v n ds where v = x z 2, 0,xz+1 and S is the surface that

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1 M D T P. erif the divergence theorem for d where is the surface of the sphere + + = a.. Calculate the surface integral encloses the solid region + +,. (a directl, (b b the divergence theorem. v n d where v =,,+ and is the surface that. A closed surface is made up of the part of the paraboloid = with, together with, the circular disc +, =. erif the divergence theorem for this closed surface for F =,,.. erif the divergence theorem for div F d d d where F =,,. 5. erif the divergence theorem for F =,, and for the closed surface = consisting of : half of a circular disc +,, =, : the part of the plane = with + and, and : the part of the clinder + =with and. 6. Calculate the surface integral F n d where F =,, and is the surface that encloses the solid region +,, (a directl, (b b the divergence theorem. 7. Let F =,, and let be the solid region below the cone + =, above the hemisphere =, and inside the clinder + =. Calculate div F d (a directl, (b b the divergence theorem. 8. Let be the pramid with the rectangular base =,, and the verte above the base (ape at (,,. erif the divergence theorem for div F d where F =,,. 9. Calculate the surface integral F n d where F =,, and is the surface that encloses the solid region + + 9,,, (a directl, (b b the divergence theorem.. Let F =,, and let D be the circular disc +, =. erif the divergence theorem for D div F d d d.

2 Divergence Theorem upplementar Problems - OLUTION KEY. erif the divergence theorem for d where is the surface of the sphere + + = a. a Using spherical coordinates with ρ = a, d = a sin φdφdθ; LH = F n d = d = (a sin φ cos θ (a cos φ a sin φdφdθ = a 6 cos θdθ sin φ cos φdφ The first integral is evaluated b the power-reducing formula: cos θ = In the second integral, we substitute u =cosφ, du = sin φdφ so that +cos θ sin φ cos φdφ = ( cos φcos φ sin φdφ = ( u u du = u u du = u5 LH = a ] 6 θ + sin θ] cos 5 φ 5 cos φ = a6 ( ( = 5 a6 5 u + C The surface is a level surface of g(,, = + +. Its unit normal vector is ± g g = ±,, = ± + + a, a, a To ensure the outward normal orientation, we pick the + sign: n = a, a, a. An eample of a vector field F such that F n = is F = a,,. ( divf = a + ( + ( = a Using spherical coordinates with d = ρ sin φdρdφdθ we have RH = divf d = a a (ρ cos φ ρ sin φdρdφdθ = a dθ cos φ sin φdφ a Using substitution in the middle integral u =cosφ, we obtain ] ] RH = a θ] ρ 5 a ( = a(( + = 5 a6 cos φ 5 a 5 5 ρ dρ

3 . Calculate the surface integral encloses the volume + +, (a directl, (b b the divergence theorem. v nd where v =,,+ and is the surface that Part (a: v nd = v nd + g(,, = + + ; ± g g = v nd ±,, + + }{{} 6 = ±,, ; choose + sign: n =,, ; v n =( +( + = + + = + Using spherical coordinates with ρ =, we have d = sin φdφdθ : v nd = / ( sin φ cos θ + cos φ ( sin φdφdθ =8 cos θdθ / sin φdφ + dθ / cos φ sin φdφ The integral sin φdφ is evaluated b substitution u =cosφ so that it becomes ( u du = u + u + C v nd =8 ] θ sin θ + =8 ( +8 ( =8 ( = 8 =;n = k; =; v n = ; ( d = (area of = LH = v nd = v nd + cos φ + cos φ ] / +θ] v nd = 8 = 6 ] / sin φ ( Part (b: divv = + ( + ( +=+ Using spherical coordinates with d = ρ sin φdρdφdθ RH = divv d = / ( + ρ sin φ cos θ ρ sin φdρdφdθ = dθ / sin φdφ / ρ dρ + cos θdθ sin φdφ ρ dρ }{{} sin θ] = =θ] cos φ]/ ] ρ =((+(8 = 6

