Surface Integrals: Apply It

Giselle Miller

Surface Integrals: Apply It

Understand and find parametric representations of surfaces
Calculate surface integrals of scalar functions
Calculate surface integrals of vector fields
Use surface integrals to solve real-world problems

Applications: Mass Flux vs. Flow Rate

In an earlier example, we computed the mass flux, which is the rate of mass flow per unit area. If we want to find the flow rate (measured in volume per time) instead, we can use flux integral [latex]\displaystyle\int \displaystyle\int_S{\bf{v}}\cdot{\bf{N}}dS[/latex], which leaves out the density.

Since the flow rate of a fluid is measured in volume per unit time, flow rate does not take mass into account. Therefore, we have the following characterization:

[latex]\text{Flow rate of fluid across } S: \displaystyle\int \displaystyle\int_S{\bf{v}}\cdot{\bf{N}}dS[/latex]

To compute the flow rate of the fluid in the previous example, we simply remove the density constant, which gives a flow rate of [latex]90\pi[/latex] m³/sec.

Both mass flux and flow rate are important in physics and engineering. Mass flux measures how much mass is flowing across a surface; flow rate measures how much volume of fluid is flowing across a surface.

Application: Heat Flow

In addition to modeling fluid flow, surface integrals can be used to model heat flow. Suppose that the temperature at point [latex](x, y, z)[/latex] in an object is [latex]T(x,y,z)[/latex].

Then the heat flow is a vector field proportional to the negative temperature gradient in the object. To be precise, the heat flow is defined as vector field [latex]\mathbf{F} = -k\nabla T[/latex], where the constant [latex]k[/latex] is the thermal conductivity of the substance from which the object is made (this constant is determined experimentally).

The rate of heat flow across surface [latex]S[/latex] in the object is given by the flux integral:

[latex]\displaystyle\iint_S{\bf{F}}\cdot{d}{\bf{S}}=\displaystyle\iint_S-k\nabla{T}\cdot{d}{\bf{S}}[/latex].

A cast-iron solid cylinder is given by inequalities [latex]x^2+y^2\leq1[/latex], [latex]1\leq{z}\leq4[/latex]. The temperature at point [latex](x, y, z)[/latex] in a region containing the cylinder is [latex]T(x,y,z)=(x^2+y^2)z[/latex]. Given that the thermal conductivity of cast iron is [latex]55[/latex], find the heat flow across the boundary of the solid if this boundary is oriented outward.

Let [latex]S[/latex] denote the boundary of the object. To find the heat flow, we need to calculate flux integral [latex]\displaystyle\iint_S-k\nabla{T}\cdot{d}{\bf{S}}[/latex]. Notice that [latex]S[/latex] is not a smooth surface but is piecewise smooth, since [latex]S[/latex] is the union of three smooth surfaces (the circular top and bottom, and the cylindrical side). Therefore, we calculate three separate integrals, one for each smooth piece of [latex]S[/latex]. Before calculating any integrals, note that the gradient of the temperature is [latex]\nabla{T}=\langle2xz,2yz,x^2+y^2\rangle[/latex].

First we consider the circular bottom of the object, which we denote [latex]S_1[/latex]. We can see that [latex]S_1[/latex] is a circle of radius 1 centered at point [latex](0, 0, 1)[/latex], sitting in plane [latex]z=1[/latex]. This surface has parameterization [latex]{\bf{r}}(u,v)=\langle{v}\cos{u},v\sin{u}1\rangle[/latex], [latex]0\leq{u}\leq2\pi, \ 0\leq{v}\leq1[/latex]. Therefore,

[latex]{\bf{t}}_u=\langle-v\sin{u},v\cos{u},0\rangle\text{ and }{\bf{t}}_v=\langle\cos{u},v\sin{u},0\rangle[/latex],

and

[latex]{\bf{t}}_u\times{\bf{t}}_v=\langle0,0,-v\sin^2u-v\cos^2u\rangle=\langle0,0,-v\rangle[/latex].

