n<\/section>This definition extends to parameter domains that are not rectangles by using the same logic. The basic idea is to chop the parameter domain into small pieces, choose a sample point in each piece, and so on. The exact shape of each piece becomes irrelevant as the areas shrink to zero.\r\n

Computing Surface Integrals<\/h3>\r\nScalar surface integrals are difficult to compute from the definition, just as scalar line integrals are. To make surface integrals easier to compute, we approximate surface areas [latex]\\Delta S_{ij}[\/latex] with small pieces of a tangent plane, as we did when calculating surface area.\r\n\r\nRecall the definition of vectors [latex]{\\bf{t}}_u[\/latex] and [latex]{\\bf{t}}_v[\/latex]:\r\n

[latex]\\mathbf{t}_u = \\left\\langle \\dfrac{\\partial x}{\\partial u}, \\dfrac{\\partial y}{\\partial u}, \\dfrac{\\partial z}{\\partial u} \\right\\rangle[\/latex] and [latex]\\mathbf{t}_v = \\left\\langle \\dfrac{\\partial x}{\\partial v}, \\dfrac{\\partial y}{\\partial v}, \\dfrac{\\partial z}{\\partial v} \\right\\rangle[\/latex]<\/p>\r\n

From our work with surface area, we know that:<\/p>\r\n

[latex]\\Delta{S}_{ij}\\approx||{\\bf{t}}_u(P_{ij})\\times{\\bf{t}}_v(P_{ij})||\\Delta{u}\\Delta{v}[\/latex]<\/p>\r\nUsing this approximation in our Riemann sum gives:\r\n

[latex]\\displaystyle\\iint_Sf(x,y,z)dS\\approx\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(P_{ij})||{\\bf{t}}_u(P_{ij})\\times{\\bf{t}}_v(P_{ij})||\\Delta{u}\\Delta{v}[\/latex].<\/p>\r\n

This approximation becomes arbitrarily close to [latex]\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(P_{ij})\\Delta{S}_{ij}[\/latex] as we increase the number of pieces [latex]S_{ij}[\/latex] by letting [latex]m[\/latex] and [latex]n[\/latex] go to infinity.<\/p>\r\n

We can now state the formula for calculating scalar surface integrals:<\/p>\r\n

[latex]\\displaystyle\\iint_Sf(x,y,z)dS=\\displaystyle\\iint_Df({\\bf{r}}(u,v)||{\\bf{t}}_u\\times{\\bf{t}}_v||dA[\/latex].<\/p>\r\n\r\n

\r\n

calculating scalar surface integrals<\/h3>\r\n

Let [latex]S[\/latex] be a smooth surface with parameterization [latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex] over parameter domain [latex]D[\/latex]. Then:<\/p>\r\n

[latex]\\displaystyle\\iint_Sf(x,y,z)dS=\\displaystyle\\iint_Df(\\mathbf{r}(u,v)) ||\\mathbf{t}_u \\times \\mathbf{t}_v|| \\, dA[\/latex]<\/p>\r\n

where [latex]\\mathbf{t}_u[\/latex] and [latex]\\mathbf{t}_v[\/latex] are the tangent vectors with respect to [latex]u[\/latex] and [latex]v[\/latex].<\/p>\r\n\r\n<\/section>

\r\n

Line Integrals vs. Surface Integrals<\/strong><\/p>\r\n

The surface integral formula transforms a surface integral into a double integral. This is analogous to the line integral formula:<\/p>\r\n

[latex]\\displaystyle\\iint_C f(x,y,z) \\, ds = \\displaystyle\\int_a^bf(\\mathbf{r}(t)) ||\\mathbf{r}'(t)|| , dt[\/latex]<\/p>\r\n

In the surface integral case, vector [latex]\\mathbf{t}_u \\times \\mathbf{t}_v[\/latex] is perpendicular to the surface. In the line integral case, vector [latex]\\mathbf{r}'[\/latex] is tangent to the curve.<\/p>\r\n\r\n<\/section>

