Lecture 18 ouble Integrals (cont d) Electrostatic field near an infinite flat charged plate Consider a thin, flat plate of infinite size that is charged, with constant charge density ρ (in appropriate units of charge/unit area). The problem is to find the electrostatic field E at a point P that is a distance d from the plate. It is convenient to place the flat infinite plate on the y-ais and to situate the origin of the 3 coordinate system so that point P lies on the z-ais at (,,d), as shown below. z d P O y Because the plate etends infinitely in the y-plane, we may argue that the electric field E at P will point in the direction of the z-ais, i.e., E = Ek. Our goal is to compute E, which may depend on the distance d. Now let da be an infinitesimal element of area that is situated at a point Q on the plate with coordinates (,y). The infinitesimal amount of charge dq in this element of area is dq = ρ da. (1) This charge will contribute to the total electrostatic field E at point P. We shall denote this infinitesimal contribution as de, acknowledging that it is a vector quantity. From Coulomb s law for electrostatic force, de = dq 1 4πǫ QP. (2) QP 3 We are now going to use a symmetry property to reduce the compleity of the calculation. For each point Q, there is a corresponding point Q situated at (, y) that produces an infinitesimal electric 136
field de. The and y components of the vectors de and de cancel each other out since Q and Q lie on opposite sides of the origin. As a result, the net electrostatic field vector E at P will point in the z direction. We need to compute only the z component of the vector de, i.e., its projection onto the z ais. This component is given by where θ is the angle between QP and the z ais, shown below: de z = de cos θ, (3) z de θ P decos θ O Q(, y) y From the diagram, cos θ = OP. (4) QP Taking the magnitude of de in (2) and using (4)), we find that de cos θ = ρ da OP 4πǫ (5) QP 3 = ρ d 1 4πǫ [ 2 + y 2 + d 2 da. ] 3/2 This is the contribution to the electostatic field vector E = Ek at point P from the charge element dq at point Q. We now integrate over all (,y) R 2 to find the total field: E = ρ d 4πǫ 1 [ 2 + y 2 + d 2 da. (6) ] 3/2 It doesn t really matter in which order we perform the integration. If we let da = dy d, then the inner integral is I 1 = 1 [ 2 + y 2 + d 2 dy. (7) ] 3/2 Yes, you can look this up in tables, but it s good for character to grind it out by hand. In order to remove the square root function, we use the trigonometric substitution y = ( 2 + d 2 ) 1/2 tan θ, π 2 < θ < π 2, (8) 137
so that y will range from to, then dy = ( 2 + d 2 ) 1/2 sec 2 θ dθ (9) and [ 2 + y 2 + d 2 ] 3/2 = ( 2 + d 2 ) 3/2 (1 + tan 2 θ) 3/2 (1) = ( 2 + d 2 ) 3/2 sec 3 θ. The inner integral then becomes I 1 = (2 + d 2 ) 1/2 ( 2 + d 2 ) 3/2 = = 1 ( 2 + d 2 ) 2 2 + d 2. π/2 π/2 π/2 π/2 sec 2 θ sec 3 θ cos θ dθ dθ (11) We insert this result into the outer integral, omitting the constant factor for the moment: Once again, one can use tables. Or, if we use the substitution 1 I 2 = 2 2 d. (12) + d2 = dtan θ, π 2 < θ < π 2, (13) then and Then d = dsec 2 dθ (14) 2 + d 2 = d 2 (1 + tan 2 θ) = d 2 sec 2 θ. (15) I 2 = 2 d d 2 π/2 = 2 d π/2 π/2 π/2 sec 2 θ sec 2 dθ (16) θ dθ = 2π d. 138
Inserting these results into Eq. (6), we arrive at Thus, at point P (,, d), the electrostatic field vector is E = ρ d 4πǫ 2π d = ρ 2ǫ. (17) E = ρ 2ǫ k. (18) Note that the field vector is independent of the distance between the point P and the plate. By convention, the charge density is also signed, i.e., ρ > if the plate is positively charged and ρ < if the plate is negatively charged. A well-known consequence is that if you would place two such infinite plates, oppositely charged with the same magnitudes of the charge densities, parallel to each other, then the electric field between the two plates would be E = ρ ǫ k. (19) This is a practical way to construct a homogeneous electric field over a region in space. Appendi: What if the plate is not infinite in etent? (not covered in lecture) Of course, no plate can be infinite in etent, but