Chapter 15 Vector Calculus

Chapter 15 Vector Calculus 151 Vector Fields 152 Line Integrals 153 Fundamental Theorem and Independence of Path 153 Conservative Fields and Potential Functions 154 Green s Theorem 155 urface Integrals 155 Flux of Vector Field across urface 155 ivergence Theorem 156 tokes Theorem 156 imply Connected, Orientation 156 Irrotational and Conservative Vector Fields

urface Integrals If the surface is given by the graph z = z(x, y), where (x, y) lies in the domain of R 2, ie = { (x, y, z(x, y)) (x, y) }, then the surface area element d = 1 + z 2 x + z 2 y dxdy If the surface is given by parametric form r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, where (u, v) lies in the domain of of uv-plane urface area element d = N dudv = r u r v dudv = (y, z) (z, x) (x, y) i + j + (u, v) (u, v) (u, v) k dudv = (y u z v z u y v ) 2 + (x u z v z u x v ) 2 + (x u y v y u x v ) 2 du dv efinition Let g = g(x, y, z) be a continuous function defined on domain containing, the surface integral of the function g on is g(x, y, z) d = g(x, y, z(x, y)) 1 + z 2 x + z 2 y dxdy, or = g(x(u, v), y(u, v), z(u, v)) r u r u du dv

efinition Let be a parameterized surface or a graph, then the surface area of is given by 1 d = 1 + z 2 x + z 2 y dxdy, or = r u r v dudv ExampleFind the area of the part of the surface 2z = x 2 that lies directly above the triangle in the xy-plane with vertices at A(, ), B(1, ) and C(1, 1) olution The surface is a graph z(x, y) = x 2 /2 defined on the region = { (x, y) x 1, y x }, then z x = x and z y = The surface area is 1 + z 2 x + x 2 y dxdy ABC

efinition Let be a parameterized surface or a graph, then the surface area of is given by 1 d = 1 + z 2 x + z 2 y dxdy, or = r u r v dudv ExampleFind the area of the part of the surface 2z = x 2 that lies directly above the triangle in the xy-plane with vertices at A(, ), B(1, ) and C(1, 1) olution The surface is a graph z(x, y) = x 2 /2 defined on the region = { (x, y) x 1, y x }, then z x = x and z y = The surface area is ABC 1 1 2 1 + z 2 x + x 2 y dxdy = 1 + x 2 d(1 + x 2 ) = 1 2 1 x 1 1 + x 2 dydx = x 1 + x 2 dx = [ (1 + x 2 ) 1/2+1 1 + 1/2 ] 1 = 1 3 (2 2 1)

Example Find the area of the part of the paraboloid : z = 9 x 2 y 2 that lies above the plane z = 5 olution For any point P(x, y, 9 x 2 y 2 ) of the graph lying above the plane z = 5, we have 9 x 2 y 2 5, so its projection point Q(x, y) on xy-plane satisfies x 2 + y 2 2 2 ince z = 9 x 2 y 2, so z x = 2x, and z y = 2y, hence z 2 x + z 2 y = 4(x 2 + y 2 ) It follows the surface area of is 1 d = 1 + z 2 x + z 2 y dxdy = x 2 +y 2 2 2 2 1 8 x 2 +y 2 2 2 1 + 4(x 2 + y 2 ) dxdy = 1 + 4r 2 d(1 + 4r 2 ) = 1 8 1 12 (93/2 1 3/2 ) = 26 12 = 13 6 2π 2 r [ (1 + 4r 2 ) 1/2+1 1 + 1/2 1 + 4r 2 drdθ = ] 2 =