Calculate surface integral [latex]\\displaystyle\\iint_S5dS[\/latex], where [latex]S[\/latex] is the surface with parameterization [latex]{\\bf{r}}(u,v)=\\langle{v},u^2,v\\rangle[\/latex] for [latex]0\\leq{u}\\leq2[\/latex] and\u00a0[latex]0\\leq{v}\\leq{u}[\/latex].[reveal-answer q=\"768642501\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"768642501\"]\r\n

Notice that this parameter domain [latex]D[\/latex] is a triangle, and therefore the parameter domain is not rectangular. This is not an issue though, because the surface integral equation\u00a0does not place any restrictions on the shape of the parameter domain.<\/p>\r\n

To use the surface integral equation\u00a0to calculate the surface integral, we first find vector [latex]{\\bf{t}}_u[\/latex] and [latex]{\\bf{t}}_v[\/latex]. Note that [latex]{\\bf{t}}_u=\\langle1,2u,0\\rangle[\/latex] and [latex]{\\bf{t}}_v=\\langle0,0,1\\rangle[\/latex]. Therefore,<\/p>\r\n

[latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ 1&2u&0 \\\\ 0&0&1\\end{vmatrix}=\\langle2u,-1,0\\rangle[\/latex]<\/p>\r\n

and<\/p>\r\n

[latex]||{\\bf{t}}_u\\times{\\bf{t}}_v||=\\sqrt{1+4u^2}[\/latex].<\/p>\r\n

By\u00a0the surface integral equation,<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\iint_S5dS&=5\\displaystyle\\iint_D\\sqrt{1+4u^2}dA \\\\\r\n&=5\\displaystyle\\int_0^2\\int_0^u\\sqrt{1+4u^2}dvdu=5\\displaystyle\\int_0^2u\\sqrt{1+4u^2}du \\\\\r\n&=5\\left[\\frac{(1+4u^2)^{3\/2}}3\\right]_0^2=\\frac{5(17^{3\/2}-1)}{12}\\approx28.79\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

\r\n

Calculate surface integral [latex]\\displaystyle\\iint_S(x+y^2)dS[\/latex], where [latex]S[\/latex] is cylinder [latex]x^2+y^2=4[\/latex], [latex]0\\leq{z}\\leq3[\/latex] (Figure 1).<\/p>\r\n\r\n[caption id=\"attachment_1484\" align=\"aligncenter\" width=\"264\"]\"A Figure 1. Integrating function [latex]f(x,y,z)=x+y^2[\/latex] over a cylinder.[\/caption][reveal-answer q=\"247755319\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"247755319\"]\r\n

To calculate the surface integral, we first need a parameterization of the cylinder. Following our previous examples, a parameterization is:<\/p>\r\n

[latex]{\\bf{r}}(u,v)=\\langle2\\cos{u},2\\sin{u},v\\rangle, \\ 0\\leq{u}\\leq2\\pi, \\ 0\\leq{v}\\leq3[\/latex].<\/p>\r\n

The tangent vectors are [latex]{\\bf{t}}_u=\\langle\\sin{u},\\cos{u},0\\rangle[\/latex] and [latex]{\\bf{t}}_v=\\langle0,0,1\\rangle[\/latex]. Then,<\/p>\r\n

[latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ -\\sin{u}&\\cos{u}&0 \\\\ 0&0&1\\end{vmatrix}=\\langle\\cos{u},\\sin{u},0\\rangle[\/latex]<\/p>\r\n

