Section 14.6 Surface Integrals
Subsection 14.6.1 Surface integrals of scalar fields
Definition 14.6.2. Surface Integral.
Let G(x,y,z) be a continuous function defined on a surface S. The surface integral of G on S is
Example 14.6.3. Finding the mass of a thin sheet.
Find the mass of a thin sheet modeled by the plane 2x+y+z=3 over the triangular region of the xy-plane bounded by the coordinate axes and the line y=2β2x, as shown in Figure 14.6.4, with density function Ξ΄(x,y,z)=x2+5y+z, where all distances are measured in cm and the density is given as gm/cm2.
We begin by parametrizing the planar surface \(\surfaceS\text{.}\) Using the techniques of the previous section, we can let \(x=u\) and \(y=v(2-2u)\text{,}\) where \(0\leq u\leq 1\) and \(0\leq v\leq 1\text{.}\) Solving for \(z\) in the equation of the plane, we have \(z=3-2x-y\text{,}\) hence \(z = 3-2u-v(2-2u)\text{,}\) giving the parametrization \(\vec r(u,v) = \langle u, v(2-2u), 3-2u-v(2-2u)\rangle\text{.}\)
We need \(dS=\snorm{\vec r_u\times \vec r_v}dA\text{,}\) so we need to compute \(\vec r_u\text{,}\) \(\vec r_v\) and the norm of their cross product. We leave it to the reader to confirm the following:
We need to be careful to not βsimplifyβ \(\snorm{\vec r_u\times \vec r_v} = 2\sqrt{6}\sqrt{(u-1)^2}\) as \(2\sqrt{6}(u-1)\text{;}\) rather, it is \(2\sqrt{6}|u-1|\text{.}\) In this example, \(u\) is bounded by \(0\leq u\leq 1\text{,}\) and on this interval \(|u-1| = 1-u\text{.}\) Thus \(dS = 2\sqrt{6}(1-u)dA\text{.}\)
The density is given as a function of \(x\text{,}\) \(y\) and \(z\text{,}\) for which we'll substitute the corresponding components of \(\vec r\) (with the slight abuse of notation that we used in previous sections):
Thus the mass of the sheet is:
Subsection 14.6.2 Flux
Let a surface S lie within a vector field βF. One is often interested in measuring the flux of βF across S; that is, measuring βhow much of the vector field passes across S.β For instance, if βF represents the velocity field of moving air and S represents the shape of an air filter, the flux will measure how much air is passing through the filter per unit time. As flux measures the amount of βF passing across S, we need to find the βamount of βF orthogonal to S.β Similar to our measure of flux in the plane, this is equal to βFβ βn, where βn is a unit vector normal to S at a point. We now consider how to find βn.Definition 14.6.6. Flux over a surface.
Let βF be a vector field with continuous components defined on an orientable surface S with normal vector βn. The flux of βF across S is
If S is parametrized by βr(u,v), which is smooth on its domain R, then
Example 14.6.7. Finding flux across a surface.
Let S be the surface given in Example 14.6.3, where S is parametrized by βr(u,v)=β¨u,v(2β2u),3β2uβv(2β2u)β© on 0β€uβ€1, 0β€vβ€1, and let βF=β¨1,x,βyβ©, as shown in Figure 14.6.8. Find the flux of βF across S.
Using our work from the previous example, we have \(\vec n = \vec r_u\times\vec r_v = \langle 4-4u,2-2u,2-2u\rangle\text{.}\) We also need \(\vec F\big(\vec r(u,v)\big) = \langle 1, u, -v(2-2u)\rangle\text{.}\)
Thus the flux of \(\vec F\) across \(\surfaceS\) is:
To make full use of this numeric answer, we need to know the direction in which the field is passing across \(\surfaceS\text{.}\) The graph in Figure 14.6.8 helps, but we need a method that is not dependent on a graph.
Pick a point \((u,v)\) in the interior of \(R\) and consider \(\vec n(u,v)\text{.}\) For instance, choose \((1/2,1/2)\) and look at \(\vec n(1/2,1/2) = \langle 2,1,1\rangle/\sqrt{6}\text{.}\) This vector has positive \(x\text{,}\) \(y\) and \(z\) components. Generally speaking, one has some idea of what the surface \(\surfaceS\) looks like, as that surface is for some reason important. In our case, we know \(\surfaceS\) is a plane with \(z\)-intercept of \(z=3\text{.}\) Knowing \(\vec n\) and the flux measurement of positive \(5/3\text{,}\) we know that the field must be passing from βbehindβ \(\surfaceS\text{,}\) i.e., the side the origin is on, to the βfrontβ of \(\surfaceS\text{.}\)
Example 14.6.10. Flux across surfaces with shared boundaries.
Let S1 be the unit disk in the xy-plane, and let S2 be the paraboloid z=1βx2βy2, for zβ₯0, as graphed in Figure 14.6.11. Note how these two surfaces each have the unit circle as a boundary.
Let βF1=β¨0,0,1β© and βF2=β¨0,0,zβ©. Using normal vectors for each surface that point βupward,β i.e., with a positive z-component, find the flux of each field across each surface.
We begin by parametrizing each surface.