efinition Let ρ(x, y, z) be the density (mass per unit area) of the a surface in space, and the mass of is given by g(x, y, z) d Example Let ρ(x, y, z) = z 2 be the density function of the upper hemisphere : x 2 + y 2 + z 2 = a 2, z Find its mass olution Write as in the graph of the function z(x, y) = a 2 x 2 y 2, defined on the region = { (x, y) x 2 + y 2 a 2 } For any point (x, y, z), we have z = a 2 x 2 y 2 x, z x =, and z y a 2 x 2 y 2 y = Then a 2 x 2 y 2 1 + z 2 x + z 2 y = 1 + x2 +y 2 a = 2 Moreover on, we have a 2 x 2 y 2 a 2 x 2 y 2 z 2 = ( a 2 x 2 y 2 ) 2 = a 2 x 2 y 2 Then the mass of is given by ρ(x, y, z) d = 2π a (a 2 x 2 y 2 ) a 2 r 2 r dr dθ = 2πa 2 = a [ = πa (a 2 r 2 ) 3/2] a = πa4 a a dx dy a 2 x 2 y2 a 2 r 2 d(a 2 r 2 )

Example Let be a parametric surface given by r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, where (u, v) lies in the domain of of uv-plane On, define a vector field i j k x y (y, z) (z, x) (x, y) z N(u, v) = r u r v = u u u = i + j + x y z (u, v) (u, v) (u, v) k v v v Proposition The vector field N is continuous on, and is normal to the tangent plane of the surface everywhere efinition A parametric surface given by r = r(u, v) is called orientable if there is a continuous unit normal vector field n on A choice of n is called an orientation of, in this case, is called an oriented surface For an oriented surface with an orientation n, one defines the unit normal vector field n(u, v) on, by n(u, v) = N N = 1 ( ) (y, z) (z, x) (x, y) i + j + N (u, v) (u, v) (u, v) k

urface Integral with respect to coordinate area elements efinition Let g = g(x, y, z) be a scalar function defined on a domain containing, define the surface integral of with respect to the coordinate x-axis, y-axis and z-axis respectively as (y, z) g(x, y, z)n i d = g(r(u, v)) du dv, (u, v) (z, x) g(x, y, z)n j d = g(r(u, v)) du dv, and (u, v) (x, y) g(x, y, z)n k d = g(r(u, v)) du dv, (u, v) n i d = 1 ( ) (y, z) N (u, v),, (1,, ) N du dv = (y, z) du dv = dx dy (u, v) Remark The jacobian factors are used to make the integral independent of parametrization of the surface

urface Integral of Vector Field Let F = (P, Q, R) be a continuous vector field defined in a domain containing a smooth oriented surface, with unit normal vector field efine the flux of F across, or the surface integral of F over as ( ) (y, z) (z, x) (x, y) F n d = (P, Q, R),, du dv (u, v) (u, v) (u, v) = P dydz + Q dzdx + R dxdy Remark If we change the orientation of, the surface integral will change by a minus sign

Example uppose that is given as a graph of the function z = z(x, y) where (x, y) is in a domain in xy-plane Prove that the flux of F = (P, Q, R) is given by F n d = ( P(x, y, z(x, y)) z ) z Q(x, y, z(x, y)) + R(x, y, z(x, y)) dxdy x y Proof Parameterize by r(x, y) = (x, y, z(x, y)), then (y, z) (x, y) = y x y y = 1 (z, x) = z x, z x z y z x z y (x, y) = z x z y = z y, x x x y (x, y) and = 1 Hence, (x, y) F n d = (P, Q, R) ( (y, z) (x, y), (z, x) (x, y) ) (x, y), (x, y) ( P(x, y, z(x, y)) z z Q(x, y, z(x, y)) + R(x, y, z(x, y)) x y dx dy = ) dxdy Remark The formula above may be used to calculate the flux of F across the graph of a function z = z(x, y)

Example Calculate the flux F n d, where F = k, and is the upper hemi-sphere z = a 2 x 2 y 2 with : x 2 + y 2 a 2, with the outer normal vector field on olution As the flux of F = (P, Q, R) = (,, 1) across is given by F n d = = ( P(x, y, z(x, y)) z x 1 dxdy = Area of = πa 2 ) z Q(x, y, z(x, y)) + R(x, y, z(x, y)) dxdy y Remark In the following, we directly compute F n and d in terms of x and y

