# AB2.5: Surfaces and Surface Integrals. Divergence Theorem of Gauss

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1 AB2.5: urfaces and urface Integrals. Divergence heorem of Gauss epresentations of surfaces or epresentation of a surface as projections on the xy- and xz-planes, etc. are For example, z = f(x, y), x = g(x, z) g(x, y, z) =. z = + a 2 x 2 y 2 or x 2 + y 2 + z 2 = a 2, z represents a hemisphere of radius a and center O. A surface can be represented by a vector function r(u, v) = [x(u, v), y(u, v), z(u, v)] = x(u, v)i + y(u, v)j + z(u, v)k, u, v his is called a parametric representation of a surface, u, v varying in a two-dimensional region are the parameters of the representation. EXAMPLE 1 Parametric representation of a cylinder A circular cylinder x 2 + y 2 = a 2, 1 z 1 has radius a, height 2, and the z-axis as the axis. A parametric representation is he components of r(u, v) are r(u, v) = [a cos u, a sin u, v] = a cos ui + a sin uj + vk, u, v in rectangle : u 2π, 1 v 1. x = a cos u, y = a sin u, z = v. EXAMPLE 2 Parametric representation of a sphere A sphere x 2 + y 2 + z 2 = a 2 has the parametric representation r(u, v) = a cos v cos ui + a cos v sin uj + a sin vk, u, v in rectangle : u 2π, π/2 v π/2.

2 he components of r(u, v) are x = a cos v cos u, y = a cos v sin u, z = a sin v. EXAMPLE 3 Parametric representation of a cone A circular cone z = + x 2 + y 2, z H has the parametric representation r(u, v) = u cos vi + u sin vj + uk, u, v in rectangle : v 2π, u H. he components of r(u, v) are x = u cos v, y = u sin v, z = u. Indeed, this yields x 2 + y 2 = z 2. angent to a surface Get a curve C on by a pair of continuous functions u = u(t), v = v(t) so that C has the position vector r(u(t), v(t)). By the chain rule, we get a tangent vector of a curve C r (t) = d r dt = r u u + r v v. Hence the partial derivatives r u and r v at a point P are tangential to at P and we assume that are linearly indepedent. hen their vector product gives a normal vector N of at P, he corresponding unit normal vector n N = r u r v. n = 1 N N = 1 r u r v r u r v. If is represented by then n = g(x, y, z) =, 1 grad g. grad g

3 EXAMPLE 4 Unit normal vector of a sphere g(x, y, z) = x 2 + y 2 + z 2 a 2 = : n = 1 grad g grad g = 1 [ x a grad g = a, y a, z ] = x a a i + y a j + z a k. EXAMPLE 5 Unit normal vector of a cone g(x, y, z) = z + x 2 + y 2 = : n = 1 grad g grad g = 1 [ 2 x x2 + y, 2 x x2 + y i + y 2 x2 + y j k. 2 y x2 + y 2, 1 ] = Definition and evaluation of surface integrals A surface integral of a vector function F(r) over a surface is defined as F nda = F(r(u, v)) N(u, v)dudv, where r(u, v) = [x(u, v), y(u, v), z(u, v)] = x(u, v)i + y(u, v)j + z(u, v)k, u, v is a parametric representation of with a normal vector and the corresponding unit normal vector Note that N = r u r v n = 1 N N. nda = n N dudv = N dudv, and it is assumed that the parameters u, v vary in a domain on the u, v-plane. In terms of components F = [F 1, F 2, F 3 ) = F 1 i + F 2 j + F 3 k, and n = [cos α, cos β, cos γ] = cos αi + cos βj + cos γk, N = [N 1, N 2, N 3 ) = N 1 i + N 2 j + N 3 k, F nda = (F 1 cos α + F 2 cos β + F 3 cos γ)da =

4 (F 1 N 1 + F 2 N 2 + F 3 N 3 )dudv. EXAMPLE 1 Flux through a surface Compute the flux of water through the parabolic cylinder : y = x 2, x 2, z 3 if the velocity vector is v = F = [3z 2, 6, 6xz]. olution. is represented by r(u, v) = [u, u 2, v] = ui + u 2 j + vk, u 2, v 3 because one can set x = u, z = v, and y = x 2 = u 2. From this r u = [1, 2u, ], r v = [,, 1]; the vector product of r u and r v gives a normal vector N of the parabolic cylinder N = r u r v = he corresponding unit normal vector 1 2u 1 = 2ui j = [2u, 1, ]. n = 1 N N = 1 (2ui j) u 2 On Hence F(r(u, v)) = F() = [3v 2, 6, 6uv] = 3(v 2 i + 2j + 2uvk). F(r(u, v)) N(u, v) = 3[v 2, 2, 2uv] [2u, 1, ] = 3(2uv 2 2) = 6(uv 2 1). he parameters u, v vary in the rectangle : u 2, v 3. Now we can write and calculate the flux integral: 3 2 F nda = 3 2 6(uv 2 1)dudv = 6( v 2 dv udu F(r(u, v)) N(u, v)dudv = 3 2 dudv) = 6( ) = 72.

