Mathematik für Physik III (WS 99/00)

1)

WegIntegral[
          f_, {t0_, 
            t1_}] := (∫_t0^t1) Apply[
              f, γ[
                t]] (Sqrt[γ'[t] . γ'[
                    t]]) \[DifferentialD]t;

a)

f[x_, y_, z_] :=  y;
γ[t_] :=  {0, 0, t};
WegIntegral[f, {0, 1}]

b)

f[x_, y_, z_] :=  x Cos[z];
γ[t_] :=  {t, (t^2), 0};
WegIntegral[f, {0, 1}]

c)

f[x_, y_, z_] :=  1  +  (y/3);
γ[t_] :=  {30 Cos[t]^3, 30 Sin[t]^3, 
          0};
WegIntegral[f, {0, (π/2)}]

2)

sol = 
      Solve[{(x^2) + (y^2) + (z^2) == 1, x + y + z == 0}, {x, 
          y}]

{t0, t1} =  z /. Solve[(Sqrt[2 - 3 (z^2)]) == 0, z]

γ1[
          t_] = ({x, y, z}  /.  sol[[1]]  /.  
            z -> t);
γ2[
          t_] = ({x, y, z}  /.  sol[[2]]  /.  
            z -> t);

gr1 = 
        ParametricPlot3D[Evaluate[γ1[t]], {t, t0, t1}];

gr2 = 
        ParametricPlot3D[Evaluate[γ2[t]], {t, t0, t1}];

Show[gr1, gr2]

ρ[x_, y_, z_] := (x^2)

(∫_t0^t1) 
              Apply[ρ, γ1[
                t]] (Sqrt[γ1'[t] . γ1'[
                    t]]) \[DifferentialD]t + (∫_t0^t1) 
              Apply[ρ, γ2[
                t]] (Sqrt[γ2'[t] . γ2'[
                    t]]) \[DifferentialD]t // Simplify

3)

KurvenIntegral[
          f_, {t0_, 
            t1_}] := (∫_t0^t1) Apply[
                f, γ[t]] . γ'[
                t] \[DifferentialD]t;

a)

f[x_, 
          y_]  :=  {(E^x), (E^y)};
γ[t_] :=  {Sqrt[t], t};

KurvenIntegral[f, {0, 1}]

b)

f[x_, y_, z_] :=  {x^2 + 5 y + 3  y z, 5 x + 3 x z - 2, 
          3 x y - 4 z};
γ[t_] =  {Sin[t], Cos[t], t};

KurvenIntegral[f, {0, 2  π}]

5)

KurvenIntegral[
          f_, {t0_, 
            t1_}] := (∫_t0^t1) Apply[
                f, γ[t]] . γ'[
                t] \[DifferentialD]t;

f[x_, y_] :=  {x, y};
γ[t_] :=  {Cos[t]^3, Sin[t]^3};

KurvenIntegral[f, {0, 2  π}]

6)

Needs["Calculus`VectorAnalysis`"];
SetCoordinates[Cartesian[x, y, z]];

f[x_, y_, z_] := 
        (1/(Sqrt[(x^2) + (y^2) + (z^2)]));
g[x_, y_, z_] = Grad[f[x, y, z]];
Clear[γ1, γ2, γ3];
γ[
          t_] := {γ1[t], γ2[t], γ3[t]};

Simplify[Apply[g, γ[t]] . γ'[t]]

Integrate[%, t]

Fazit: Mathematica kann das (so) nicht...

7)

KurvenIntegral[
          f_, {t0_, 
            t1_}] := (∫_t0^t1) Apply[
                f, γ[t]] . γ'[
                t] \[DifferentialD]t;

PolygonIntegral[f_, P_, Q_] := 
        Block[{}, 
γ[t_] = 
            P  +  t (Q - P); 
KurvenIntegral[
            f, {0, 1}]
];
PolygonIntegral[f_, P_, Q_, R__] := 
        PolygonIntegral[f, P, Q] + PolygonIntegral[f, Q, R];

a)

f[x_, y_] :=  {x y, y (E^x)};