Example Calculate the flux F n d, where F = k, and is the upper hemi-sphere x 2 + y 2 + z 2 = a 2, z of radius a centered at (,, ), with the outer normal vector field n on olution It follows from the gradient that n(x, y, z) = (x2 +y 2 +z 2 ) (x 2 +y 2 +z 2 ) = (x,y,z) a, and F n = a z o one can rewrite as the graph of the function z = a 2 x 2 y 2 defined on = { (x, y) x 2 + y 2 a 2 } Then it follows from d = 1 + z 2 x + z 2 y dxdy that z a F n d = a d = 2 x 2 y 2 1 + z 2 x + z 2 y dxdy a a = 2 x 2 y 2 a a a x 2 y dxdy = dxdy = πa 2 2

Example Find the flux of F(x, y, z) = (x, y, 3) out of the region bounded by the paraboloid z = x 2 + y 2 and the plane z = 4 olution Let 1 be the circular top { (x, y, 4) x 2 + y 2 2 2 }, and 2 be the parabolic part parameterized by z(x, y) = x 2 + y 2 with x 2 + y 2 4 as shown in the diagram In this case, the outward unit normal vector field is given by on 1 : n 1 (x, y, z) = k, and on 2 by n 2 (x, y, z) = ( z + x2 + y 2 ) (2x, 2y, 1) ( z + x 2 + y 2 = In fact, one can ) 4x 2 + 4y 2 + 1 check n 2 (,, ) = 1, so F 1 n 1 = (x, y, 3) (,, 1) = 3, and (2x,2y, 1) F 2 n 2 = (x, y, 3) = 2x2 +2y 2 3 On 4x 2 +4y 2 +1 4x 2, the surface area 2 +4y 2 +1 element d = 1 + z 2 x + z 2 y dxdy = 4x 2 + 4y 2 + 1 dxdy for = { (x, y) x 2 + y 2 4} Then F n d = F n 1 d + F n 2 d = 3 dxdy + (2x 2 + 2y 2 3) dxdy 1 2 2π 2 [ 2r = (2x 2 + 2y 2 ) dxdy = 2r 2 4 ] 2 r dr dθ = 2π = 16π 4

Example uppose that a body has temperature u(x, y, z) at its point (x, y, z) The flow of heat in the body is described by the heat-flow vector q(x, y, z) = k u(x, y, z) for some constant k efinition Let be a closed surface within the body bounding the solid region and let n denote the outer unit normal vector for Then the net heat flow across the sphere out of the region across its boundary is defined by q n d = k u n d

Example uppose that a uniform solid ball B of radius R is centered at the origin, and at its point (x, y, z) the temperature u(x, y, z) = c(r 2 x 2 y 2 ) Find the rate of the flow of heat across the sphere of radius a < R centered at O(,, ) olution The heat flow across the sphere = { (x, y, z) x 2 + y 2 + z 2 = a 2 } is given by q n d = k u n d = k (R 2 x 2 y 2 z 2 ) n d = (x, y, z) 2(x k ( 2x, 2y, 2z) d = 2 + y 2 + z 2 ) k d = a a 2ak d = 2ak(4πa 2 ) = 8kπa 3

ivergence Theorem efinitions A surface is called piecewise smooth, if it consists of a finite number of smooth parametric surfaces A surface is called closed if it is the boundary of a bounded solid region in space ivergence Theorem uppose that is a closed piecewise smooth surface that bounds a space region Let n be the outer unit normal vector field, which is continuous on each smooth piece of If F = (P, Q, R) is continuously differentiable on T, then F n d = F dv, ie ( P P dydz + Q dzdx + R dxdy = x + Q y + R ) dv z Remark The condition that the surface is closed plays a crucial roles in divergence theorem