4 . A closed surface is made up of the part of the paraboloid = with, together with, the circular disc +, =. erif the divergence theorem for this closed surface for F =,,. LH = F nd = F nd + F nd g(,, = + g + ; g = ±,, ; choose + : n =,, ; F n = Projectingontothe-plane, then switching to polar coordinates we have F nd = ] =θ] r + r = =; n = k; F n = ; / n {}}{ + +d d = (r +rdrdθ F nd = ( d = (area of = LH = = divf = (+ (+ ( = Using clindrical coordinates with d = rddrdθ, we obtain RH = divf d = r rddrdθ= r ]= r = dr dθ = ] r( r dr dθ =θ] r r =( ( =

5 . erif the divergence theorem for div F d d d where F =,,. 5 LH = F nd = F nd + F nd + F nd + F nd + F nd 5 g(,, = g =; g = ±,,,,. Choose : n = =,, ; F n = ; project onto the -plane: + / n { }}{ ( ( F nd = ] +d d = + d d =] = =; n = k; F n =; F nd = =; n = i;. F n =9; F nd = 9 d d = 9]= = d = 9 d = ] = =; n = i; F n =; F nd = 5 =; n = j; F n =; F nd =(Area of the rectangle 5 = (( = 8 5 LH = ++++8= ( div F = + (+ ( = + RH = div F d d d = ( + d d d = ( + ( ( = ( + d d = + ] ] = =(5( 8 = = ]= = d d

6 5. erif the divergence theorem for F =,, and for the closed surface = consisting of : half of a circular disc +,, =, : the part of the plane = with + and, and : the part of the clinder + =with and. F nd + F nd + LH = F nd = F nd =;n = k; F n =; F nd = ; + =; n =,, ; F n = ; Projectingontothe-plane, then switching to polar coordinates ields / n {}}{ F nd = ( d d = / r cos θrdrdθ / = sinθ] / / ] r = = ( ( g(,, = + ; Integrating directl in clindrical coordinates with d = rddθ we have / F nd = / LH = + = g g = ±,, = ±,,. Choose + :n =,, ; F n = + + =. cosθ d dθ = / / cos θdθ= sin θ] / / = ( = div F = (+ (+ ( = Using clindrical coordinates with d = rddrdθ, we obtain RH = divf d = / = ] r / r cos θ rddrdθ= / / sin θ]/ / = ( ( = r cos θdrdθ

7 6. Calculate the surface integral solid region +,, (a directl, (b b the divergence theorem. F nd where F =,, and is the surface that encloses the Part (a: F nd = F nd + F nd + F nd + =;n = k; F n = ; F n d = (Area of = =;n = k; F n =; F n d =(Area of = F nd satisfies + =, i.e., g(,, = + =; g g = ±,, ; + + choose + :n =,, ; F n = ; Denote D to be the projection of onto the -plane. witching to polar coordinates ields F nd = D ( cos θ sin θdθ ( = ( ( ( r6 dr =;n = j; F n =; / n {}}{ + +d d = ( ( dθ rdr = cos θ ] = 58 F nd = F nd = + + ( 58 += 58 ( r cos θrsin θr rdrdθ ] ] r 7 θ] r 7 Part (b: div F = (+ ( + ( = Using clindrical coordinates with d = rdrddθ, we obtain divf d = r sin θrdrddθ= sin θ r =( ( ( ( = ] r= r= d dθ = cos θ] 7/]

8 7. Let F =,, and let be the solid region below the cone + =, above the hemisphere =, and inside the clinder + =. Calculate div F d (a directl, (b b the divergence theorem. Part (a: div F = (+ (+ (= Using clindrical coordinates with d = rdrddθ, we obtain divf d = r r r d dr dθ = r(r ( r dr dθ =θ] r ] =r = r dr dθ = ] r r =((8 8 + = (Note that divf d can also be evaluated using spherical coordinates with d = ρ sin φdρdφdθ as / / / sin φ ρ cos φρ sin φdρdφdθ Part (b: F nd = F nd + F nd + F nd g(,, = + + ± g =; g = ±,, = ± + +,, ; choose sign:n =,, ; F n = = r Using spherical coordinates with ρ =, we have d = sin φdφdθ : F nd = / ( sin φ ( cos φ ( sin φdφdθ = 6 θ] / g(,, = + =; ± g g = sin φ ] / / = 6(( = 6 6 ±,,,, ; choose sign:n = F n = = r + + ρ = (ρ sin (ρ cos ρ = ρ. Using spherical coordinates with φ =, we have d = ρ sin dρ dθ : ( ( ] F nd = ρ ρ dρ dθ = ρ θ] = ((6 = 6 g(,, = + =; ± g g = ±,, + + = ±,, ; choose + sign:n =,, ; F n = + = Using clindrical coordinates with r =, we have d =d dθ : F nd = (( d dθ ] =θ] =((8 = 6 F nd = =