PolygonIntegral[f, {0, 0}, {2, 0}, {2, 1}, {0, 1}, {0, 0}]

b)

f[x_, y_] :=  {y, y  - x};

PolygonIntegral[f, {0, 0}, {1, 0}, {1, 1}]

PolygonIntegral[f, {0, 0}, {0, 1}, {1, 1}]

PolygonIntegral[f, {0, 0}, {1, 1}]

8)

Needs["Calculus`VectorAnalysis`"];
SetCoordinates[Cartesian[x, y, z]];

Rot[
          f_] := (∂_y) 
            f[[1]]  - (∂_x) f[[2]];

a)

f[x_, y_] := {12 x y + 3, 6 x^2};
Rot[f[x, y]] == 0

F = ∫f[x, y][[1]] \[DifferentialD]x + 
        Const[y]

F = 
      F  /.  Flatten[
          DSolve[(∂_y) F == f[x, y][[2]], Const[y], 
            y]]

{(∂_x) F, (∂_y) F} == f[x, y]

b)

f[x_, y_] := {x y, y};
Rot[f[x, y]] == 0

c)

f[x_, y_] := {3  (x^2) y, (x^3)};
Rot[f[x, y]] == 0

F = ∫f[x, y][[1]] \[DifferentialD]x + 
        Const[y]

F = 
      F  /.  Flatten[
          DSolve[(∂_y) F == f[x, y][[2]], Const[y], 
            y]]

{(∂_x) F, (∂_y) F} == f[x, y]

d)

f[x_, y_, z_] := {x , y, z};
Curl[f[x, y, z]] == {0, 0, 0}

F = ∫f[x, y, z][[1]] \[DifferentialD]x + 
        Constz[y]

F = (F  /.  
            Flatten[DSolve[(∂_y) 
                    F == f[x, y, z][[2]], Constz[y], 
                y]] ) /.  C[1] -> Const[z]

F = (F  /.  
          Flatten[DSolve[(∂_z) F == f[x, y, z][[3]], 
              Const[z], z]] )

Grad[F] == f[x, y, z]

e)

f[x_, y_, z_] := {(x^2) y , z (E^x), 
          x y Log[z]};
Curl[f[x, y, z]] == {0, 0, 0}

f)

f[x_, y_, z_] := {(y^2) Cos[x] , 
          2  y Sin[x]  + (E^(2  z)), 
          2  y (E^(2  z))};
Curl[f[x, y, z]] == {0, 0, 0}

F = ∫f[x, y, z][[1]] \[DifferentialD]x + 
        Constz[y]

F = (F  /.  
            Flatten[DSolve[(∂_y) 
                    F == f[x, y, z][[2]], Constz[y], 
                y]] ) /.  C[1] -> Const[z]

F = (F  /.  
          Flatten[DSolve[(∂_z) F == f[x, y, z][[3]], 
              Const[z], z]] )

Grad[F] == f[x, y, z]

10)

a)

(∫_0^2)(∫_3^4)(2  x + 
              3  y) \[DifferentialD]y \[DifferentialD]x

b)

(∫_0^(π/2))(∫_0^(π/2))Sin[
            x + y] \[DifferentialD]y \[DifferentialD]x

11)

a)

(∫_0^1)(∫_0^(1 - x))((x^2) + 
              (y^2)) \[DifferentialD]y \[DifferentialD]x

b)

(∫_(-2)^2)(∫_0^((Sqrt[4 - (x^2)])))(x^2) 
          y \[DifferentialD]y \[DifferentialD]x

12)

a)

(∫_0^1)(∫_1^((E^x)))(x + 
                y) \[DifferentialD]y \[DifferentialD]x  // 
      Simplify

b)

(∫_0^1)(∫_((x^3))^((x^2)))y \[DifferentialD]y \[DifferentialD]x

c)

(∫_(-3)^2)(∫_0^((y^2)))((x^2) + 
              y) \[DifferentialD]x \[DifferentialD]y

d)

(∫_0^1)(∫_((y^2))^y)((x^n) + 
              (y^m)) \[DifferentialD]x \[DifferentialD]y

FullSimplify[%]