5 EXAMPLE 2 urface integral Compute the surface integral for being a portion of the plane : x + y + z = 1, x, y, z 1. for F = [x 2,, 3y 2 ]. olution. etting x = u and y = v, we have z = 1 u v, so that can be represented by From this a normal vector r(u, v) = [u, v, 1 u v], v 1, u 1 v. N = r u r v = he corresponding unit normal vector r u = [1,, 1], r v = [, 1, 1]; = i + j + k = [1, 1, 1]. n = 1 N N = 1 3 (i + j + k). On Hence F(r(u, v)) = F() = [u 2,, 3v 2 ] = u 2 i + 3v 2 k). F(r(u, v)) N(u, v) = [u 2,, 3v 2 ] [1, 1, 1] = u 2 + 3v 2.dudv. he parameters u, v vary in the triangle : v 1, u 1 v. Now we can write and calculate the surface integral: F nda = F(r(u, v)) N(u, v)dudv = (u 2 + 3v 2 )dudv = = (1/3) 1 1 v 1 (u 2 + 3v 2 )dudv = (1 v) 3 dv v 1 1 v dv u 2 du + 3 v 2 dv du = v 2 (1 v)dv = (1/3) 1 t 3 dt + 3 (1/3) (1/4) + 3(1/3 1/4) = 1/3. 1 (v 2 v 3 )dv =

6 Divergence theorem of Gauss ecall that if v(x, y, z) is a differentiable vector function, v(x, y, z) = v 1 (x, y, z)i + v 2 (x, y, z)j + v 3 (x, y, z)k, then the function div v = v 1 x + v 2 y + v 3 z is called the divergence of v. Formulate the divergence theorem of Gauss. Let be a closed bounded region in space whose boundary is a piecewise smooth orientable surface (consists of finitely many smooth surfaces). Let F(x, y, z) be a vector function that is continuous and have continuous first partial derivatives everywhere in some domain containing. hen div FdV = F nda. Here n is an outer unit normal vector of pointing to the outside of. In components, ( F 1 x + F 2 y + F ) 3 dxdydz = (F 1 cos α + F 2 cos β + F 3 cos γ)da. z or ( F 1 x + F 2 y + F ) 3 dxdydz = z (F 1 dydz + F 2 dzdx + F 3 dxdy). EXAMPLE 1 Evaluation of a surface integral by the divergence theorem Evaluate I = (x 3 dydz + x 2 ydzdx + x 2 zdxdy), where is a piecewise smooth surface consisting of the cylinder x 2 + y 2 = a 2 ( z b) and the circular disks z = and z = b (x 2 + y 2 a 2 ) ( consists of three parts of smooth surfaces). olution. We have Hence the divergence of F = [F 1, F 2, F 3 ] is Introducing the polar coordinates F 1 = x 3, F 2 = x 2 y, F 3 = x 2 z. div F = F 1 x + F 2 y + F 3 z = 3x2 + x 2 + x 2 = 5x 2. x = r cos θ, y = r sin θ (cylindrical coordinates r, θ, z)

7 we have dxdydz = rdrdθdz, and by the divergence theorem, the surface integral is transformed to a triple integral over the closed region in space whose boundary is the surface of the cylinder, (x 3 dydz + x 2 ydzdx + x 2 zdxdy) = 5b div FdV = b a 2π 5 r 2 cos 2 θrdrdθdz = z= r= θ= a 2π 5b a4 8 r 3 cos 2 θdrdθ = 5b a4 4 2π 2π (1 + 2 cos θ)dθ = 5 4 πba4. EXAMPLE 2 Verification of the divergence theorem Evaluate I = F nda), F = 7xi zk cos 2 θdθ = over the sphere : x 2 + y 2 + z 2 = 4 by the divergence theorem and directly. 5x 2 dxdydz = olution. We have he divergence of F is F = [F 1,, F 3 ], F 1 = 7x, F 3 = z. div F = F 1 x + F 2 y + F 3 z = = 6, and, by the divergence theorem, I =, ball div FdV = 6, ball dxdydz = π23 = 64π. Now we will calculate the surface integral over directly. Use the parametric representation of the sphere of radius 2 hen : r(u, v) = 2 cos v cos ui + 2 cos v sin uj + 2 sin vk, u, v in rectangle : u 2π, π/2 v π/2. r u = [ 2 sin u cos v, 2 cos v cos u, ], r v = [ 2 sin v cos u, 2 sin v sin u, 2 cos v], N = r u r v = 2 sin u cos v 2 cos v cos u = [4 cos 2 v cos u, 4 cos 2 v sin u, 4 cos v sin v]. 2 sin v cos u 2 sin v sin u 2 cos v