and [latex]||{\\bf{t}}_u\\times{\\bf{t}}_v||=\\sqrt{\\cos^2u+\\sin^2u}=1[\/latex]. By the surface integral equation,<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\iint_Sf(x,y,z)dS&=\\displaystyle\\iint_Df({\\bf{r}}(u,v))||{\\bf{t}}_u\\times{\\bf{t}}_v|| \\ dA \\\\\r\n&=\\displaystyle\\int_0^3\\displaystyle\\int_0^{2\\pi}(2\\cos{u}+4\\sin^2u)dydv \\\\\r\n&=\\displaystyle\\int_0^3[2\\sin{u}+2u-\\sin(2u)]_0^{2\\pi}dv=\\displaystyle\\int_0^34\\pi{d}v=12\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Calculate surface integral [latex]\\displaystyle\\iint_Sf(x,y,z)dS[\/latex], where [latex]f(x, y, z)=z^{2}[\/latex] and [latex]S[\/latex] is the surface that consists of the piece of sphere [latex]x^{2}+y^{2}+z^{2}=4[\/latex] that lies on or above plane [latex]z=1[\/latex] and the disk that is enclosed by intersection plane [latex]z=1[\/latex] and the given sphere (Figure 2).[caption id=\"attachment_1486\" align=\"aligncenter\" width=\"407\"]\"A Figure 2. Calculating a surface integral over surface [latex]S[\/latex].[\/caption][reveal-answer q=\"223579284\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"223579284\"]\r\n
\r\n

Notice that [latex]S[\/latex] is not smooth but is piecewise smooth; [latex]S[\/latex] can be written as the union of its base [latex]S_1[\/latex] and its spherical top [latex]S_2[\/latex], and both [latex]S_1[\/latex] and [latex]S_2[\/latex] are smooth. Therefore, to calculate [latex]\\displaystyle\\iint_Sz^2dS[\/latex], we write this integral as [latex]\\displaystyle\\iint_Sz^2d{S_1}+\\displaystyle\\iint_{S_2}z^2dS[\/latex] and we calculate integrals [latex]\\displaystyle\\iint_{S_1}z^2dS[\/latex] and [latex]\\displaystyle\\iint_{S_2}z^2dS[\/latex].<\/span><\/h2>\r\n
\r\n

First, we calculate [latex]\\displaystyle\\iint_{S_1}z^2dS[\/latex]. To calculate this integral we need a parameterization of [latex]S_1[\/latex]. This surface is a disk in plane [latex]z=1[\/latex] centered at [latex](0, 0, 1)[\/latex]. To parameterize this disk, we need to know its radius. Since the disk is formed where plane [latex]z=1[\/latex] intersects sphere [latex]x^{2}+y^{2}+z^{2}=4[\/latex], we can substitute [latex]z=1[\/latex] into equation [latex]x^{2}+y^{2}+z^{2}=4[\/latex]:<\/p>\r\n

[latex]x^2+y^2+1=4\\Rightarrow{x}^2+y^2=3[\/latex].<\/p>\r\n

Therefore, the radius of the disk is [latex]\\sqrt3[\/latex] and a parameterization of [latex]S_1[\/latex] is [latex]{\\bf{r}}(u,v)=\\langle{u}\\cos{v},u\\sin{v},1\\rangle[\/latex], [latex]0\\leq{u}\\leq\\sqrt3[\/latex],[latex]0\\leq{v}\\leq2\\pi[\/latex]. The tangent vectors are [latex]{\\bf{t}}_u=\\langle\\cos{v},\\sin{v},0\\rangle[\/latex] and [latex]{\\bf{t}}_v=\\langle-u\\sin{v},u\\cos{v},0\\rangle[\/latex], and thus<\/p>\r\n

[latex]{\\bf{t}}_u\\times{\\bf{t}}_v=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ \\cos{v}&\\sin{v}&0 \\\\-u\\sin{v}&u\\cos{v}&0\\end{vmatrix}=\\langle0,0,u\\cos^2v+u\\sin^2v\\rangle=\\langle0,0,u\\rangle[\/latex].<\/p>\r\n

The magnitude of this vector is [latex]u[\/latex]. Therefore,<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\iint_{S_1}z^2dS&=\\displaystyle\\int_0^{\\sqrt3}\\displaystyle\\int_0^{2\\pi}f({\\bf{r}}(u,v))||{\\bf{t}}_u\\times{\\bf{t}}_v|| \\ dvdu \\\\\r\n&=\\displaystyle\\int_0^{\\sqrt3}\\displaystyle\\int_0^{2\\pi}u \\ dvdu \\\\\r\n&=2\\pi\\displaystyle\\int_0^{\\sqrt3} \\ udu \\\\\r\n&=3\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n

