EOS 250 Geophysical Fields and Fluxes Volume Integrals

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1 EOS 25 Geophysical Fields and Fluxes olume Integrals February 17, 21 Overview These notes cover the following: Quantities defined per unit volume : volume integrals olume Integrals We have seen in the example of pressure variations in the atmosphere that mass is not concentrated in point particles (except at the atomic level), but distributed in space. This why we can define density ρ as mass per unit volume: as a single point has not volume, it also has no mass. In general, however, density is not a constant, but can vary in space and possibly in time (recall the example of pressure and density in the atmosphere). This means that we have to reconsider how the density field ρ(x, y, z) and the mass of a body are related. We have already touched on this in previous examples, when we considered the mass of a slice of the atmosphere, in which we assumed that ρ depends only on height. Here, we want to generalize these ideas. Before we do so, note that instead of mass, we could also consider other physical quantities that are not generally concentrated at points but distributed in space: for instance, momentum in the atmosphere or thermal energy in the Earth s interior are distributed rather than concentrated at points. We will see later that we can also describe these quantites using densities : for instance, we can define thermal energy density as heat content per unit volume of the Earth s interior, in the same way that we are about to define mass density. In general, when we have a medium in which mass, momentum, energy etc. are distributed in space, we call it a continuum. The vector calculus you will learn in this course is the essential tool for formulating physics in continua. If the density of a material varies in space, how is density defined in the first place? Clearly, we cannot take an arbitrarily-sized and shaped sample of the material, 1

2 measure its mass and volume, and define density as mass over volume: the answer would depend on precisely what the size and shape of the sample are. To define density at a point (x, y, z), consider instead the mass m of a small volume centered on (x, y, z). For small enough volumes, we expect m to be proportional to. This allows us to define density as the constant of proportionality, ρ m. Implicit here is, of course, the limit of small. So how do we go backwards: given a density field ρ(x, y, z), how do we calculate the mass M of a body occupying a known volume? As with the surface S in the mass transfer calculation, we split the total volume into small pieces of size. Then we calculate the mass of each piece as m ρ, and sum over the masses of all the small pieces that make up the whole volume : M m ρ. Again, the limit of small is implied, and we can again recognize a Riemann sum that we denote by an integral: M ρ d, where the subscript indicates the volume the integral is taken over. To calculate an integral like this, we adapt the idea of a Riemann sum in one dimension. The obvious way of splitting into small bits is into cuboids of side lengths x y z (figure 1), x y z. The mass M of the body is then the sum of ρ over all the cuboids contained in the volume (again, in the limit of small x, y and z). To keep track of all these cuboids, we can use indices i, j and k to identify their x-, y- and z-positions respectively, letting x i, y j and z k be the coordinates of the centre of the cuboid at the kth level above the base of the vertical column of cuboids with position i, j in the xy-plane. To sum, we have to sum over all indices (i, j, k) that correspond to cuboids in the volume, ρ d ρ(x i, y j, z k ) x y z. i,j,k 2

3 z zhmax( x,y) y zh y x,y min( ) yymax( x) x min yy x min( ) x max x k... k2 k1 z z index k index j j1 j... y y ( xyz i, j, k) i1 i2 i... index i. x sum over k sum over j ( xy, ) i j y sum over i x Figure 1: Breaking into small cuboids. 3