8 On we have and, correspondingly, x = 2 cos v cos u, z = 2 sin v, Hence F(r(u, v)) = F() = [7x,, z] = [14 cos v cos u,, 2 sin v]. F(r(u, v)) N(u, v) = (14 cos v cos u)4 cos 2 v cos u + ( 2 sin v)(4 cos v sin v) = 56 cos 3 v cos 2 u 8 cos v sin 2 u. he parameters u, v vary in the rectangle : u 2π, π/2 v π/2. Now we can write and calculate the surface integral: 8 F nda = F(r(u, v)) N(u, v)dudv = 2π π/2 8 (7 cos 3 v cos 2 u cos v sin 2 v)dudv = π/2 { 7 2π (1 + cos 2u)du 2 8π { π/2 56π 7 π/2 π/2 π/2 π/2 π/2 } π/2 cos 3 vdv 2π cos v sin 2 vdv = π/2 π/2 cos 3 vdv 16π cos vdv sin 2 vdv = π/2 } (1 sin 2 v)d sin v 2 8π { 1 7 (1 t 2 )dt 2 1 he values obtained by both methods coincide. π/2 π/2 1 1 dv sin 2 vd sin v } t 2 dt = 8π[7 (2 2/3) 4/3] = 8π 4/3 6 = 64π. EXAMPLE 2 Applications of the divergence theorem By the mean value theorem for triple integrals, f(x, y, z)dv = f(x, y, z )V ( ) = where (x, y, z ) is a certain point in and V ( ) is the volume of. By the divergence theorem, div F(x, y, z ) = 1 V ( ) div FdV = 1 V ( ) ( ) F nda. Choosing a fixed point P : (x 1, y 1, z 1 ) in and shrinking down to onto P so that the maximum distance d( ) of the points of from P tends to zero, we obtain div F(x 1, y 1, z 1 ) = lim d( ) 1 V ( ) ( ) F nda,

9 which is sometimes used as the definition of the divergence. From this expression, it follows that the divergence is independent of a particular choice of Cartesian coordinates. EXAMPLE 4 A basic property of solutions of Laplaces equation ecall that we can transform the double integral of the Laplacian of a function into a line integral of its normal derivative. In the same manner, by the divergence theorem, we can transform the triple integral of the the Laplacian of a function into a surface integral of its normal derivative. Indeed, setting F = grad f we have On the other hand, div F = div grad f = 2 f x f y f z 2 = 2 f. F n = grad f n is, by definition, the normal derivative of f (the directional derivative in the direction of the outer normal vector to, the boundary of ), f. hus, by the divergence theorem, the n desired formula for the integral of the Laplacian of f becomes 2 fdxdydz = f n da. hus, if f(x, y, z) is a harmonic function in ( 2 f = in ), then the integral of the normal derivative of this function over any piecewise smooth orientable surface in whose entire interior belongs to is zero.

10 POBLEM Find the normal vector to the xy-plane r(u, v) = [u, v] = ui + vj and the parameter curves u = const and v = const. olution. ecall that the vector product a b of two vectors a = [a 1, a 2, a 3 ] and b = [b 1, b 2, b 3 ] is a vector v = a b perpendicular to both a and b so that a, b, v form a right-handed triple: v = [v 1, v 2, v 3 ] = a b = a 1 a 2 a 3 = v 1 i + v 2 j + v 3 k, b 1 b 2 b 3 or For the xy-plane v 1 = a 2 a 3 b 2 b 3, v 2 = a 3 a 1 b 3 b 1, v 3 = r(u, v) = [u, v, ] = ui + vj; r u = [1,, ] = i, r v = [, 1, ] = j, a 1 a 2 b 1 b 2 and the vector product of r u and r v gives a normal vector N of the xy-plane N = r u r v = 1 = k. 1 he corresponding unit normal vector n = 1 N N = 1 1 k = k. he parameter curves u = const and v = const are straight lines. POBLEM Find the normal vector to the cone r(u, v) = u cos vi + u sin vj + cuk = [u cos v, u sin v, cu] and the parameter curves u = const and v = const. olution. he cone is given by the representation z = c x 2 + y 2. We have r u = [cos v, sin v, c], r v = [ u sin v, u cos v, ], and a normal vector N of the cone N = r u r v = cos v sin v c = cu cos vi cu sin vj + uk = u[c cos v, c sin v, 1]. u sin v u cos v