Now we calculate [latex]\\displaystyle\\iint_{S_2} \\ dS[\/latex]. To calculate this integral, we need a parameterization of [latex]S_2[\/latex]. The parameterization of full sphere [latex]x^{2}+y^{2}+z^{2}=4[\/latex] is<\/p>\r\n

[latex]{\\bf{r}}(\\phi,\\theta)=\\langle2\\cos\\theta\\sin\\phi,2\\sin\\theta\\sin\\phi,2\\cos\\phi\\rangle, \\ 0\\leq\\theta\\leq2\\pi, \\ 0\\leq\\phi\\leq\\pi[\/latex].<\/p>\r\n

Since we are only taking the piece of the sphere on or above plane [latex]z=1[\/latex], we have to restrict the domain of [latex]\\phi[\/latex]. To see how far this angle sweeps, notice that the angle can be located in a right triangle, as shown in\u00a0Figure 3\u00a0(the [latex]\\sqrt3[\/latex] comes from the fact that the base of [latex]S[\/latex] is a disk with radius [latex]\\sqrt3[\/latex]). Therefore, the tangent of [latex]\\phi[\/latex] is [latex]\\sqrt3[\/latex], which implies that [latex]\\phi[\/latex] is [latex]\\pi\/6[\/latex]. We now have a parameterization of [latex]S_2[\/latex]:<\/p>\r\n

[latex]{\\bf{r}}(\\phi,\\theta)=\\langle2\\cos\\theta\\sin\\phi,2\\sin\\theta\\sin\\phi,2\\cos\\phi\\rangle, \\ 0\\leq\\theta\\leq2\\pi, \\ 0\\leq\\phi\\leq\\pi\/3[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_1487\" align=\"aligncenter\" width=\"422\"]\"A Figure 3. The maximum value of [latex]\\phi[\/latex] has a tangent value of [latex]{\\sqrt3}[\/latex][\/caption]\r\n

The tangent vectors are<\/p>\r\n[latex]{\\bf{t}}_\\phi=\\langle2\\cos\\theta\\cos\\phi,2\\sin\\theta\\cos\\phi,-2\\sin\\phi\\rangle\\text{ and }{\\bf{t}}_\\theta=\\langle-2\\sin\\theta\\sin\\phi,u\\cos\\theta\\sin\\phi,0\\rangle[\/latex],\r\n

and thus<\/p>\r\n

[latex]\\begin{aligned}\r\n{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta&=\\begin{vmatrix}\r\n{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\\r\n2\\cos\\theta\\cos\\phi&2\\sin\\theta\\cos\\phi&-2\\sin\\phi \\\\\r\n-2\\sin\\theta\\sin\\phi&2\\cos\\theta\\sin\\phi&0\r\n\\end{vmatrix} \\\\\r\n&=\\langle4\\cos\\theta\\sin^2\\phi,4\\sin\\theta\\sin^2\\phi,4\\cos^2\\theta\\sin\\phi+4\\sin^2\\theta\\cos\\phi\\sin\\phi\\rangle \\\\\r\n&=\\langle4\\cos\\theta\\sin^2\\phi,4\\sin\\theta\\sin^2\\phi,4\\cos\\phi\\sin\\phi\\rangle\r\n\\end{aligned}[\/latex].<\/p>\r\n

The magnitude of this vector is<\/p>\r\n

[latex]\\begin{aligned}\r\n||{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta||&=\\sqrt{16\\cos^2\\theta\\sin^4\\phi+16\\sin^2\\theta\\sin^4\\phi+16\\cos^2\\phi\\sin^2\\phi} \\\\\r\n&=4\\sqrt{\\sin^4\\phi+\\cos^2\\phi\\sin^2\\phi}\r\n\\end{aligned}[\/latex].<\/p>\r\n