4 When taking a sum over multiple indices, the order of summing does not matter. We can therefore sum over k first, then over j and finally over i, ρ d { [ ] } ρ(x i, y j, z k ) z y x. i j k Of course, as z, the sum over k is the sum over cuboids in the (i, j)th vertical column, where we can treat i and j temporarily as constants. This sum turns into an integral over z: ρ(x i, y j, z k ) z ρ(x i, y j, z) dz. k As we just stated, the sum is formed with i and j treated as constants, so the integral over z is take, with x i and y j treated as constants. Next, the sum over j is then the sum over a row of such columns with fixed i, and turns into an integral over y taken at constant x i, so [ ] [ ρ(x i, y j, z k ) z y j k ] ρ(x i, y, z) dz dy. Here, the integral with respect to z is computed first, treating y as well as x i as constants. Subsequently, the integral with respect to y is computed, treating x i as constant. Lastly the sum over i is finally the sum over all thee rows of vertical columns, and turns into an integral over x, so ρ d { [ ] } ρ(x i, y j, z k ) z y x i j k { [ ] } ρ(x, y, z) dz dy dx As above, the integral with respect to z is computed first, treating y as well as x as constants, then the integral with respect to y is computed, treating x as constant, and lastly, the integral with respect to x is calculated. For obvious reasons, the integral over x, y and z is also known as a triple integral. Importantly, the volume occupies a finite and prescribed portion of space, so the Riemann sums above turn into definite integrals: for each index (i, j and k) we sum only over those cubes that lie within the volume (see figure 1). The hard part is therefore often to figure out what limits to use. Example 1 Let ρ 1+x 2 +y 2 +z 2, and let be the unit cube, for which < x < 1, < y < 1, < z < 1. Compute ρ d. 4

5 In this case, the limits of integration are clear as each coordinate ranges from to 1. Computing each integral in turn, 1 { 1 [ 1 ] } ρ d 1 + x 2 + y 2 + z 2 dz dy dx 1 { 1 [ z + x 2 z + y 2 z + z 3 /3 ] } z1 dy dz z 1 { 1 } 4/3 + x 2 + y 2 dy dx [ 4y/3 + x 2 y + y 3 /3 ] y1 y dx 5/3 + x 2 dx The tricky issue is therefore the question of limits of integration for irregularlyshaped bodies. We consider only a body that has a single upper surface of the form z h max (x, y), and a lower surface z h min (x, y) (figure 1). Also, consider the projection xy of this body onto the xy-plane, and assume that this projection has an upper boundary in the xy-plane of the form y y max (x), a lower boundary y y min (x), a left-hand edge at x x min and a right-hand edge at x x max. The limits here come once more from viewing the integrals as sums. When taking the sum over k at fixed i and j, we are summing from the bottom of a vertical column to the top. This means that the range of k will depend on x i and y j, as z k must lie between h min (x i, y j ) and h max (x i, y j ). In other words, when taking the limit z, the integral is from z h min (x i, y j ) to h max (x i, y j ). The next sum, over j, must capture all the columns in a row at fixed i, that is, the integral must range from y min (x i ) to y max (x i ). Lastly, the sum over i must go over all these rows, and the x-integral must therefore range from x min to x max (figure 1). Hence [ xmax ( ymax ) ] hmax(x,y) ρ d ρ(x, y, z) dz dy dx. x min y min h min (x,y) We give an example of how to use this formula next. Example 2 Let be the tetrahedron with vertices (,, ), (1,, ), (, 1, ) and (,, 1) (figure 2), and let ρ x + y + z. Compute ρ d. From the diagram, we have h min (x, y), y min (x), x min, x max 1. We still need to find y max (x) and h max (x, y). Let us find y max (x) first. Clearly, this is the straight line connecting the vertices (, 1, ) to (1,, ). Let us write this in the form y ax+b. If the line passes through 5

6 z 1 n 1 y n 1 x Figure 2: The volume in example 2. (Ignore the vectors labelled n for now.) (, 1, ) then y 1 when x, so 1 a + b, or b 1. Similarly, y when x 1 if the line passes through (1,, ), so a 1 + b, or a b 1. Hence y max 1 x. Next, we need to write the slanted upper surface in the form z h max (x, y). This works the same way as above. We know that the triangle is part of a plane, and hence we expect that z ax + by + c. (1) describes the surface for some set of coefficients a, b and c (with a and b potentially different from above). How do we find them? We know that the plane must pass through the points (x, y, z) (1,, ), (, 1, ) and (,, 1). Hence these coordinates must satisfy the equation (1). Substituting the coordinates into (1), we get, for each of these points in turn, a + c b + c 1 c Hence c 1, a b 1 and z 1 x y. Therefore This allows us to write ρ d h max (x, y) 1 x y. 1 [ 1 x ( 1 x y 6 ) ) (x + y + z) dz dy dx.