11 he parameter curves u = const are circles x 2 + y 2 = u 2, z = cu and v = const are straight lines y = x tan v. POBLEM Find a parametric representation of the plane 3x + 4y + 6z = 24. olution. We have z = 4 (1/2)x (2/3)y. herefore, setting x = 8u and y = 6v, we obtain a parametric representation r(u, v) = [8u, 6v, 4(1 u v)] = 8ui + 6vj + 4(1 u v)k. Another parametric representation can be obtained by setting x = u and y = v r(u, v) = [u, v, 4 (1/2)u (2/3)v] = ui + vj + (4 (1/2)u (2/3)v)k. ake a parametric representation of the plane 3x + 4y + 6z = 24. hen and a normal vector N r(u, v) = [8u, 6v, 4(1 u v)] r u = [8,, 4], N = r u r v = r v = [, 6, 4], = 24i + 32j + 48k = 8(3i + 4j + 6k) = 8[3, 4, 6]. he corresponding unit normal vector n = 1 N N = 1 61 (3i + 4j + 6k). POBLEM

12 Find a parametric representation of the ellipsoid x 2 + y 2 + (1/4)z 2 = 1. olution. etting x = cos v cos u, y = cos v sin u, z = 2 sin v. we see that x 2 + y 2 + (1/4)z 2 = 1, which yields the parametric representation of the ellipsoid r(u, v) = cos v cos ui + cos v sin uj + 2 sin vk, hen r u = cos v sin ui + cos v cos uj, r v = sin v sin ui sin v cos uj + 2 cos vk. he normal vector N N = r u r v = cos v sin u cos v cos u = 2 cos 2 v cos ui+2 cos 2 v sin uj+sin v cos vk. sin v sin u sin v cos u 2 cos v POBLEM Find the unit normal vector to the ellipsoid 4x 2 + y 2 + 9z 2 = 36. hen olution. We have g(x, y, z) = 4x 2 + y 2 + 9z 2 36 =. Find the partial derivatives g x = 8x, and the unit normal vector is given by g y = 2y, g z = 18z. grad g = 2[4x, y, 9z], grad g = 2 16x 2 + y z 2, n = 1 grad g grad g = x 2 + y z grad g = x2 + y z [4x, y, 9z] = 1 (4xi + yj + 9zk). 2 16x2 + y z2 POBLEM Find the unit normal vector of the plane 4x 4y + 7z = 3. olution. We have z = 1/7( 3 4x + 4y). herefore, setting x = u and y = v, we obtain a parametric representation r(u, v) = [u, v, 1/7( 3 4u + 4v)] = ui + vj + 1/7( 3 4u + 4v)k. hen r u = [1,, 4/7], r v = [, 1, 4/7],

13 and a normal vector N N = r u r v = 1 4/7 1 4/7 = he corresponding unit normal vector (4/7)i (4/7)j + k = (1/7)(4i 4j + 7k) = (1/7)[4, 4, 7]. n = 1 N N = 1 (4i 4j + 7k). 9 On the other hand, we have the representation of the plane in the form g(x, y, z) = 4x 4y + 7z + 3 =. Find the partial derivatives hus g x = 4, and the unit normal vector is given by g y = 4, g z = 7. grad g = [4, 4, 7], grad g = = 9, n = which coincides with the previous result. POBLEM grad g grad g = 1 (4i 4j + 7k) 9 Compute the surface integral for F = [3x 2, y 2, ] and being a portion of the plane r(u, v) = [u, v, 2u + 3v], u 2, 1 v 1. olution. We have a normal vector N = r u r v = he corresponding unit normal vector r u = [1,, 2], r v = [, 1, 3]; = 2i 3j + k = [ 2, 3, 1]. n = 1 N N = 1 14 ( 2i 3j + k). On Hence F(r(u, v)) = F() = [3u 2, v 2, ] = 3u 2 i + v 2 j). F(r(u, v)) N(u, v) = [3u 2, v 2, ] [ 2, 3, 1] =