Therefore,<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\iint_{S_2}z^2dS&=\\displaystyle\\int_0^{\\pi\/3}\\displaystyle\\int_0^{2\\pi}f({\\bf{r}}(\\phi,\\theta))||{\\bf{t}}_\\phi\\times{\\bf{t}}_\\theta|| \\ d\\theta{d}\\phi \\\\\r\n&=\\displaystyle\\int_0^{\\pi\/3}\\displaystyle\\int_0^{2\\pi}16\\cos^2\\phi\\sqrt{\\sin^4\\phi+\\cos^2\\phi\\sin^2\\phi}d\\theta{d}\\phi \\\\\r\n&=32\\pi\\displaystyle\\int_0^{\\pi\/3}\\cos^2\\phi\\sqrt{\\sin^4\\phi+\\cos^2\\phi\\sin^2\\phi}d\\phi \\\\\r\n&=32\\pi\\displaystyle\\int_0^{\\pi\/3}\\cos^2\\phi\\sin\\phi\\sqrt{\\sin^2\\phi+\\cos^2\\phi}d\\phi \\\\\r\n&=32\\pi\\displaystyle\\int_0^{\\pi\/3}\\cos^2\\phi\\sin\\phi{d}\\phi \\\\\r\n&=32\\pi\\left[-\\frac{\\cos^3\\phi}3\\right]_0^{\\pi\/6}=32\\pi\\left[\\frac13-\\frac{\\sqrt3}8\\right]=\\frac{28\\pi}3\r\n\\end{aligned}[\/latex].<\/p>\r\nSince\r\n

[latex]\\displaystyle\\iint_Sz^2 \\ dS=\\displaystyle\\iint_{S_1}z^2 \\ dS+\\displaystyle\\iint_{S_2}z^2 \\ dS=3\\pi\\frac{28\\pi}3=\\frac{37\\pi}3[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n

\r\n

Analysis<\/span><\/strong><\/p>\r\n

In this example we broke a surface integral over a piecewise surface into the addition of surface integrals over smooth subsurfaces. There were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

Scalar surface integrals have several real-world applications. Recall that scalar line integrals can be used to compute the mass of a wire given its density function. In a similar fashion, we can use scalar surface integrals to compute the mass of a sheet given its density function. If a thin sheet of metal has the shape of surface [latex]S[\/latex] and the density of the sheet at point [latex](x, y, z)[\/latex]\u00a0is [latex]\\rho(x,y,z)[\/latex], then mass [latex]m[\/latex] of the sheet is:<\/p>\r\n

[latex]m=\\displaystyle\\iint_S\\rho(x,y,z)dS[\/latex].<\/p>\r\n\r\n

A flat sheet of metal has the shape of surface [latex]z=1+x+2y[\/latex] that lies above rectangle [latex]0\\leq{x}\\leq4[\/latex] and [latex]0\\leq{y}\\leq2[\/latex]. If the density of the sheet is given by [latex]\\rho(x,y,z)=x^2yz[\/latex], what is the mass of the sheet?[reveal-answer q=\"689236251\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"689236251\"]\r\n

Let [latex]S[\/latex] be the surface that describes the sheet. Then, the mass of the sheet is given by [latex]m=\\displaystyle\\iint_Sx^2yzdS[\/latex]. To compute this surface integral, we first need a parameterization of [latex]S[\/latex]. Since [latex]S[\/latex] is given by the function [latex]f(x,y)=1+x+2y[\/latex], a parameterization of [latex]S[\/latex] is [latex]{\\bf{r}}(x,y)=\\langle{x},y,1+x+2y\\rangle[\/latex], [latex]0\\leq{x}\\leq4[\/latex], [latex]0\\leq{y}\\leq2[\/latex].<\/p>\r\n

The tangent vectors are [latex]{\\bf{t}}_x=\\langle1,0,1\\rangle[\/latex] and [latex]{\\bf{t}}_y=\\langle1,0,2\\rangle[\/latex]. Therefore, [latex]{\\bf{t}}_x\\times{\\bf{t}}_y=\\langle-1,-2,1\\rangle[\/latex] and [latex]||{\\bf{t}}_x\\times{\\bf{t}}_y||=\\sqrt6[\/latex]. By the scalar line integral equation,<\/p>\r\n