7 But we have already stated that we will take the z-integral first, treating x and y as constant. This is done at follows: 1 x y (x + y + z) dz [ xz + yz + z 2 /2 ] z1 x y z x(1 x y) + y(1 x y) + (1 x y) 2 /2 Note that when computing the integral with respect to z we treat x and y as constants not only in the integrand, but also in the limits. Once we have completely computed the integral with respect to z, including applying the limits, we compute the integral with respect to y, treating x as constant. Last, the integral with respect to x is computed: 1 [ 1 x ( ρ d x(1 x y) + y(1 x y) + (1 x y) 2 /2 ) ] dy dx [ x(1 x)y xy 2 /2 + (1 x)y 2 /2 y 3 /3 (1 x y) 3 /6 ] y1 x y dx x(1 x) x/2 + (1 x)/2 + (1 x) 3 /6 dx [ 1/2 x 2 + (1 x) 3 /6 ] dx [ x/2 x 3 /3 (1 x) 4 /24 ] 1 1/8 Exercise 1 Let be a cuboid with rectangular base area A in the xy-plane and height h. Let density ρ be a function of height z only. Show that ρ d A h ρ(z) dz. Exercise 2 Let be the tetrahedron with vertices (,, ), (1,, ), (1, 1, ) and (,, 1). Let ρ 1 + xy + z. Compute the mass of the body. Exercise 3 What do you think the physical meaning of the integral 1 d is? Confirm your answer by computing this integral for the tetrahedron in the previous ex- ercise, and compare that answer with the volume of the tetrahedron given by 1/3 base area height. If ρ ρ constant, evaluate ρ d, if also denotes the volume of the body. (Comment?) We have seen how to relate total mass M to density in cases where density is not constant, where M ρ d. As a result, it is more useful not to think of density as being mass over volume. In fact, if we take mass over volume, we get a mean density ρ M ρ d, 7

8 but this is value depends on the particular shape of the volume and does not give any information about the local density at a point (x, y, z). An analogous concept would be to say that the definition of velocity as distance over time gives a mean velocity but not velocity at any instant in time. Take x(t) be position as a function of time, and put v x(t 2) x(t 1 ) t 2 t 1. This is the mean velocity between times t 1 and t 2, and can again be written as an integral similar to the formulat for ρ above, v t2 v(t) dt, t 2 t 1 t 1 but v does not give information about instantaneous velocity at any point in the interval from t 1 to t 2. A better way to think of the density field ρ(x, y, z) is as a measure of how concentrated mass is near a point (x, y, z). Of course, a single point (x, y, z) does not have any mass, and only a small but finite volume around that point will contain a finite amount of mass m.to get a measure of how concentrated mass is near (x, y, z), it is then natural to take the ratio of m to. The more mass is concentrated, the bigger m will be for a fixed volume, and therefore the biger density field ρ(x, y, z) m will be. 1 As we mentioned at the beginning, there are other quantities that are similar to density: for instance, the concentration of a chemical measured in moles per cubic metre, the heat content of a substance, measured in Joules per cubic metre, or charge density, measured in Coulombs per cubic metre. All of these quantities have in common that they measure some physical entity per unit volume they are essentially generalized densities, with amounts of a chemical, energy and charge taking the place of mass in the definition of ordinary density (which we could also think of as mass density ). The theory developed above can also be applied to these other densities. For instance, we would define a chemical concentration field c(x, y, z) in terms of the the number of moles n in a small volume around the point (x, y, z), c(x, y, z) n. 1 The analogue of this for velocity is of course to take velocity v(t) to be the rate of change of x with respect to t at a given point in time 8

9 Following the same steps as above, it is then clear that the number of moles N of the chemical in some fixed volume is given by N c d. Similarly, if q(x, y, z) is the charge density field, the total charge Q in a given volume is Q q d. 9