13)

a)

f[x_] := (Sqrt[1 - (x^2)]);  g[x_] := 1;  x0 = 0;  
    x1 = ((Sqrt[3])/2);

Plot[{f[x], g[x]}, {x, x0, x1}, PlotRange -> {0, 1.5}, 
        Frame -> True, 
        PlotStyle -> {RGBColor[0, 0, 1], RGBColor[0, 1, 0]}];

π (∫_x0^x1)(g[x]^2 - 
              f[x]^2) \[DifferentialD]x

b)

f[x_] := (Sqrt[x]);  g[x_] := 1;  x0 = 0;  x1 = 1;

Plot[{f[x], g[x]}, {x, x0, x1}, PlotRange -> {0, 1.5}, 
        Frame -> True, 
        PlotStyle -> {RGBColor[0, 0, 1], RGBColor[0, 1, 0]}];

π (∫_x0^x1)(g[x]^2 - 
              f[x]^2) \[DifferentialD]x

14)

(∫_0^(2  a))(((2 
                 (x^2)))^2) \[DifferentialD]x

15)

(∫_0^1)((∫_0^x)((-1)/(x^2)) \[DifferentialD]y + (∫_x^1) 
                    (1/(y^2)) \[DifferentialD]y) \[DifferentialD]x

(∫_0^1)((∫_0^y) 
                    (1/(y^2)) \[DifferentialD]x + (∫_y^1)((-1)/(x^2)) \[DifferentialD]x) \[DifferentialD]y

16)

y[x_] =  
      y  /.  Solve[((((x/a)))^2) + ((((y/b)))^2) == 1, 
            y][[2]]

(∫_(-a)^a)(∫_(-y[x])^(y[x]))1 \[DifferentialD]y \[DifferentialD]x

Simplify[%, a > 0]

17)

y[x_] =  
      y  /.  Solve[(x^2) + (y^2) == z, y][[2]]

(∫_0^10)((∫_(-(Sqrt[z]))^((Sqrt[z])))(∫_(-y[x])^(y[x]))1 \[DifferentialD]y \[DifferentialD]x ) \[DifferentialD]z

Bemerkung: Das innere Doppelintegral ist wegen der Rotationssymmetrie um die z-Aachse natuerlich  πz .

18)

(∫_(-π)^π)(∫_(-π)^π) (E^Sin)[
                x + y] \[DifferentialD]x \[DifferentialD]y

Simplify[
      (1/E)  ≤  (E^Sin)[x + y]  ≤  E, Element[{x, y}, Reals]]

Simplify[-1  ≤ Sin[x + y]  ≤  1, 
      Element[{x, y}, Reals]]

{(∫_(-π)^π)(∫_(-π)^π) 
                (1/E) \[DifferentialD]x \[DifferentialD]y, (∫_(-π)^π)(∫_(-π)^π) E \[DifferentialD]x \[DifferentialD]y}

19)

(∫_0^2)(∫_0^2)(∫_0^(x (y^2) + (y^3)))1 \[DifferentialD]z \[DifferentialD]y \[DifferentialD]x

20)

z[x_, y_] =  
      z  /.  Solve[((((x/a)))^2) + ((((y/b)))^2) + ((((z/c)))^2) == 
              1, z][[2]]

y[x_] =  
      y  /.  Solve[((((x/a)))^2) + ((((y/b)))^2) == 1, 
            y][[2]]

(∫_(-a)^a)(∫_(-y[x])^(y[x]))(∫_(-z[x, y])^(z[x, y]))1 \[DifferentialD]z \[DifferentialD]y \[DifferentialD]x

21)

(∫_0^1)(∫_0^(1 - x))(∫_0^(x y))1 \[DifferentialD]z \[DifferentialD]y \[DifferentialD]x

22)

Limit[(3/(4  π (ϵ^3))) (∫_0^ϵ)(∫_(-π/2)^(π/2))(∫_0^(2  π))f[
                  r, θ, φ] (r^2) 
                Cos[θ] \[DifferentialD]φ \[DifferentialD]θ \[DifferentialD]r, ϵ -> 0]

23)