9 8. Let be the pramid with the rectangular base =,, and the verte above the base (ape at (,,. erif the divergence theorem for div F d where F =,,. 5 LH = F nd = F nd + F nd + F nd + F nd + F nd =; n = k; F n = ; =; n = j; F n =; F nd = (Area of the rectangle = F nd = part of the plane containing A(,,, B(,,, D(,, ; normal vector AB AD =,,. The plane equation is ( + =, which simplifies to + =. The outward unit normal vector is n =,, ; F n = + =. F nd = (Area of triangle = ( (( = (same result could also be calculated b projecting onto the -plane / n F nd = {}}{ ( ( d d =(Area of triangle =; n = i; F n = ; F nd = (Area of triangle = 5 part of the plane containing B(,,, C(,,, D(,, ; normal vector CB CD =,,. The plane equation is ( =, which simplifies to + =. The outward unit normal vector is n =, 5, 5 ; F n = + 5. Projectingontothe-plane we obtain F nd = 5 = ( + 5 ( + + / n {}}{ ( 5 d = F nd = = 5 d d = ] = 9 5 ( ( + 5 d d = ( + ] = = d div F = ( + ( + (= RH = divf d = = ( d d = = 6 5 5] = 5 d d d = ] = = d d ] = = d = ( + d

10 9. Calculate the surface integral F n d where F =,, and is the surface that encloses the solid region + + 9,,,. (a directl, (b b the divergence theorem. Part (a: F n d = F n d + F n d + F n d + F n d =;n = i; F n =; F n d =(Area of = 9 =;n = j; F n =; F n d = =;n = k; F n =; F n d = g(,, = + + =9; ± g g = ±,, = ± + +,, ; choose + : n =,, F n = + ; Using spherical coordinates with ρ =, we have d = sin φdφdθ F n d = / ( / =9 / cosθ sin φ+( cos φ ( sinφdφdθ cos θdθ ( ( ( / / sin φ dφ+8 dθ }{{} / cos(φ ] = 9 sin θ] / φ sin(φ / F n d = 9 8 θ] / = 8 8 cos φ sin φdφ } {{ } substitute u=cos φ ] cos φ =9( ( / 8( ( 99 = 8 Part (b: div F = ( + ( + ( = =ρcos φ divf d = / / ρ cos φρ sin φdρdφdθ ( ( ( / = dθ cos φ sin φdφ ρ dρ / }{{} =θ] / sin φ substitute u=sin φ ] ] ρ / =( ( 8 8 ( = 8

11 . Let F =,, and let D be the circular disc +, =. erif the divergence theorem for div F d d d. D D= LH = F n d = F n d + F n d + F n d =;n = k; F n = =; g(,, = + =; ± g g = F n d = ±,, + + F n = + = r =; using clindrical coordinates with r =, we have d =d dθ : F n d = =θ sinθ +cosθ] = F n d = = ±,, ; choose + : n =,, ; cosθ sinθ (( d dθ = ( cosθ sinθ dθ part of the plane =, i.e., + + =; anormalvectoris±,, ; the outward unit normal is n = we project onto the -plane / n {}}{ F n d =+ +6 = ( d d =(Area of = 6 8,, ; F n = ++ = div F = (+ (+ ( = RH = divf d = r cos θ r sin θ rddrdθ = r ( r cos θ r sin θ dr dθ = 6r r cos θ r sin θ ] r= r= = ( 8cosθ 8sinθ dθ =θ 8sinθ +8cosθ] = 8