the plate can be sufficiently large so that the infinite approimation is quite good near the center. But one could certainly acknowledge the finiteness of the plate by performing the integration over a finite region of R 2. For eample (as was brought to the instructor s attention by some students who were concurrently taking the Physics E&M course), we might wish to consider a place of dimensions 2L 2L, with the point P situated a distance d from the center of the plate. The total electric field would then be E = ρ d L L 1 4πǫ L L [ 2 + y 2 + d 2 da. (2) ] 3/2 The inner integral can be evaluated using the trigonometric substitution y = ( 2 + d 2 ) 1/2 tan θ: L L Inserting this result into the outer integral yields 1 [ 2 + y 2 + d 2 ] 3/2 dy = 2L 1 ( 2 + d 2 ) (21) 2 + d 2 + L2. E = ρld L 2πǫ L 1 ( 2 + d 2 ) d. (22) 2 + d 2 + L2 139
This definite integral can be evaluated with the help of MAPLE. If we make the substitution u 2 = 2 + d 2 so that d = u du, then The definite integral then becomes d = u du = d 2 +L 2 2 d u du. (23) 2 u2 1 u u 2 d 2 du, (24) L 2 + u2 which can now be handled by MAPLE. There are three cases to consider: (i) L < d, (ii) L = d and (iii) L > d. We present the results for the third case L > d since it would be perhaps the most relevant one: the plate is not infinite but large with respect to the distance d between the point P and the plate. E = ρ [ ( ) ( 2L Tan 1 2 + d 2 Tan 1 d(d 2 + 3L 2 ) ) 2πǫ d (L 2 d 2 ) + π ]. (25) 2L 2 + d 2 2 Quite a complicated epression, n est-ce pas? This is the price that must be paid for integrating over a square region, which is not a natural shape for the inherent circular symmetry of the force field due to a charge. One should epect that in the limit L, we retrieve the result in Eq (18), and indeed this is the case. Looking at (25), as L, the first Tan 1 term will approach π/2 and the second term will approach. This implies that the entire epression in the large square brackets will approach π as L. Thus, in the limit L, we obtain the infinite plate result. Here is an interesting question: What is the strength of the field E when d is very large in comparison to L, i.e., the ratio d L is small, but not zero? In this case, we epect that the square plate will behave roughly as a point particle with total charge Q = ρ(2l) 2 so that the field will be roughly E ρ πǫ ( ) L = ρ(4l2 ) d 4πǫ d 2 = Q 4πǫ d2. (26) See if you can derive this result from Eq. (25). Hint: Linearize the epressions in the Tan 1 functions and then linearize the Tan 1 functions themselves. 14
ouble integrals in planar polar coordinates (Relevant section from Stewart, Calculus, Early Transcendentals, Sith Edition: 15.4) Often, because of the particular symmetry of a physical problem, it is much more convenient to perform an integration in a coordinate system that is suited to the symmetry. Planar polar coordinates are well suited for problems that ehibit circular symmetry, e.g., circular regions, circular sectors. As such, we should eamine the problem of integrating a function f over a region in this coordinate system. Note that in many applications, the function f may still be epressed as a function of Cartesian coordinates and y. There is no problem we merely have to transform and y into r and θ. There is, however, a complication regarding the infinitesimal element of area da that is produced by the infinitesimal changes dr and dθ. In Cartesian coordinates, there was no such problem the infinitesimal changes d and dy in, respectively, and y, produce an element of area da = d dy, independent of and y, i.e., independent of where the element is located in the plane, as sketched below. y d dy da = ddy, independent of and y This is not the case for polar coordinates, as the figure below should make clear. Here we have indicated the area element A at a point (r,θ) that is produced by letting r increase by r and θ increase by θ. r A θ (r, θ) θ r The area of this region can be obtained by subtracting the area of the smaller, inner sector with 141