The boundary of the unit disk in the \(xy\)-plane is the unit circle, which can be described with \(\langle \cos u,\sin u,0\rangle\text{,}\) \(0\leq u\leq 2\pi\text{.}\) To obtain the interior of the circle as well, we can scale by \(v\text{,}\) giving
As the boundary of \(\surfaceS_2\) is also the unit circle, the \(x\) and \(y\) components of \(\vec r_2\) will be the same as those of \(\vec r_1\text{;}\) we just need a different \(z\) component. With \(z = 1-x^2-y^2\text{,}\) we have
where \(0\leq u\leq 2\pi\) and \(0\leq v\leq 1\text{.}\)
We now compute the normal vectors \(\vec n_1\) and \(\vec n_2\text{.}\)
For \(\vec n_1\text{:}\) \(\vec r_{1u}= \langle -v\sin u, v\cos u,0\rangle\text{,}\) \(\vec r_{1v} = \langle \cos u,\sin u,0\rangle\text{,}\) so
As this vector has a negative \(z\)-component, we instead use
Similarly, \(\vec n_2\text{:}\) \(\vec r_{2u}= \langle -v\sin u, v\cos u,0\rangle\text{,}\) \(\vec r_{2v} = \langle \cos u,\sin u,-2v\rangle\text{,}\) so
Again, this normal vector has a negative \(z\)-component so we use
We are now set to compute flux. Over field \(\vec F_1=\langle 0,0,1\rangle\text{:}\)
These two results are equal and positive. Each are positive because both normal vectors are pointing in the positive \(z\)-directions, as does \(\vec F_1\text{.}\) As the field passes through each surface in the direction of their normal vectors, the flux is measured as positive.
We can also intuitively understand why the results are equal. Consider \(\vec F_1\) to represent the flow of air, and let each surface represent a filter. Since \(\vec F_1\) is constant, and moving βstraight up,β it makes sense that all air passing through \(\surfaceS_1\) also passes through \(\surfaceS_2\text{,}\) and vice-versa.
If we treated the surfaces as creating one piecewise-smooth surface \(\surfaceS\text{,}\) we would find the total flux across \(\surfaceS\) by finding the flux across each piece, being sure that each normal vector pointed to the outside of the closed surface. Above, \(\vec n_1\) does not point outside the surface, though \(\vec n_2\) does. We would instead want to use \(-\vec n_1\) in our computation. We would then find that the flux across \(\surfaceS_1\) is \(-\pi\text{,}\) and hence the total flux across \(\surfaceS\) is \(-\pi + \pi = 0\text{.}\) (As \(0\) is a special number, we should wonder if this answer has special significance. It does, which is briefly discussed following this example and will be more fully developed in the next section.)
We now compute the flux across each surface with \(\vec F_2=\langle 0,0,z\rangle\text{:}\)
Over \(\surfaceS_1\text{,}\) \(\vec F_2 = \vec F_2\big(\vec r_2(u,v)\big) = \langle 0,0,0\rangle\text{.}\) Therefore,
\begin{align*} \amp = \iint_R\langle 0,0,0\rangle\cdot\langle 0,0,v\rangle\, dA\\ \amp = \int_0^1\int_0^{2\pi} (0)\, du\, dv\\ \amp = 0\text{.} \end{align*}Over \(\surfaceS_2\text{,}\) \(\vec F_2 = \vec F_2\big(\vec r_2(u,v)\big) = \langle 0,0,1-v^2\rangle\text{.}\) Therefore,
\begin{align*} \amp = \iint_R\langle 0,0,1-v^2\rangle\cdot\langle 2v^2\cos u,2v^2\sin u,v\rangle\, dA\\ \amp = \int_0^1\int_0^{2\pi} (v^3-v)\, du\, dv\\ \amp = \pi/2\text{.} \end{align*}This time the measurements of flux differ. Over \(\surfaceS_1\text{,}\) the field \(\vec F_2\) is just \(\vec 0\text{,}\) hence there is no flux. Over \(\surfaceS_2\text{,}\) the flux is again positive as \(\vec F_2\) points in the positive \(z\) direction over \(\surfaceS_2\text{,}\) as does \(\vec n_2\text{.}\)
Exercises 14.6.3 Exercises
Terms and Concepts
1.
In the plane, flux is a measurement of how much of the vector field passes across a ; in space, flux is a measurement of how much of the vector field passes across a .
2.
When computing flux, what does it mean when the result is a negative number?
3.
When S is a closed surface, we choose the normal vector so that it points to the of the surface.
4.
If S is a plane, and βF is always parallel to S, then the flux of βF across S will be .
In the following exercises, a surface S that represents a thin sheet of material with density Ξ΄ is given. Find the mass of each thin sheet.
In the following exercises, a surface S and a vector field βF are given. Compute the flux of βF across S. (If S is not a closed surface, choose βn so that it has a positive z-component, unless otherwise indicated.)
7.
S is the plane f(x,y)=3x+y on 0β€xβ€1, 1β€yβ€4; βF=β¨x2,βz,2yβ©.
8.
S is the plane f(x,y)=8βxβy over the triangle with vertices at (0,0), (1,0) and (1,5); βF=β¨3,1,2β©.
9.
S is the paraboloid f(x,y)=x2+y2 over the unit disk; βF=β¨1,0,0β©.
10.
S is the unit sphere; βF=β¨yβz,zβx,xβyβ©.
11.
S is the square in space with corners at (0,0,0), (1,0,0), (1,0,1) and (0,0,1) (choose βn such that it has a positive y-component); βF=β¨0,βz,yβ©.
12.
S is the disk in the yz-plane with radius 1, centered at (0,1,1) (choose βn such that it has a positive x-component); βF=β¨y,z,xβ©.
13.
S is the closed surface composed of S1, whose boundary is the ellipse in the xy-plane described by x225+y29=1 and S2, part of the elliptical paraboloid f(x,y)=1βx225βy29 (see graph); βF=β¨5,2,3β©.
14.
S is the closed surface composed of S1, part of the unit sphere and S2, part of the plane z=1/2 (see graph); βF=β¨x,βy,zβ©.