Simplify[2  ≤ f + (1/f) , f > 0]

(1/2) (∫_a^b)(∫_a^b) 
              2 \[DifferentialD]x \[DifferentialD]y  // 
      Simplify

24)

(∫_0^1)(∫_0^x) Sin[
                x]/x \[DifferentialD]y \[DifferentialD]x

25)

y[x_] =  
      y  /.  Solve[((((x/a)))^2) + ((((y/b)))^2) == 1, 
            y][[2]]

(∫_(-a)^a)(∫_(-y[x])^(y[x]))(∫_0^(c x  +  d y  + h))1 \[DifferentialD]z \[DifferentialD]y \[DifferentialD]x

Simplify[%, a > 0]

26)

8 (∫_0^((a/2)))(∫_0^x)(∫_0^(h (1 - ((2  x)/a))))1 \[DifferentialD]z \[DifferentialD]y \[DifferentialD]x // Simplify

27)

Polarkoordinaten:

(∫_(-r)^r)(∫_(R - (Sqrt[(r^2) - (z^2)]))^(R + (Sqrt[(r^2) - (z^2)])))(∫_0^(2  π))ρ \[DifferentialD]φ \[DifferentialD]ρ \[DifferentialD]z

Simplify[%, r > 0]

28)

a)

(∫_0^1)(∫_0^1)(∫_0^1) 
              z (E^(x + 
                    y)) \[DifferentialD]x \[DifferentialD]y \[DifferentialD]z // Simplify

b)

(∫_0^π)(∫_0^1)(∫_0^(1 - y))x Cos[z] \[DifferentialD]x \[DifferentialD]y \[DifferentialD]z

29)

(∫_0^1)(∫_0^(2  x))(∫_((x^2) + (y^2))^(x + y))1 \[DifferentialD]z \[DifferentialD]y \[DifferentialD]x

32)

a)

(∫_0^a)(∫_0^((Sqrt[a x  - (x^2)])))((x^2) + 
              (y^2)) \[DifferentialD]y \[DifferentialD]x

Das ist falsch!! So stimmt das Ergebnis (warum?):

(∫_0^(a/2))(∫_0^((Sqrt[a x  - (x^2)])))((x^2) + (y^2)) \[DifferentialD]y \[DifferentialD]x + (∫_(a/2)^a)(∫_0^((Sqrt[a x  - (x^2)])))((x^2) + 
                  (y^2)) \[DifferentialD]y \[DifferentialD]x // 
      Simplify

Simplify[%, a > 0]

b)

(∫_0^1)(∫_((x^2))^x) 
              (1/(Sqrt[(x^2) + (y^2)])) \[DifferentialD]y \[DifferentialD]x

33)

a)

(∫_0^h)(∫_0^(2  π))(∫_0^a) ρ (r^2) 
            r \[DifferentialD]r \[DifferentialD]φ \[DifferentialD]z

b)

(∫_0^h)(∫_0^(2  π))(∫_0^a) ρ ((r^2) Cos[φ]^2 + (z^2)) 
              r \[DifferentialD]r \[DifferentialD]φ \[DifferentialD]z  // Simplify

34)

(3/(4  π)) (∫_0^1)(∫_0^π)(∫_0^(2  π))(E^(-r Cos[θ])) (r^2) Sin[θ] \[DifferentialD]φ \[DifferentialD]θ \[DifferentialD]r

35)

ParametricPlot[{(1 + Sin[t]) Cos[t], (1 + Sin[t]) 
            Sin[t]}, {t, 0, 2  π}];

(∫_0^(2  π))((∫_0^(1 + Sin[φ]))r \[DifferentialD]r) \[DifferentialD]φ

36)

(∫_r1^r2)(∫_0^π)(∫_0^(2  π))(r^(-m/2)) (r^2) 
              Sin[θ] \[DifferentialD]φ \[DifferentialD]θ \[DifferentialD]r  // Simplify

37)

Ohne Beschraenkung der Allgemeinheit a=(0,0,R)