radius r from the area of the larger sector with radius r + r. First we must determine the area of a sector of radius R and subtended angle φ: The ratio φ/(2π) will be the ratio of the area of the sector to the area of a circle with radius R. Thus Returning to the diagram above, A = φ 2π πr2 = 1 2 R2 φ. (27) A = 1 2 θ[(r + r)2 r 2 ] = 1 [2r + r] r θ 2 = 1 [ 2 r 1 + r ] r θ. (28) r As r, the ratio r/r will become negligible with respect to 1. Therefore, in the infinitesimal limit, r dr and θ dθ, we have the result, da = r dr dθ. (29) Once again, note the presence of r in the epression. This is due to the fact that the farther out we are from the origin, i.e., the larger r is, the larger the area that is produced by letting the radial vector sweep through a given angle. This is clearly indicated in the figure below, where we see how area elements A get smaller as r approaches zero. da = r dr dθ (r, θ) Thus, if we wish to integrate a function g(r,θ) in polar coordinates, we must use the formula g da = g(r,θ) r dr dθ. (3) 142
Lecture 19 ouble integrals in planar polar coordinates (cont d) In the previous lecture, we showed that the infinitesimal element of area da at a point (r,θ) is given by da = r dr dθ. (31) Thus, if we wish to integrate a function g(r,θ) in polar coordinates, we must use the formula g da = g(r,θ) r dr dθ. (32) We now consider some eamples. Eample 1: Let us first check this result by computing the area of a circular region of radius R. In this case, in order to scan the entire circular region, we must let θ range from to 2π and r range from to R. Thus The inner integral is 2π R R Inserting this result into the outer integral gives 1 2π 2 R2 r dr dθ. (33) r dr = 1 2 R2. (34) dθ = πr 2, (35) the correct result. We would obtain the same result using the other order of integration for the iterated integral: R 2π R r dθ dr = 2π r dr = 2π 1 2 R2 = πr 2. (36) Eample 2: The integral unit circle. y da, where is the region in the first quadrant enclosed by the Notice that the integrand is epressed in terms of Cartesian coordinates we must convert y into polar coordinates if we intend to perform the integration over polar coordinates. = r cos θ, y = r sin θ. (37) 143
The region can be scanned by letting r range from to 1 and θ from to π/2: π/2 1 y da = (r cos θ)(r sinθ)r dr dθ (38) The inner integral is = π/2 1 1 Inserting this result into the outer integral gives 1 4 π/2 sin θ cos θr 3 dθ. r 3 dr = 1 4. (39) sin θ cos θ dθ = 1 4 1 2 sin2 θ π/2 = 1 8. (4) Note: The above integral could also have been evaluated, without much difficulty, using Cartesian coordinates (Eercise). Eample 3: Find the center of mass of a thin homogeneous metal plate (constant density) which has the shape of a semicircle of radius R. It is convenient to introduce a coordinate system so that the plate is located in the upper half of the plane and its base sits on the -ais between = R and = R. Since the plate is homogeneous, i.e., ρ is constant, it follows that, the coordinate of the center of mass, is zero. It remains to compute the y coordinate, which is given by ȳ = y dm dm = yρ da. (41) ρ da The denominator integral is simply the mass of the plate: M = ρ da = ρ da = ρ (area of plate) = 1 2 πr2 ρ, (42) To compute the numerator integral in polar coordinates, π R y da = r sin θ r dr dθ. (43) The inner integral is The numerator integral then becomes Thus R r 2 dr = 1 3 R3. (44) 1 π 3 R3 sin θ dθ = 2R3 3. (45) ȳ = 2R3 3 2 πr 2 = 4 R.42R. (46) 3π 144
Another derivation of the area element da = r dr dθ in terms of the Jacobian (Relevant section of tetbook by Stewart: 15.9) In the previous lecture, we used a geometric argument to show that the infinitesimal element of area in polar coordinates at a point (r,θ) is given by da = r dr dθ. (47) This result is dependent upon r, for reasons mentioned in the previous lecture, which is in stark contrast to the Cartesian coordinate case, where da = d dy. In what follows, we outline another method to derive Eq. (47), which employs the eplicit relationships between the Cartesian coordinates, y and the polar coordinates r, θ. This method can be used in general to derive changes in area