Example Let be the surface (with outer unit normal vector n) of the region bounded by the planes z =, y =, y = 2 and the parabolic cylinder z = 1 x 2 Apply the divergence theorem to compute F n d, where F(x, y, z) = (x + cos y)i + (y + sin z)j + (z + e x )k olution For any point (x, y, z) in, we have z 1 x 2, hence 1 x 2, so x 2 1, ie 1 x 1, hence we have = { (x, y, z) 1 x 1, y 2, z 1 x 2 } F(x, y, z) = (x + cos y) x + (y + sin z) y + (z + e x ) z = 1 + 1 + 1 = 3 Instead of evaluating the surface integral directly, we can apply the divergence theorem that F n d = F dv = 3 dv = 1 2 1 x 2 1 3 dz dy dx = 3 2 (1 x 2 ) dx = 6(2 2 3 ) = 8 1 1 Remark Though we had not determined the outer normal vector field n, but it is necessary to know that n is pointing outward on the boundary of the solid region, before we apply divergence theorem

Example Let be the surface of the solid cylinder bounded by the planes z =, z = 3 and the cylinder x 2 + y 2 = 4 Calculate the outward flux F n d, where F(x, y, z) = (x 2 + y 2 + z 2 )(xi + yj + zk) olution Note that the divergence F(x, y, z) = (x 3 + xy 2 + xz 2 ) x + (x 2 y + y 3 + yz 2 ) y + (x 2 z + y 2 z + z 3 ) z = 5(x 2 + y 2 + z 2 ) It follows from the divergence theorem and then use cylindrical coordinates that F n d = F dv = 5(x 2 + y 2 + z 2 ) dv = 2π 2 3 = 1π = 1π 2 5(r 2 + z 2 ) r dz dr dθ [ r 3 z + 1 ] 3 3 rz3 [ 3 4 r4 + 9 2 r2 ] 2 2 dr = 1π (3r 3 + 9r) dr = 3π

Example Let be the sphere x 2 + y 2 + z 2 = 4, which bounds a solid ball Calculate the outward flux F n d, where F(x, y, z) = (x 2 + y 2 + z 2 )(xi + yj + zk) olution Note that the divergence F(x, y, z) = (x 3 + xy 2 + xz 2 ) x + (x 2 y + y 3 + yz 2 ) y + (x 2 z + y 2 z + z 3 ) z = 5(x 2 + y 2 + z 2 ) It follows from the divergence theorem and then use spherical coordinates that F n d = F dv = 5(x 2 + y 2 + z 2 ) dv = = 2π π a a 5ρ 4 dρ 5ρ 2 ρ 2 sin ϕ dρ dϕ dθ π sin ϕ dϕ 2π = a 5 [1 ( 1)] 2π = 4πa 5 dθ

Example Let be a solid region bounded by a smooth parametric closed surface, with the outer unit normal vector field n to Prove that the volume of is 1 x dydz + y dzdx + z dxdy 3 olution Let F(x, y, z) = (x, y, z), and F(x, y, z) = (x, y, z) = (x) x + (y) y + (z) z = 3 Then it follows from the definition of flux integral and divergence theorem that 1 x dydz + y dzdx + z dxdy = 3 1 F n d = 1 F dv = 1 3 dv = volume of 3 3 3 Example Let V and A be the volume and the surface area of the sphere of radius a centered at (,, ) Prove that 3V = aa olution Let be the sphere given by { (x, y, z) x 2 + y 2 + z 2 = a 2 }, (x, y, z) hence the outer normal vector n(x, y, z) = Then 3V a = 1 x dydz + y dzdx + z dxdy = (x, y, z) n d = 3 (x, y, z) (x, y, z) d = 1 (x 2 + y 2 + z 2 ) d = a d = aa a a

Example how that the divergence of the continuously differentiable 1 vector field F at the point P is given by F(P) = lim F n d, r B r r where r is the sphere of radius r centered at P and B r = 3 4 πr3 is the volume of the ball B r that the sphere bounds olution As F is continuously differentiable at P(a, b, c), so F is continuous at P Then for any ε >, there exists δ > such that F(x, y, z) F(a, b, c) < ε for all point (x, y, z) in the open ball centered at P(a, b, c) with radius δ > Then F(P) 1 F n d B r = 1 r B r B r F(P) F n d r = 1 B r F(P)dV F d B r B r = 1 F(P) F d < 1 ε dv = B r ε = ε, and B r B r B r B r B r hence the result follows Remark Let F be a vector field defined on a region containing a point P, then P is called a source if F(P) >, and sink if F(P) <