(∫_r1^r2)(∫_0^π)(∫_0^(2  π))(1/(Sqrt[(r^2) Sin[θ]^(2 ) + (((r Cos[θ] - R))^2)])) (r^2) 
              Sin[θ] \[DifferentialD]φ \[DifferentialD]θ \[DifferentialD]r  // Simplify

39)

a)

(1/8) (∫_(-1)^1)(∫_(-1)^1)(∫_(-1)^1)((x^2) + (y^2) + 
                  (z^2)) \[DifferentialD]x \[DifferentialD]y \[DifferentialD]z

40)

(∫_1^2)(∫_1^2)(∫_1^2)(1 + x) 
            y (E^z) \[DifferentialD]x \[DifferentialD]y \[DifferentialD]z

41)

M = (∫_0^1)(∫_0^1) (E^(x + 
                    y)) \[DifferentialD]x \[DifferentialD]y // 
        Simplify

(1/M) (∫_0^1)(∫_0^1) {x, 
                y} (E^(x + 
                    y)) \[DifferentialD]x \[DifferentialD]y // 
      Simplify

47)

Needs["Calculus`VectorAnalysis`"];
SetCoordinates[Spherical[r, θ, φ]];

Grad[f[r, θ, φ]]

Laplacian[f[r, θ, φ]]

Div[{f1[r, θ, φ], f2[r, θ, φ], 
        f3[r, θ, φ]}]

48)

Needs["Calculus`VectorAnalysis`"];
SetCoordinates[Cartesian[x, y, z]];

r = (Sqrt[(x^2) + (y^2) + (z^2)]);

Grad[(r^n)]

Laplacian[(1/r)] // Simplify

Laplacian[(r^n)] // Simplify

Div[(r^n) {x, y, z}] // Simplify

%  /.  n -> -3

50)

Needs["Calculus`VectorAnalysis`"];
SetCoordinates[Cartesian[x, y, z]];

G[x_, y_, z_] := {G1[x, y, z], G2[x, y, z], 
          G3[x, y, z]};

Div[f[x, y, z] G[x, y, z]]  ==  
        f[x, y, z] Div[G[x, y, z]] +  G[x, y, z] . Grad[f[x, y, z]] // 
      Simplify

a[x_, y_, z_] := {a1[x, y, z], a2[x, y, z], 
          a3[x, y, z]};
b[x_, y_, z_] := {b1[x, y, z], b2[x, y, z], 
          b3[x, y, z]};

Achtung: Vorzeichenfehler in der Angabe!

Div[CrossProduct[a[x, y, z], b[x, y, z]]] == 
        b[x, y, z] . Curl[a[x, y, z]] - a[x, y, z] . Curl[b[x, y, z]] // 
      Simplify

51)

Φ[u_, v_] := {u Cos[v],  u Sin[v],  
          (u^2) + (v^2)};

T[u_, v_] :=  Φ[u0, v0] +  
          Transpose[{(∂_u) Φ[u, 
                    v], (∂_v) Φ[u, v]}] . {{u - 
                  u0}, {v - v0}};

T[u, v] /.  {u0 -> 1, v0 -> 0} // MatrixForm

53)

Needs["Calculus`VectorAnalysis`"];
Φ[u_, v_]  :=  {x[u, v], y[u, v], 
          z[u, v]};
kreuz[u_, v_]  :=  
        CrossProduct[(∂_u) Φ[u, 
              v], (∂_v) Φ[u, 
              v]];

fdet[{f_, g_}]  := 
      Det[{{(∂_u) f, (∂_v) f}, {(∂_u) 
              g, (∂_v) g}}]

DotProduct[kreuz[u, v], kreuz[u, v]] ==  
        fdet[{x[u, v], y[u, v]}]^2  + 
fdet[{y[u, v], 
                z[u, v]}]^2  +  fdet[{x[u, v], z[u, v]}]^2  // 
      Simplify

DotProduct[kreuz[u, v], kreuz[u, v]] ==  
        DotProduct[(∂_u) Φ[u, 
                  v], (∂_u) Φ[u, 
                  v]]*
DotProduct[(∂_v) Φ[u, v], (∂_v) Φ[u, v]]  -  
          DotProduct[(∂_u) Φ[u, 
                  v], (∂_v) Φ[u, 
                  v]]^2 //Simplify