and volume elements due to coordinate transformations. Recall that the relationships between Cartesian and polar coordinates are given by = (r,θ) = r cos θ, y = y(r,θ) = r sin θ. (48) We could have also gone the other way, i.e., epressing r and θ as functions of and y, but the relations in Eq. (48) will be easier to work with. Now consider the total differentials of and y: Using the Chain Rule, we have d = dy = r y r dr + θ dθ dr + y θ dθ. (49) which we can write in matri notation as d dy = r y r θ y θ dr dθ. (5) This is a linear transformation relating the differentials d and dy to dr and dθ. There is an important consequence of this transformation. Let s consider general linear transformations in the plane having the form y = C, (51) where = ( 1, 2 ), y = (y 1,y 2 ) and C is a 2 2 matri. You may recall the following fact from linear algebra. Let be a region in R 2. Take each point ( 1, 2 ) and transform it to a point 145
(y 1,y 2 ) R 2. Let the set of all such (y 1,y 2 ) define the new region. And let A() and A( ) denote the areas of and, respectively. Then A( ) = det C A(). (52) In other words, the magnitude of the determinant of the matri C representing the linear transformation determines whether there is a magnification, reduction or no change in the area of region after it is transformed to : 1. If det C < 1, then there is a reduction in area. 2. If det C > 1, then there is a magnification in area. 3. And if det C = 1, then there is no change in area. We now go back to Eq. (5). In this equation, we consider region to be a rectangular bo with sides of infinitesimal length dr and dθ. Of course, the area of this bo is dr dθ. The matri in Eq. (5) we ll call it C then transforms this bo into a bo, region, with sides of infinitesimal length d and dy. The area of this bo is d dy. The conversion factor between the areas of and is given by the determinant of the transformation C, i.e., d dy = det C dr dθ. (53) This determinant is so important that we denote it with a special symbol, as follows: det C = (,y) (r,θ) = r θ. (54) y r y θ We now compute this determinant eplicitly: (, y) (r,θ) = From Eq. (53), we have the result, cos θ r sin θ sin θ r cos θ = r (55) which agrees with the result derived in the previous lecture. d dy = r dr dθ, (56) Finally, we mention that this method applies to general coordinate transformations of the form = (u,v), y = y(u,v). (57) 146
The result is where d dy = (, y) (u, v) du dv, (58) (, y) (u,v) = u y u v y v. (59) In practical applications, however, it is not necessary, or even advisable, to memorize the above formula. It is probably easier to start with the differentials associated with the transformations from,y space to u,v space, for eample, Eq. (49) in the case of polar coordinates. We then let the math do the work remembering, however, that it is the magnitude of the determinant of the Jacobian transformation matri J that determines the proportionality constant between d dy and du dv. 147
Lecture 2 Integration in R 3 Triple Integrals (Relevant section from Stewart, Calculus, Early Transcendentals, Sith Edition: 15.6) Triple integration is a natural etension of double integration. Here, we wish to integrate a scalar-valued function of three variables, f(,y,z), over a region R 3. Once again in the Spirit of Calculus, the region is broken up into tiny cubes V ijk for convenience, we assume that they all have the same volume, V = y z. For each block V ijk, we choose a sample point ( i,y j,z k ) and form the Riemann sum S n1 n 2 n 3 = n 1 i=1 n 2 j=1 n 3 k=1 f( i,y j,z k ) V. The primes indicate that the summation is performed only over those cubes V ijk that contain points in ). In the limit n 1,n 2,n 3 (plus some details!!!), S n1 n 2 n 3 = f dv }{{} triple integral of f over Practically, we compute triple integrals as iterated integrals ddydz or dydzd or 4 other possibilities. As in the double integral case, we employ whichever form is convenient to the calculation. Motivating eamples: If ρ(,y,z) is the charge density at at point (,y,z), then dq = ρdv is the infinitesimal element of charge at (,y,z). The total charge (or mass) over the region is then Q = dq = ρ dv. (6) 148