[latex]\\begin{aligned}\r\nm&=\\displaystyle\\iint_Sx^2yzdS \\\\\r\n&=\\sqrt6\\int_0^4\\displaystyle\\int_0^2x^2y(1+x+2y)dydx \\\\\r\n&=\\sqrt6\\displaystyle\\int_0^4\\frac{22x^2}3+2x^3 \\ dx \\\\\r\n&=\\frac{2,560\\sqrt6}9 \\\\\r\n&\\approx696.74\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

Surface Integral of a Scalar-Valued Function<\/h2>\n

Now that we can parameterize surfaces and calculate their surface areas, we can define surface integrals. Let’s start with the surface integral of a scalar-valued function.<\/p>\n

A surface integral of a scalar-valued function is similar to a scalar line integral, but in one higher dimension. For a line integral, the domain of integration is a parameterized curve (one-dimensional). For a surface integral, the domain of integration is a parameterized surface (two-dimensional).<\/p>\n

The definition follows the same logic as line integrals. For scalar line integrals, we chop the domain curve into tiny pieces, choose a point in each piece, evaluate the function at that point, and take a limit of the corresponding Riemann sum. For scalar surface integrals, we chop the domain surface into tiny pieces and proceed in the same fashion.<\/p>\n

Defining the Surface Integral<\/h3>\n

Let [latex]S[\/latex] be a piecewise smooth surface with parameterization [latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex] with parameter domain [latex]D[\/latex]. Let [latex]f(x,y,z)[\/latex] be a function with a domain that contains [latex]S[\/latex].<\/p>\n

For simplicity, assume the parameter domain [latex]D[\/latex] is a rectangle (we can extend this to any parameter domain). Divide rectangle [latex]D[\/latex] into subrectangles [latex]D_{ij}[\/latex] with horizontal width [latex]\\Delta u[\/latex] and vertical length [latex]\\Delta v[\/latex].<\/p>\n

Suppose [latex]i[\/latex] ranges from [latex]1[\/latex] to [latex]m[\/latex] and [latex]j[\/latex] ranges from [latex]1[\/latex] to [latex]n[\/latex], so [latex]D[\/latex] is subdivided into [latex]mn[\/latex] rectangles. This division of [latex]D[\/latex] into subrectangles gives a corresponding division of [latex]S[\/latex] into pieces [latex]S_{ij}[\/latex].<\/p>\n

Choose point [latex]P_{ij}[\/latex] in each piece [latex]S_{ij}[\/latex], evaluate [latex]f[\/latex] at [latex]P_{ij}[\/latex], and multiply by the area of [latex]S_{ij}[\/latex] to form the Riemann sum:<\/p>\n

[latex]\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n f(P_{ij}) \\Delta S_{ij}[\/latex]<\/p>\n

\n

surface integral of a scalar-valued function<\/h3>\n

The\u00a0surface integral of a scalar-valued function<\/span><\/strong> of [latex]f[\/latex] over a piecewise smooth surface [latex]S[\/latex] is:<\/p>\n

[latex]\\large{\\displaystyle\\iint_Sf(x,y,z)dS=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(P_{ij})\\Delta{S_{ij}}}[\/latex].<\/p>\n<\/section>\n

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Notice the similarities between this definition and the definition of a scalar line integral. In the line integral, we chop a curve into pieces, evaluate a function at a point in each piece, and let the length of the pieces shrink to zero. In the surface integral, we chop a surface into pieces, evaluate a function at a point in each piece, and let the area of the pieces shrink to zero.<\/p>\n<\/section>\n

This definition extends to parameter domains that are not rectangles by using the same logic. The basic idea is to chop the parameter domain into small pieces, choose a sample point in each piece, and so on. The exact shape of each piece becomes irrelevant as the areas shrink to zero.<\/p>\n