54)

Needs["Calculus`VectorAnalysis`"];

Φ[r_, φ_]  :=  {r Cos[φ], 
        r Sin[φ], φ}

k[
        r_, φ_]:=CrossProduct[(∂_r) Φ[
            r, φ], (∂_φ) Φ[
            r, φ]]

f[r_, φ_]  =  
      Sqrt[DotProduct[k[r, φ], k[r, φ]]]  // 
        Simplify

(∫_0^(2 π))(∫_0^1) f[
            r, φ] \[DifferentialD]r \[DifferentialD]φ

55)

Needs["Calculus`VectorAnalysis`"];
Φ[φ_, ψ_] = {(R + 
              r Cos[φ]) Cos[ψ], (R + 
              r Cos[φ]) Sin[ψ], 
        r Sin[φ]}

k[φ_, ψ_] = CrossProduct[(∂_φ) Φ[φ, ψ], (∂_ψ) Φ[φ, ψ]] //Simplify

f[φ_, ψ_]  =  
      Sqrt[DotProduct[k[φ, ψ], k[φ, ψ]]]  // 
        Simplify

(∫_0^(2 π))(∫_0^(2 π))f[φ, ψ] \[DifferentialD]φ \[DifferentialD]ψ

Simplify[%, {r > 0, R > 0}]

58)

g[x_, y_]  =  x^2  +  y

(∫_0^1)(∫_0^1) 
          x  (Sqrt[1 + ((∂_x) 
                      g[x, y])^2  +  ((∂_y) 
                      g[x, y])^2]) \[DifferentialD]x \[DifferentialD]y

59)

Needs["Calculus`VectorAnalysis`"];
Φ[φ_, θ_] = {Cos[φ] Sin[θ], Sin[φ] Sin[θ], Cos[θ]}

k[φ_, θ_] = 
      CrossProduct[(∂_φ) Φ[φ, θ], (∂_θ) Φ[φ, θ]]  // 
        Simplify

f[φ_, θ_]  =  
      Sqrt[DotProduct[k[φ, θ], k[φ, θ]]]  // 
        Simplify

(∫_0^(2 π))(∫_0^π) Φ[φ, θ][[3]]^2 f[φ, θ] \[DifferentialD]φ \[DifferentialD]θ

60)

Needs["Calculus`VectorAnalysis`"];

Φ[r_, φ_]  :=  {r Cos[φ], 
        r Sin[φ], φ}

k[
        r_, φ_]:=CrossProduct[(∂_r) Φ[
            r, φ], (∂_φ) Φ[
            r, φ]]

f[r_, φ_]  =  
      Sqrt[DotProduct[k[r, φ], k[r, φ]]]  // 
        Simplify

ρ[x_, y_, z_]  =  
      2 (Sqrt[(x^(2  )) +  (y^(2   ))])

(∫_0^(2 π))(∫_0^1) 
        Apply[ρ, Φ[r, φ]]  f[
            r, φ] \[DifferentialD]r \[DifferentialD]φ

61)

Needs["Calculus`VectorAnalysis`"];
Φ[φ_, θ_] = {R Cos[φ] Sin[θ], R Sin[φ] Sin[θ], R Cos[θ]}

a)

k[φ_, θ_] = 
      CrossProduct[(∂_φ) Φ[φ, θ], (∂_θ) Φ[φ, θ]]  // 
        Simplify

f[φ_, θ_]  =  
      Sqrt[DotProduct[k[φ, θ], k[φ, θ]]]  // 
        Simplify

Simplify[%, {R > 0, 0 < θ < π}]

b)

r[x_, y_, z_]  =  {x, y, z}

(∫_0^(2 π))(∫_0^π) DotProduct[
              Apply[r, Φ[φ, θ]], 
              k[φ, θ]] \[DifferentialD]θ \[DifferentialD]φ  // Simplify

62)

a  =  {a1, a2, a3};
b = {b1, b2, b3};

A =  Transpose[{a, b}]

c={{c11,  c12}, {c21,  c22}}

prod  =  A .c

α  =  Transpose[prod][[1]]

