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Triple Integrals in Cylindrical Coordinates

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Triple Integrals in Cylindrical Coordinates Triple Integrals in Cylindrical Coordinates Many applications involve densities for solids that are best expressed in non- Cartesian coordinate systems. In particular, there are many applications in which the use of triple integrals is more natural in either cylindrical or spherical coordinates. For example, suppose that f (r; ) g (r; ) in polar coordinates and that U (x; y; z) is a continuous function. If S is the solid between z = f (x; y) and z = g (x; y) over a region R in the xy-plane, then f(x;y) U (x; y; z) dV = U (x; y; z) dz dA ZZZS ZZR "Zg(x;y) # Let’s suppose now that in polar coordinates, R is bounded by = ; = ; r = p () ; and r = q () : Since dA = rdrd in polar coordinates, a change of variables into cylindrical coordinates is given by q() g(r;) U (x; y; z) dV = U (r cos ; r sin () ; z) r dzdrd (1) ZZZS Z Zp() Zf(r;) In practice, however, it is often more straightforward to simply evaluate the …rst integral in z and then transform the resulting double integral into polar coordinates. EXAMPLE 1 Evaluate the following 1 p1 x2 2xy x2 + y2 dzdydx Z0 Z0 Z0 Solution: Rather than employ (1) directly, let’s …rst evaluate the integral in z. That is, 1 p1 x2 2xy 1 p1 x2 2 2 2 2 2xy x + y dzdydx = zx + zy 0 dydx Z0 Z0 Z0 Z0 Z0 1 p1 x2 = 2xy x2 + y2 dydx Z0 Z0 = 2xy x2 + y2 dA ZZR 1 where R is the quarter of the unit circle in the 1st quadrant. In polar coordinates, R is bounded by = 0, = =2; r = 0; and r = 1: Thus, 1 p1 x2 2xy =2 1 x2 + y2 dzdydx = 2r cos () r sin () r2 rdrd Z0 Z0 Z0 Z0 Z0 =2 1 = sin (2) r5 drd Z0 Z0 =2 r6 1 = sin (2) d 6 Z0 0 =2 1 = sin (2) d 6 Z0 1 = 6 Check your Reading: Where did the r come from in (1)? Triple Integrals in Spherical Coordinates If U (r; ; z) is given in cylindrical coordinates, then the spherical transformation z = cos () ; r = sin () transforms U (r; ; z) into U ( sin () ; ; cos ()) : Similar to polar coordinates, we have @ (z; r) = dd @ (; ) so that a triple integral in cylindrical coordinates becomes q() g(r;) q() g( sin();) U (r; ; z) r dzdrd = U ( sin () ; ; cos ()) r dd d Z Zp() Zf(r;) Z Zp() Zf( sin();) However, r = sin () ; which leads to the following: 2 Triple Integrals in Spherical Coordinates: If S is a solid bounded in spherical coordinates by = , = ; = p () ; = q () ; = f (; ) ; and = g (; ) ; and if U (; ; ) is contin- uous on S; then q() g(;) U (; ; ) dV = U (; ; ) 2 sin () ddd ZZZS Z Zp() Zf(;) (2) In particular, it is important to notice that dV = 2 sin () ddd and it is acceptable to use dV = rddd since r = sin () : It is also important to remember the relationships given by the two right triangles relating (x; y; z) to (; ; ) : In particular, x2 + y2 = r2 implies x2 + y2 = 2 sin2 () and r2 + z2 = 2 implies that x2 + y2 + z2 = 2 (3) 3 Also remember that ranges over [0; 2] ; while ranges over [0; ]. EXAMPLE 2 Use spherical coordinates to evaluate the triple in- tegral 1 p1 x2 p1 x2 y2 x2 + y2 + z2 dzdydx 1 p1 x2 p1 x2 y2 Z Z Z p Solution: To begin with, we notice that this iterated integral re- duces to x2 + y2 + z2dV unit sphere ZZZ p As a result, in spherical coordinates it becomes 2 1 2 2 sin () ddd 0 0 0 Z Z Z p since x2 + y2 + z2 = 2: Thus, we have 2 1 2 1 3 sin () ddd = 3 sin () ddd Z0 Z0 Z0 Z0 Z0 Z0 2 4 1 = sin () dd 4 Z0 Z0 0 2 1 = sin () dd 4 0 0 Z 2 Z 1 = cos () 0 d 4 0 j = Z 4 EXAMPLE 3 Find the volume of the solid above the cone z2 = x2 + y2 and below the plane z = 1: Solution: The cone z2 = x2 + y2 corresponds to = =4 in spher- ical. Moreover, z = 1 corresponds to cos () = 1; or = sec () : Thus, 2 =4 sec() V = dV = 2 sin () ddd ZZZS Z0 Z0 Z0 2 =4 3 sec() = sin () dd 3 Z0 Z0 0 2 =4 1 = sec 3 () sin () dd 3 Z0 Z0 However, sec () sin () = tan () ; so that 1 2 =4 V = tan () sec2 () dd 3 Z0 Z0 Thus, u = tan () ; du = sec2 () d, u (0) = tan (0) = 0 and u (=4) = tan (=4) = 1 yields 1 2 1 V = udud = 3 Z0 Z0 1 2 1 = d 3 0 2 Z = 3 5 Check your Reading: Why does the cone z2 = x2+y2 correspond to = =4? Applications in Spherical and Cylindrical Coordinates Triple integrals in spherical and cylindrical coordinates occur frequently in ap- plications. For example, it is not common for charge densities and other real- world distributions to have spherical symmetry, which means that the density is a function only of the distance . ( Note: Scientists and engineers use both to denote charge density and also to denote distance in spherical coordinates. The context in which appears will indicate how it is being used). EXAMPLE 4 The charge