Computing Surface Integrals<\/h3>\n

Scalar surface integrals are difficult to compute from the definition, just as scalar line integrals are. To make surface integrals easier to compute, we approximate surface areas [latex]\\Delta S_{ij}[\/latex] with small pieces of a tangent plane, as we did when calculating surface area.<\/p>\n

Recall the definition of vectors [latex]{\\bf{t}}_u[\/latex] and [latex]{\\bf{t}}_v[\/latex]:<\/p>\n

[latex]\\mathbf{t}_u = \\left\\langle \\dfrac{\\partial x}{\\partial u}, \\dfrac{\\partial y}{\\partial u}, \\dfrac{\\partial z}{\\partial u} \\right\\rangle[\/latex] and [latex]\\mathbf{t}_v = \\left\\langle \\dfrac{\\partial x}{\\partial v}, \\dfrac{\\partial y}{\\partial v}, \\dfrac{\\partial z}{\\partial v} \\right\\rangle[\/latex]<\/p>\n

From our work with surface area, we know that:<\/p>\n

[latex]\\Delta{S}_{ij}\\approx||{\\bf{t}}_u(P_{ij})\\times{\\bf{t}}_v(P_{ij})||\\Delta{u}\\Delta{v}[\/latex]<\/p>\n

Using this approximation in our Riemann sum gives:<\/p>\n

[latex]\\displaystyle\\iint_Sf(x,y,z)dS\\approx\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(P_{ij})||{\\bf{t}}_u(P_{ij})\\times{\\bf{t}}_v(P_{ij})||\\Delta{u}\\Delta{v}[\/latex].<\/p>\n

This approximation becomes arbitrarily close to [latex]\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(P_{ij})\\Delta{S}_{ij}[\/latex] as we increase the number of pieces [latex]S_{ij}[\/latex] by letting [latex]m[\/latex] and [latex]n[\/latex] go to infinity.<\/p>\n

We can now state the formula for calculating scalar surface integrals:<\/p>\n

[latex]\\displaystyle\\iint_Sf(x,y,z)dS=\\displaystyle\\iint_Df({\\bf{r}}(u,v)||{\\bf{t}}_u\\times{\\bf{t}}_v||dA[\/latex].<\/p>\n

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calculating scalar surface integrals<\/h3>\n

Let [latex]S[\/latex] be a smooth surface with parameterization [latex]\\mathbf{r}(u,v) = \\langle x(u,v), y(u,v), z(u,v) \\rangle[\/latex] over parameter domain [latex]D[\/latex]. Then:<\/p>\n

[latex]\\displaystyle\\iint_Sf(x,y,z)dS=\\displaystyle\\iint_Df(\\mathbf{r}(u,v)) ||\\mathbf{t}_u \\times \\mathbf{t}_v|| \\, dA[\/latex]<\/p>\n

where [latex]\\mathbf{t}_u[\/latex] and [latex]\\mathbf{t}_v[\/latex] are the tangent vectors with respect to [latex]u[\/latex] and [latex]v[\/latex].<\/p>\n<\/section>\n

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Line Integrals vs. Surface Integrals<\/strong><\/p>\n

The surface integral formula transforms a surface integral into a double integral. This is analogous to the line integral formula:<\/p>\n

[latex]\\displaystyle\\iint_C f(x,y,z) \\, ds = \\displaystyle\\int_a^bf(\\mathbf{r}(t)) ||\\mathbf{r}'(t)|| , dt[\/latex]<\/p>\n

In the surface integral case, vector [latex]\\mathbf{t}_u \\times \\mathbf{t}_v[\/latex] is perpendicular to the surface. In the line integral case, vector [latex]\\mathbf{r}'[\/latex] is tangent to the curve.<\/p>\n<\/section>\n

Calculate surface integral [latex]\\displaystyle\\iint_S5dS[\/latex], where [latex]S[\/latex] is the surface with parameterization [latex]{\\bf{r}}(u,v)=\\langle{v},u^2,v\\rangle[\/latex] for [latex]0\\leq{u}\\leq2[\/latex] and\u00a0[latex]0\\leq{v}\\leq{u}[\/latex].<\/p>\n