Of course, charge density could be replaced by mass density. As in the case of double integrals, the main challenge is to determine the limits of integration so that we perform a complete scan of the region without any duplications, i.e., counting any points more than once. In Cartesian coordinates, the simplest case is a rectangular bo, 1 2, y 1 y y 2, z 1 z z 2. (61) There are, in principle, si differenent ways to conduct the scanning of the region. One or more may be more convenient that the others. Eample 1: Suppose that a cube of material is placed in the first octant of R 3. The sides of the cube are of unit length and the mass density function is given by ρ(,y,z) = 2 + y 2 + z 2. Find the center of mass of the cube. First compute the total mass of the cube: M = ρ dv (62) The inner integral is 1 = 1 1 1 ( 2 + y 2 + z 2 ) dzdyd. ( 2 + y 2 + z 2 ) dz = 2 z + y 2 z + 1 3 z3 z=1 z= (63) = 2 + y 2 + 1 3 Now insert this result into the middle integral: 1 ( 2 + y 2 + 1 3 ) dy = 2 y + 1 3 y3 + 1 3 y y=1 y= (64) = 2 + 2 3. We insert this result into the outer integral to arrive at the final result, i.e., the total mass M is M = 1 ( 2 + 2 )d = = 1. (65) 3 Because of the rather special form of the integrand, there are other ways to evaluate the integral which involve less work. First of all, we can epand the integral as follows, ( 2 + y 2 + z 2 )dv = 2 dv + y 2 dv + z 2 dv. (66) 149
Now consider the first integral for the moment, 2 dv = = = 1 1 1 ( 1 2 dv ) ( 1 2 d ( ) 1 (1)(1) 3 ) ( 1 ) dy dz = 1 3. (67) where we have eploited the fact that the integrand is separable. The other two integrals have the same values. Adding up the three results produces the final result M = 1. Let us now determine the center of mass of this cube. The -coordinate of the center of mass will be given by The numerator is given by the integral The inner integral is ρ dv = = dm dm. = ρ dv M. (68) 1 1 1 ( 2 + y 2 + z 2 ) dzdyd. (69) 1 ( 2 + y 2 + z 2 ) dz = [ 3 z + y 2 z + 1 3 z3 ] z=1 = 3 + y 2 + 1 3 z= (7) Now insert this result into the middle integral: 1 ( 3 + y 2 + 1 [ 3 ) dy = 3 y + 1 3 y3 + 1 ] y=1 3 y y= (71) = 3 + 2 3. We finally insert this result into the outer integral: We find that 1 ( 3 + 2 3 )d = 7 12. (72) Therefore, = 7 12. Similar integrals occur for the other coordinates and we have ( 7 (,ȳ, z) = 12, 7 12, 7 ). (73) 12 (We could have made an argument that = ȳ = z by symmetry.) 15
If the mass density function of the cube were constant, then the center of mass would be located at (1/2,1/2,1/2). However, the density function considered above, ρ(,y,z) = 2 + y 2 + z 2, is very small near (,,) (in fact, it s zero at the origin) and increases as we move outward radially. It assumes a maimum value of 3 at (1,1,1). The result is that the center of mass is displaced radially outward from (1/2,1/2,1/2) by a shift of 1/12 in each coordinate. Eample 2: This eample reveals how the boundaries of a region determine the limits of integration. Consider the tetrahedron in the first octant (,y,z ) that is pictured below. It is the region bounded by the y, yz and z planes and the plane a + y b + z c = 1, The tetrahedron has vertices at (,, ), (a,, ), (, b, ) and (,, c). z c a + y b + z c = 1 a b y We wish to set up an iterated triple integral that could integrate a function f(,y,z) over this region. There are si possibilities we shall consider only one:??? f dv = f d dy dz, (74)??? where the limits are to be determined. The limits for the outer integration in z are easy: All points in the region have z values that lie between and c. Thus: Outer integration: z c. This outer integration in z will produce a series of slices of the region that are parallel to the y plane. For each z [,c], there corresponds a slice of the region we ll call it S(z) as shown below. 151