14 6u 2 3v 2 = 3(2u 2 + v 2 ). he parameters u, v vary in the rectangle : u 2, 1 v 1. Now we can write and calculate the surface integral: F nda = F(r(u, v)) N(u, v)dudv = 3 (2u 2 + v 2 )dudv = = 6 dv u 2 du 3 v 2 dv du == 12 u 2 du 6 v 2 dv = POBLEM [2 (8/3) + 2/3] = 32 4 = 36. Compute the surface integral for F = [x z, y x, z y] and being a portion of the cone r(u, v) = u cos vi + u sin vj + uk, u, v in rectangle : v 2π, u 3. olution. We have and a normal vector N of the cone N = r u r v = r u = [cos v, sin v, 1], r v = [ u sin v, u cos v, ], cos v sin v 1 u sin v u cos v = u cos vi u sin vj + uk = u[cos v, sin v, 1]. On Hence F(r(u, v)) = F() = [u cos v u, u sin v u cos v, u u sin v] = u[(cos v 1)i + (sin v cos v)j + (1 sin v)k]. F(r(u, v)) N(u, v) = u[cos v 1, sin v cos v, 1 sin v] ( u)[cos v, sin v, 1] = u 2 [cos v(cos v 1) + sin v(sin v cos v) + sin v 1] = u 2 (1 cos v sin v cos v + sin v 1) = u 2 (sin v cos v sin v cos v). he parameters u, v vary in the rectangle : v 2π, u 3. Now we can write and calculate the surface integral: F nda = = 2π F(r(u, v)) N(u, v)dudv = u 2 (sin v cos v sin v cos v)dudv = 3 (sin v cos v sin v cos v)dv u 2 du =

15 2π 2π 2π ( 1/3)( sin vdv cos vdv sin v cos vdv) = ( 1/3)( + + ) =. POBLEM Find the total mass of the mass distribution of the density in the box : x 1, y 3, z 2. σ = x 2 + y 2 + z 2 olution. he required total mass M is given by the triple integral M = 2 2 dz 3 POBLEM σ(x, y, z)dxdydz = 2 z= 2 dz 3 y= 3 1 dy (x 2 + y 2 + z 2 )dx = x= dy x 2 dx + dz y 2 dy dx + z 2 dz dy dx = = = = Evaluate the surface integral for F = [x 2,, z 2 ] over the surface of the box : x 1, y 3, z 2. olution. We have Hence the divergence of F = [F 1, F 2, F 3 ] is F 1 = x 2, F 2 =, F 3 = z 2. div F = F 1 x + F 2 y + F 3 z = 2x + 2z. By the divergence theorem, the desired surface integral equals a triple integral over the box F nda = div FdV = 2 (x + z)dxdydz = dz z= 2 3 dz 1 2 dy xdx + zdz 1 2 y= dy (x + z)dxdydz = x= 1 3 dy 1 1 dx =. POBLEM

16 Evaluate the surface integral for F = [cos y, sin x, cos z] over the surface of the cylinder consisting of the cylinder x 2 + y 2 = 4 ( z 2) and the circular disks z = 2 and z = 2 (x 2 + y 2 4) olution. We have Hence the divergence of F = [F 1, F 2, F 3 ] is Introducing the polar coordinates we have F 1 = cos y, F 2 = sin x, F 3 = cos z. div F = F 1 x + F 2 y + F 3 z = sin z. x = r cos θ, y = r sin θ (cylindrical coordinates r, θ, z) dxdydz = rdrdθdz, and by the divergence theorem, the surface integral is transformed to a triple integral over the closed region in space whose boundary is the surface of the cylinder of radius 2 and height 4, F nda = div FdV = 2 2 2π sin zdxdydz = sin zdz rdrdθ =. z= 2 r= θ= POBLEM Verify basic property of solutions of Laplaces equation for f(x, y, z) = 2z 2 x 2 y 2 and the surface of the box : x 1, y 2, z 4. olution. et hen F = grad f = [ 2x, 2y, 4z]. div F = div grad f = 2 f x f y f z 2 = 2 f = =, so that f(x, y, z) = 2z 2 x 2 y 2 is a harmonic function. Now calculate directly the surface integral over the sixe successive sides of the surface of the box beginning from its upper side parallel to the x, y-plane and situated in the plane z = 4, then taking the opposite side in the plane z = etc.: f n da = f f dxdy dxdy+ z z=4 z z= f f dydz dydz+ x x=1 x x=

17 f dxdz y y=2 f dxdz = y y= ( 2) 8 + ( 4) 4 =. POBLEM Evaluate I = F nda, F = [x, z, y] over the hemisphere : x 2 + y 2 + z 2 = 4, z by the divergence theorem. olution. We have he divergence of F is F = [x, z, y], F 1 = x, F 2 = z, F 3 = y. div F = F 1 x + F 2 y + F 3 z = = 1, and, by the divergence theorem, I =,one half of ball div FdV = dxdydz = π23 = 16π 3.

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