Example The outward flux of vector field F(x, y, z) = xi + yj + zk across the sphere x 2 + y 2 + z 2 2x 2y 2z = 3 is olution Let be the solid bounded by the sphere, then from divergence theorem we have the outward flux of F is F n d = F dv = ( x x + x x + x x ) dv = 3 dv = 3 volume of = 3 4( 6) 3 π/3 = 24π 6

Example Let F(x, y, z) = (x, y, z), and T be the surface defined by { (x, y, z) R 3 x + y + z = 1 }, with outward pointing unit normal n on T Evaluate the surface integral F nd T olution Let be the solid bounded by the surface T, ie = { (x, y, z) R 3 x + y + z 1 } One can check that the condition of the divergence theorem holds, and hence F nd = divfdv = ( x T x + x x + x x ) dv = 3 dv = 3Vol() = 3 8 1 3! 1 1 1 = 4

Example Let T be the solid cylinder defined by the inequalities x 2 + y 2 1 and z 4 Let be the entire boundary of T Let F(x, y, z) = (5z + 3)k Find the outward flux of F through olution I Let 1 = { (x, y, 4) x 2 + y 2 4 } be the top of T, and 2 = { (x, y, ) x 2 + y 2 4 } be the bottom of T The outer unit normal vector field on 1 and 2 are n 1 = k, and n 2 = k respectively The outward flux of F through 1 = F 1 n d = (5 1 4 + 3)k k d = 23 Area of 1 = 23π The outward flux of F through 2 = F 2 n d = (5 2 + 3)k ( k) d = 3 Area of 2 = 3π The flux through the curved part of the boundary is since n is horizontal there, so F n =, o the total flux of F across is 23π 3π = 2π olution II We have div F = + + 5 = 5, so by the divergence theorem, the flux equals F n d = F dv = 5 Volume of T = 5 π 1 4 = 2π T

tokes Theorem efinition A piecewise smooth surface in space is called an oriented surface, if one chooses a continuous unit normal vector field on each smooth piece A positive orientation of the boundary C of an oriented surface is a unit tangent vector T such that n T always points into Remark Think of a man walks along C as T, and heads up in n, then the region is always on the left hand side of the man tokes Theorem Let be an oriented, bounded, and piecewise smooth surface in space with positive oriented boundary C with respect to the unit normal vector field n on uppose that T is a positively oriented unit vector field tangent to C in the right orientation of C If F is a continuously differentiable vector field defined in a region containing, then F T ds = (curl F) n d C Remark The condition that the positive orientation of C agrees with the choice of unit normal vector field n is essential

Example Let F(x, y, z) = (2y, 2x, sin z) Let be the upper half of the sphere x 2 + y 2 + z 2 = 9 Evaluate the outward (ie, upward) flux of curl F through olution The surface is given by { (x, y, z) x 2 + y 2 + z 2 = 9, z }, then its boundary C is the circle x 2 + y 2 = 9 in the xy-plane oriented counterclockwise o C is parameterized by r(t) = (3 cos t, 3 sin t, ) for t [, 2π] The flux is curlf n d = F T ds = = 2π 2π C (2 3 sin t, 2 3 cos t, ) ( 3 sin t, 3 cos t, ) dt 18(cos 2 t + sin 2 t) dt = 36π Remark As n is already fixed in the question, you can use the same idea of a man walking along the boundary C so that the surface appears in his left hand side to determine the positive orientation of C with respect to n of