β  =  Transpose[prod][[2]]

Needs["Calculus`VectorAnalysis`"];

CrossProduct[α, β]  ==  
        CrossProduct[a, b] Det[c]  // Simplify

63)

Nach 61 b):

\(\(\(c\ q\)\/R\^3\) \(∫x dσ\)\  = \ \(\(c\ q\)\/R\^3\ 4\ π\ R\^3\  = \ 4\ π\ c\ q\)\)

64)

a)

v[x_, y_, z_]  =  {2 z,  x + y, 0}

g[x_, y_]  =  (4/3) - (x/3) - ((2 y)/3)

w[x_, y_]  =  v[x, y, z]  /.  z ->  g[x, y]

(∫_0^4)(∫_0^(2 - (x/2)))(w[x, 
                    y][[3]]  -  (∂_x) 
                    g[x, y] w[x, y][[1]]  -  (∂_y) 
                    g[x, y] w[x, 
                      y][[2]]) \[DifferentialD]y \[DifferentialD]x  // Simplify

b)

vgl. 61:

Needs["Calculus`VectorAnalysis`"];
Φ[φ_, θ_] = {r Cos[φ] Sin[θ], r Sin[φ] Sin[θ], r Cos[θ]}

k[φ_, θ_] = 
      CrossProduct[(∂_φ) Φ[φ, θ], (∂_θ) Φ[φ, θ]]  // 
        Simplify

(∫_0^(2 π))(∫_0^π) Apply[
                v, Φ[φ, θ]]  .  
              k[φ, θ] \[DifferentialD]θ \[DifferentialD]φ  // Simplify

65)

Needs["Calculus`VectorAnalysis`"];

F[x_, y_, z_]  =  {z, x, -3 (y^2) z}

Parametrisierung von S durch Zylinderkoordinaten:

Φ[φ, z]  =  {4 Cos[φ],  
        4 Sin[φ],  z}

k[φ_, z_] = 
      CrossProduct[(∂_φ) Φ[φ, 
              z], (∂_z) Φ[φ, z]]  // 
        Simplify

(∫_0^5)(∫_0^((π/2)))Apply[
                F, Φ[φ, z]]  .  
              k[φ, 
                z] \[DifferentialD]φ \[DifferentialD]z  // 
      Simplify

67)

WegIntegral[P_, 
          Q_, γ_, {t0_, 
            t1_}] := (∫_t0^t1)(Apply[
                    P, γ[t]] γ'[t][[1]]  +  
                Apply[Q, γ[t]] γ'[
                      t][[2]]) \[DifferentialD]t;

P[x_, y_]  =  x y   +  (y^2);

Q[x_, y_]  =  (x^2);

γ1[t_]  =  {t, (t^2)};

γ2[t_]  =  {1 - t, 1 - t};

WegIntegral[P, Q, γ1, {0, 1}]  +  
      WegIntegral[P, Q, γ2, {0, 1}]

(∫_0^1)(∫_((x^2))^x)((∂_x) 
                Q[x, y]  -  (∂_y) 
                P[x, y]) \[DifferentialD]y \[DifferentialD]x

68)

WegIntegral[P_, 
          Q_, γ_, {t0_, 
            t1_}] := (∫_t0^t1)(Apply[
                    P, γ[t]] γ'[t][[1]]  +  
                Apply[Q, γ[t]] γ'[
                      t][[2]]) \[DifferentialD]t;

P[x_, y_]  =  2 (x^3)  -  (y^3);

Q[x_, y_]  =  (x^3)  +  (y^3);

γ1[t_]  =  {a Cos[t],  a Sin[t]};

γ2[t_]  =  {b Cos[t],  b Sin[t]};

WegIntegral[P, Q, γ2, {0, 2 π}]  -   WegIntegral[P, 
          Q, γ1, {0, 2 π}] // Simplify

Φ[r_, φ_]  =  {r Cos[φ],  
        r Sin[φ]}

f[x_, 
        y_]  =  (∂_x) Q[x, y]  -  (∂_y) 
          P[x, y]