Since the surface is oriented outward and [latex]S_1[/latex] is the bottom of the object, it makes sense that this vector points downward. By [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], the heat flow across [latex]S_1[/latex] is

[latex]\begin{aligned} \displaystyle\iint_{S_1}-k\nabla{T}\cdot{d}{\bf{S}}&=-55\displaystyle\int_0^{2\pi}\displaystyle\int_0^1\nabla{T}(u,v)\cdot({\bf{t}}_u\times{\bf{t}}_v)dvdu \\ &=-55\displaystyle\int_0^{2\pi}\displaystyle\int_0^1\langle2v\cos{u},2v\sin{u},v^2\cos^2u+v^2\sin^2u\rangle\cdot\langle0,0,-v\rangle{d}vdu \\ &=-55\displaystyle\int_0^{2\pi}\displaystyle\int_0^1\langle2v\cos{u},2v\sin{u},v^2\rangle\cdot\langle0,0,-v\rangle{dvdu} \\ &=-55\displaystyle\int_0^{2\pi}\displaystyle\int_0^1-v^3dvdu=-55\displaystyle\int_0^{2\pi}-\frac14du=\frac{55\pi}2 \end{aligned}[/latex].

Now let’s consider the circular top of the object, which we denote [latex]S_2[/latex]. We see that [latex]S_2[/latex] is a circle of radius 1 centered at point [latex](0, 0, 4)[/latex], sitting in plane [latex]z=4[/latex]. This surface has parameterization [latex]{\bf{r}}(u,v)=\langle{v}\cos{u},v\sin{u},4\rangle[/latex], [latex]0\leq{u}\leq2\pi, \ 0\leq{v}\leq1[/latex]. Therefore,

[latex]{\bf{t}}_u=\langle-v\sin{u},v\cos{u},0\rangle\text{ and }{\bf{v}}\langle\cos{u},v\sin{u},0\rangle[/latex],

and

[latex]{\bf{t}}_u\times{\bf{t}}_v=\langle0,0,-v\sin^2{u}-v\cos^2u\rangle=\langle0,0,-v\rangle[/latex].

Since the surface is oriented outward and [latex]S_1[/latex] is the top of the object, we instead take vector [latex]{\bf{t}}_u\times{\bf{t}}_v=\langle0,0,v\rangle[/latex]. By [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], the heat flow across [latex]S_1[/latex] is

[latex]\begin{aligned} \displaystyle\iint_{S_2}-k\nabla{T}\bullet{d}{\bf{S}}&=-55\displaystyle\int_0^{2\pi}\displaystyle\int_0^1\nabla{T}(u,v)\bullet({\bf{t}}_u\times{\bf{t}}_v)dvdu \\ &=-55\displaystyle\int_0^{2\pi}\displaystyle\int_0^1\langle8v\cos{u},8v\sin{u},v^2\cos^2u+v^2\sin^2u\rangle\bullet\langle0,0,v\rangle{d}vdu \\ &=-55\displaystyle\int_0^{2\pi}\displaystyle\int_0^1\langle8v\cos{u},8v\sin{u},v^2\rangle\bullet\langle0,0,v\rangle{d}vdu \\ &=55\displaystyle\int_0^{2\pi}\displaystyle\int_0^1v^3dvdu=-\frac{55\pi}2 \end{aligned}[/latex].

Last, let’s consider the cylindrical side of the object. This surface has parameterization [latex]{\bf{r}}(u,v)=\langle\cos{u},\sin{u},v\rangle[/latex], [latex]0\leq{u}\leq2\pi, \ 1\leq{v}\leq4[/latex]. By Example “Calculating the Surface Integral of a Cylinder”, we know that [latex]{\bf{t}}_u\times{\bf{t}}_v=\langle\cos{u},\sin{u},0\rangle[/latex]. By [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],

[latex]\begin{aligned} \displaystyle\iint_{S_2}-k\nabla{T}\bullet{d}{\bf{S}}&=-55\displaystyle\int_0^{2\pi}\displaystyle\int_0^1\nabla{T}(u,v)\bullet({\bf{t}}_u\times{\bf{t}}_v)dvdu \\ &=-55\displaystyle\int_0^{2\pi}\displaystyle\int_1^4\langle2v\cos{u},2v\sin{u},\cos^2u+\sin^2u\rangle\bullet\langle\cos{u},\sin{u},0\rangle{d}vdu \\ &=-55\displaystyle\int_0^{2\pi}\displaystyle\int_0^1\langle2v\cos{u},2v\sin{u},1\rangle\bullet\langle\cos{u},\sin{u},0\rangle{d}vdu \\ &=-55\displaystyle\int_0^{2\pi}\displaystyle\int_0^1(2v\cos^2u+2v\sin^2u)dvdu \\ &=-55\displaystyle\int_0^{2\pi}\displaystyle\int_0^12vdvdu=-55\displaystyle\int_0^{2\pi}du=-110\pi \end{aligned}[/latex].

Therefore, the rate of heat flow across [latex]S[/latex] is [latex]\frac{55\pi}2-\frac{55\pi}2-110\pi=-110\pi[/latex].