Example Let F(x, y, z) = 3zi + 5xj 2yk Evaluate the line integral F T ds where C is the ellipse in which the plane z = y + 3 C intersects the cylinder x 2 + y 2 = 1 Orient C counterclockwise as view from above olution Let be the region on the plane z = y + 3 bounded by the cylinder x 2 + y 2 = 1 Then = { (x, y, z) z y = 3, x 2 + y 2 1 } is part of a level surface z y = 3, with a unit normal vector field (z y) (, 1, 1) n(x, y, z) = = on Next the (z y) 2 curl F(x, y, z) = i j k x y z = 2i + 3j + 5k, hence 3z 5x 2y curl F n = ( 2, 3, 5) (, 1, 1)/ 2 = ( 3 + 5)/ 2 = 2 F T ds = (curl F) n d = 2 d = 2Area of () = C 2 1 2π = 2π Remark can be parameterized by r(x, y) = (x, y, y + 3) defined on = { (x, y) x 2 + y 2 1 }

Example Let F(x, y, z) = (y, z, 2x) Let C be a simple closed curve contained in the plane x + y + z = 1 how that F T ds = C olution Let By tokes theorem, F T ds = (curl F) n d Here n is perpendicular to everywhere, and hence perpendicular to the (x + y + z) (1, 1, 1) plane with the unit normal vector n = = to (x + y + z) 3 the plane On the other hand, curl F(x, y, z) = curl (y, z, 2x) = ( 1, 2, 1), so curl F(x, y, z) n =, thus F T ds = ds = C C C

Example Evaluate the surface integral ( F) n d, where F(x, y, z) = 3zi + 5xj 2yk, and is the parabolic surface z = x 2 + y 2 that lie below the plane z = 4 and whose orientation is given by the upper unit normal vector olution The boundary of is the circle C parameterized by r(t) = (2 cos t, 2 sin t, 4), where t 2π It follows from the tokes theorem that ( F) n d = F T ds = 3z dx + 5x dy 2y dz = = 2π 2π C 3 4( 2 sin t dt) + 5 (2 cos t) (2 cos t dt) + 2 (2 sin t) ( dt) ( 24 sin t + 2 cos 2 t) dt = 2π C 1(1 + cos 2t) dt = 2π

Conservative and Irrotational Fields efinition A differentiable vector field F is called irrotational if F = yi + xj Example The vector field F(x, y) = x 2 + y 2 is irrotational on = R 2 \ {(, )}, but we had proved that F is not conservative on efinition Let be a region in space, is called simply connected, if every simple closed curve in can be continuously shrunk to a point while staying inside Examples (a) The entire plane, entire space are simply connected; (b) Rectangle, the sphere and ball are simply connected; (c) The interior of a torus, and the punctured plane are not a simply connected

Proposition Let F be a continuous vector field defined on a region, prove that the line integral of F is independent of path if and only if F T ds = for any piecewise smooth closed curve C in C Proof uppose that line integral of F is independent of path, then let C be any closed curve with the same starting and terminal point A, then the constant path C with A for all t is also a curve with the same starting and terminal point A It follows from the path independence of C the line integral of F that F T ds = F T ds = F ds = C C Conversely, suppose C 1 and C 2 are two paths, both of them starts from the same point A, and terminates at point B Let C = C 1 ( C 2 ) be a closed path from A to B via C 1, and back from B to A via C 2 (in reverse direction of C 2 ) Then C is a piecewise smooth closed curve in, hence one has = F T ds = F T ds + F T ds = F T ds F T ds C C 1 C 2 C 1 C 2 Hence F T ds = F T ds o the line integral of F is C 1 C 2 independence of path