z c slice S(z) a b y We must now scan over all points in the slice S(z). It might be easier to view the slice from the top: O S(z) b y a + y b = 1 a a + y b = 1 z c From our work on double integrals, we should be able to scan this triangular region. But we need to determine the equation of the hypotenuse of this triangle. How do we do this? Well, we need to think about where this hypotenuse comes from it comes from the intersection of the plane z = constant with the plane in (). For each z [,c], this line is given by a + y b = 1 z c. (75) Before we go on, let us check this result. When z =, this line becomes a + y b = 1, (76) which is indeed the line on the y-plane that connects the two points (a,,) and (,b,). When z = c, the line reduces to the point = y =, which is the ape of the tetrahedron. The net integration the middle one is over y. Clearly, the lower limit of y is, since we start at the -ais. But how far do we go with y? The answer is the y-intercept of the line in (75), which is obtained by setting = and solving for y. The result is y = b b c z. 152
Middle integration: y b b c z. Now we must determine the final integration the inner integration over. For each z [,c] and y [,b b z], we must let run from to the hypotenuse. This means solving for in Eq. (75): c = a a b y a z. Thus c Inner integration: a a b y a c z. Thus the net result is f dv = c b b c z a a b y a c z f d dy dz. (77) Special case: If we set f(,y,z) = 1, then the integral in (77) should give us the volume of the tetrahedral region. 1. Inner integration: a a b y a c z d = a a b y a z. (78) c 2. Middle integration: b b c z a a b y a c z dy = [ ay a 2b y2 a c yz ] b b c z (79) = a (b b ) c z a ( b b ) 2 2b c z a (b b ) c c z z 3. Outer integration: = c [ a (b bc ) z a ( b b ) 2 2b c z a (b bc ) c z ac 2b ( b b ) 2 c z ac ( 6b 2 b b ) 3 c z ab 2c z2 + ab 3c 2z3 = + abc 2 + abc 3 + abc 2 abc 6 = 1 6 abc. z dz (8) ] c (81) Thus the volume of the tetrahedron is V = 1 6 abc. Now suppose that we wanted to find the center of mass of this tetrahedron, assuming that it was homogeneous with constant density ρ. The coordinates (, ȳ, z) of its center of mass would be given 153
by = ρ dv ρv, ȳ = ρ y dv ρv, z = ρ z dv ρv, (82) where we have taken the constant ρ out of the integration. The pairs of ρ s in the three integrals will cancel each other out. We leave these integrations as an eercise and simply state the results. For the integral, f dv = c b b c z a a b y a c z d dy dz = 1 24 a2 bc. (83) Similarly, The net result is y dv = 1 24 ab2 c, (,ȳ, z) = z dv = 1 24 abc2. (84) ( a 4, b 4, c ). (85) 4 Spherical regions and integration in spherical polar coordinates (Relevant section from Stewart, Calculus, Early Transcendentals, Sith Edition: 15.8) Spherical regions, for eample the sphere of radius R sketched below, are often encountered in physical problems, so we should look at them in terms of multiple integration. Cartesian coordinates are not very well suited here. It is generally more convenient to use spherical polar coordinates. 2 + y 2 + z 2 = R 2 enclosing region : {(,y,z) 2 + y 2 + z 2 R 2 } Sphere with radius R 154
The spherical polar coordinate system The spherical polar coordinate system, as its name suggests, is a method of scanning points in R 3 in a spherical fashion. Imagine that you have a radar-type device at the origin of a coordinate system, which can send out a beam of light in any direction. The beam will be able to detect how far away an object is from the origin the polar coordinate r and then specify where it is on a sphere of radius r akin to the angles of latitude and longitude on the surface of the earth. z r: radial distance OP, O θ φ P (,y, z) P y r, φ: angle between OP and z-ais, φ π, θ: angle between the plane, or swinging door, determined by the points O, P and P, and the -ais. θ 2π. To obtain the relationships between spherical and Cartesian coordinates, we begin with z, which is the easiest one: z = r cos φ. (86) As for and y: Since OP = r sin φ, we have = OP cos θ, y = OP sin θ. (87) = r sin φcos θ, y = r sin φsin θ. (88) It is easy to verify that 2 + y 2 + z 2 = r 2. Note: In older tets (including the ones used by your instructor!), the roles of θ and φ were reversed: θ was the angle between the position vector OP and the z-ais, and φ the angle in the y-plane. 155