(∫_0^(2 π))(∫_a^b) Apply[
            f, Φ[
              r, φ]] r \[DifferentialD]r \[DifferentialD]φ

70)

WegIntegral[P_, 
          Q_, γ_, {t0_, 
            t1_}] := (∫_t0^t1)(Apply[
                    P, γ[φ]] γ'[φ][[1]]  +  
                Apply[Q, γ[φ]] γ'[φ][[2]]) \[DifferentialD]φ;

P[x_, y_]  =  -y

Q[x_, y_]  =  x

γ[φ_]  =  {r[φ] Cos[φ], 
        r[φ] Sin[φ]}

WegIntegral[P, Q, γ, {a, b}]  // Simplify

71)

r[φ_]  =  3 Sin[2 φ]

(1/2) (∫_0^π) 
                r[φ]^2 \[DifferentialD]φ

73)

Needs["Calculus`VectorAnalysis`"];
SetCoordinates[Cartesian[x, y, z]];
F[x_, y_, z_]  =  {y z,  x z,  x y}

Curl[F[x, y, z]]

Nach Satz von Stokes ist das Integral daher 0.

76)

Needs["Calculus`VectorAnalysis`"];
SetCoordinates[Cartesian[x, y, z]];

v[x_, y_, z_]  =  {-(y^3), (x^3), -(z^3)};

γ[t_]  =  {Cos[t], Sin[t], 
          1 - Sin[t] - Cos[t]};

(∫_0^(2 π))Apply[v, γ[t]]  .  γ'[
            t] \[DifferentialD]t

F[x_, y_, z_]  =  Curl[v[x, y, z]]

g[x_, y_]  =  1 - x - y

w[x_, y_]  =  F[x, y, z]  /.  z ->  g[x, y]

w1[x_, 
        y_]  =  w[x, y][[3]]  -  (∂_x) 
            g[x, y] w[x, y][[1]]  -  (∂_y) 
            g[x, y] w[x, y][[2]]

Φ[r_, φ_]  =  {r Cos[φ],  
        r Sin[φ]}

w2[r_, φ_]  =  
      Apply[w1, Φ[r, φ]]

(∫_0^(2 π))(∫_0^1) w2[
            r, φ] r \[DifferentialD]r \[DifferentialD]φ

80)

Needs["Calculus`VectorAnalysis`"];

SetCoordinates[Cartesian[x, y, z]];

F[x_, y_, z_]  =  {4 x,  -2 (y^2), (z^2)}

divF[x_, y_, z_]  =  Div[F[x, y, z]]

Zylinderkoordinaten:

Φ[r_, φ_, z_]  =  {r Cos[φ],  
        r Sin[φ],  z}

divFzyl[r_, φ_, z_]  =  
      Apply[divF, Φ[r, φ, z]]

(∫_0^3)(∫_0^2)(∫_0^(2 π))divFzyl[r, φ, z] 
            r \[DifferentialD]φ \[DifferentialD]r \[DifferentialD]z

Oberfläche von Ω besteht aus Boden S1, Decke S2 und Mantel S3

S1:

ν  =  {0, 0, -1}

F[x, y, z] . ν  /.  z ->  0

S2:

ν  =  {0, 0, 1}

F[x, y, z] . ν /.  z -> 3

(∫_0^2)(∫_0^(2 π))9 r \[DifferentialD]φ \[DifferentialD]r

S3:

Ψ[φ_, z_]  =  {2 Cos[φ],  
        2 Sin[φ], z}

(Sqrt[CrossProduct[(∂_φ) Ψ[φ, z], (∂_z) Ψ[φ, z]] . CrossProduct[(∂_φ) Ψ[φ, z], (∂_z) Ψ[φ, z]]]) // Simplify

n[φ_, z_] = {Cos[φ], 
        Sin[φ], 0}

a[φ_, z_]  =  
      Apply[F, Ψ[φ, z]]

(∫_0^3)(∫_0^(2 π))a[φ, z] . 
            n[φ, 
              z] 2 \[DifferentialD]φ \[DifferentialD]z

Gesamt: Wert des Oberflächenintegrals:

36  π  +  48  π