Theorem Let F be a continuously differentiable vector field in a simply connected region in space Then the vector field F is irrotational if and only if it is conservative; that is F = in if and only if F = f for some scalar function f defined on olution If F = f, then F = f = on Conversely, suppose that F is irrotational on, one wants to define a function f (x, y, z) such that f = F on Following the same idea in the planar case, Let A(a, b, c) and B(x, y, z) be two points in, and C be a path from A to B If one can show that the line integral of F is path independent, then define f (x, y, z) = (x,y,z) (a,b,c) F T ds, which depends only on the end points of C Then it follows from continuity of the vector field F that f = F on It remains to show that the line integral of F is independent of path, which is equivalent to the fact that F T ds = for any closed path C in As is simply connected, it C follows one can shrink the curve C continuously to a path in such way that C bounds a surface via the shrinking, then F T ds = ( F) n d = n d = C

ummary of Line Integral of Vector Field Let F be a vector field defined on a region Then we can summary our important result as follows: F is conservative on F = f for some function f ( ) xy = ( ) yx F T ds = F = on C for any closed path in imply connected = tokes Thm Remark Conservative vector field is irrotational; but the converse is not true In fact, it depends on the domain of the vector field ( ( y, x) compare the lower horizontal arrow) An example is F(x, y) = x 2 +y 2 on R 2 \ {(, )}

Example (a) how that the vector field F(x, y, z) = 3x 2 i + 5z 2 j + 1yzk is irrotational on R 3 (b) Find a potential function f (x, y, z) such that f = F olution (a) One needs to show that F(x, y, z) = (3x 3, 5z 2, 1yz) = (1z 1z)i = (b) One can use the line integral of F along the segment from (,, ) to (x, y, z) by r(t) = (tx, ty, tz) for t 1 as follows: f (x, y, z) = = = 1 1 (x,y,z) (,,) F T ds ( 3(tx) 2, 5(tz) 2, 1(ty)(tz) ) (x, y, z) dt (3x 3 + 15yz 2 )t 2 dt = x 3 + 5yz 2 + C

Example Let T be the solid bounded by the paraboloids 1 : z = x 2 + 2y 2 and 2 : z = 12 2x 2 y 2 Evaluate the outward flux of F(x, y, z) = xi + yj + zk across the boundary of T olution Project the solid onto the xy-plane with its shadow R Let Q(x, y, z) be the intersection of 1 and 2, then z = x 2 + 2y 2 = z = 12 2x 2 y 2, so 3(x 2 + y 2 ) = 12, then x 2 + y 2 = 4 Then the image Q (x, y) of Q in R satisfies the equation x 2 + y 2 = 2 2, ie Q lies on a circle Inside the circular disc R, we have x 2 + y 2 2, so it follows that x 2 + 2y 2 12 2x 2 y 2, and hence T = { (x, y, z) x 2 + y 2 2 2, x 2 + 2y 2 z 12 2x 2 y 2 } It follows from divergence theorem that 12 2x 2 y 2 F n d = F dv = (1 + 1 + 1) dv 2π 2 = 3 (12 3r 2 )r drdθ = 6π T x 2 +y 2 2 2 x 2 +2y 2 [ 6r 2 3 4 r4 ] 2 = 6π(24 12) = 72π

Example Let F(x, y, z) = (x + y)i + (y x)j + zk be vector field Let be the hemisphere x 2 + y 2 + z 2 = 4, z, and n be the outward pointing unit normal vector field of Then curlf ndσ is olution The boundary of is the circle given by C : r(t) = ( x(t), y(t), z(t) ) = (2 cos t, 2 sin t, ), t 2π From the tokes theorem, we have curlf ndσ = F Tds = = = 2π t= 2π 2π C [ (x(t) + y(t))x (t) + (y(t) x(t))y (t) + z(t)z (t) ]dt 4(cos t + sin t) sin t + 4(sin t cos t) cos tdt ( 4 sin 2 t 4 cos 2 t)dt = 8π

Example Let F(x, y, z) = f (x, y, z)k, where f (x, y, z) is a differentiable function defined in R 3 Then the outward flux curlf n d of curlf across the upper hemisphere = { (x, y, z) x 2 + y 2 + z 2 = 1, z } is A B f (,, ) C grad f (,, ) None of the above olution Let C be the unit circle x 2 + y 2 = 1 in the usual counterclockwise direction, one can apply tokes theorem to obtain the following curlf n d = F T ds = ds = C C