density for a certain charge cloud con- tained in a sphere of radius 10 cm centered at the origin is given by C (x; y; z) = 100 x2 + y2 + z2 cm3 What is the total charge containedp within a sphere? ( C = micro- coulombs ) Solution: If denotes the solid sphere of radius 10 cm centered at the origin, then the total charge is Q = 100 x2 + y2 + z2 dV ZZZ p However, x2 + y2 + z2 = 2 leads to 2 10 Q = 100 2 sin () ddd Z0 Z0 Z0 6 : 2500 sin (i.e., charge density is proportional to ). Evaluation of the integral leads to 2 4 10 Q = 100 sin () ddd 4 Z0 Z0 0 2 = 100 2500 sin () dd 0 0 Z 2 Z = 100 5000d Z0 = 1; 000; 000 C which is Q = coulombs. Triple integrals in spherical and cylindrical coordinates are common in the study of electricity and magnetism. In fact, quantities in the …elds of electricity and magnetism are often de…ned in spherical coordinates to begin with. EXAMPLE 5 The power emitted by a certain antenna has a power density per unit volume of P p (; ; ) = 0 sin4 () cos2 () 2 where P0 is a constant with units in Watts. What is the total power within a sphere of radius 10 m? Solution: The total power P will satisfy P0 4 2 P = 2 sin () cos () dV ZZZ2 10 P0 4 2 2 = 2 sin () cos () sin () ddd 0 0 0 Z Z2 Z 10 4 2 = P0 sin () sin () cos () ddd 0 0 0 Z 2Z Z 2 2 2 = 10P0 1 cos () sin () cos () dd Z0 Z0 Let us now let u = cos () ; du = cos () d; u (0) = 1; and u () = 1: Then 2 1 2 2 2 P = 10P0 1 u cos () dud Z0 Z1 2 16 = 10P cos2 d 0 15 Z0 7 However, 2 cos2 () = cos (2) + 1; so that 80 2 32 P = P (cos (2) + 1) d = P W atts 25 0 5 0 Z0 Check your Reading: Why does the cone z2 = x2+y2 correspond to = =4? The Inverse Square Law Suppose two point masses with masses m and M respectively are located a dis- tance r apart. Sir Isaac Newton’s inverse square law states that the magnitude F of the gravitational force between the two point masses is j j Mm F = G (4) j j r2 where G is the universal gravitational constant. However, as Newton realized and struggled with for some time, objects in the real world are not point-masses and instead, the law (4) might need to be modi…ed. In particular, let’ssuppose that one of the bodies is not a ”point-mass,”but instead is a sphere of radius R with uniform mass density : For r > R constant, let’ssuppose that the sphere is centered at (0; 0; r) : If the other body is a point- mass ”satellite”of mass m located at the origin, then the gravitational force is directed along the z-axis. Suppose now that a small ”piece” of the sphere is located at a point (; ; ) (in spherical coordinates), and suppose that it has a small mass dM: Then the 8 distance between the small piece and the origin is : so that by (4) the ”small”magnitude d F of the gravitational force between the small ”piece”and the satellite is j j Gm dM d F = (5) j j 2 The amount of d F in the vertical direction is then given by cos () d F (see above). j j j j Thus, the total gravitational force in the vertical direction is Gm cos () F = cos () d F = dM j j j j 2 ZZZS ZZZS where S is the sphere corresponding to the ”planet”. If dV denotes the volume of a small ”piece”of the sphere, then dM = dV; which leads to cos () F = Gm dV (6) j j 2 ZZZS In Cartesian coordinates, the sphere S is given by x2 + y2 + (z r)2 = R2 or x2 + y2 + z2 2rz + r2 = R2 In spherical coordinates this becomes 2 2r cos () + r2 R2 = 0 which by the quadratic formula leads to = r cos () R2 r2 (1 cos2 ()) = r cos () pR2 r2 sin2 () q 9 Thus, the sphere is contained between = r cos R2 r2 sin2 () and = r cos + R2 r2 sin2 () 1 2 q q 1 Let us also note that ranges from 0 to sin (R=r) while ranges over [0; 2] : Evaluating (6) in spherical coordinates leads to 1 2 sin (R=r) 2 cos () F = Gm 2 sin () ddd j j 2 Z0 Z0 Z1 1 2 sin (R=r) 2 = Gm cos () sin () ddd Z0 Z0 Z1 1 2 sin (R=r) = Gm ( ) cos () sin () ddd 2 1 Z0 Z0 1=2 Since = 2 R2 r2 sin2 () ; this in turn leads to 2 1 1 2 sin (R=r) 1=2 F = 2Gm R2 r2 sin2 () sin () cos () dd j j Z0 Z0 If we let u () = R2 r2 sin2 () ; then the limits of integration become 2 2 1 R 2 2 R u (0) = R and u sin = R r = 0 r r2 10 Moreover, du = 2r22 sin () cos () d; so that 1 2 sin (R=r) Gm 1=2 F = R2 r2 sin2 () 2r2 sin () cos () dd j j r2 Z0 Z0 2 0 Gm 1=2 = 2 u du d r 0 R2 Z Z 0 Gm 2 u3=2 = d r2 3=2 Z0 R2 2 3 Gm 2R = d r2 3 Z0 Gm 4R3 = r2 3 However, the volume of the sphere is V = 4R3=3; so that the mass of the sphere is M = V = 4R3=3: Thus, we have shown that GMm F = j j r2 That is, a uniformly-dense spherical ”planet” of mass M and a point-mass of mass M at the center of the sphere have the same gravitational attraction on a ”satellite”point mass outside the sphere.
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