SF1626 - Övning 8 - Code

Göran Andersson goeran@kth.se, 110414

6.34

(Underscript[∫∫, (ℝ^2)])(x^2) (e^(-(Sqrt[(x^2)+(y^2)]))) dx dy

I detta fall är inte direkt integration möjlig så vi inför polära koordinater. Först en numerisk integration för att kunna jämföra.

ClearAll["`*"]

f[{x_,y_}]:=(x^2)(E^(-(Sqrt[(x^2)+(y^2)])));
𝕩={x,y};

NIntegrate[(x^2) (E^(-(Sqrt[(x^2)+(y^2)]))), {x, -∞,∞}, {y, -∞,∞} ]

Polära koordinater införs och funktionaldeterminanten beräknas.

Clear[𝕩];𝕣={r,θ};
𝕩[{r_,θ_}]:={r Cos[θ],r Sin[θ]};
J[{r_,θ_}]=FullSimplify[Abs[Det[(∂_({𝕣}))𝕩[𝕣]]],r>0]

g[{r_,θ_}]=FullSimplify[f[𝕩[𝕣]],r>0]

(∫_0^∞)(∫_0^(2 π))g[𝕣] J[𝕣]\[DifferentialD]θ\[DifferentialD]r

N[%]

6.35

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(∫_1^2)(∫_0^∞)(1/(1+(x^2) (y^2)))\[DifferentialD]y\[DifferentialD]x

6.39

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𝕩={x,y};

𝒟[{x_,y_}]:=x≥0∧y≥0∧x+y≤1

Integrate[(((1-x-y))^α) Boole[𝒟[𝕩]],{x,-∞,∞},{y,-∞,∞},Assumptions->{α>-1}]

6.40

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(∫_(-∞)^∞)(∫_(-∞)^∞)((E^(-(Sqrt[(x^2)+(y^2)])))/(Sqrt[(x^2)+(y^2)]))\[DifferentialD]y\[DifferentialD]x

7.1

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

7.3

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

7.6

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RegionPlot3D[x+y+z≤1,{x,0,1},{y,0,1},{z,0,1},Mesh->None,PlotStyle->Opacity[0.5],AxesLabel->Automatic]

(∫_0^∞)(∫_0^∞)(∫_0^∞)(1/(1+(((x+y+z))^3))) Boole[x+y+z≤1]\[DifferentialD]z\[DifferentialD]y\[DifferentialD]x

8.3

ClearAll["`*"]

𝒟[{x_,y_,z_}]:=(x^2)+(y^2)≤z&&z≤2-((x^2)+(y^2));
𝕩={x,y,z};

RegionPlot3D[𝒟[𝕩],{x,-1,1},{y,-1,1},{z,0,2},AxesLabel->Automatic]

V=(∫_(-∞)^∞)(∫_(-∞)^∞)(∫_(-∞)^∞)Boole[𝒟[𝕩]]\[DifferentialD]z\[DifferentialD]y\[DifferentialD]x

8.8

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𝒟[{x_,y_,z_}]:=(x^2)≤4-4z&&(y^2)≤4-4z&&z≥0;
𝕩={x,y,z};

RegionPlot3D[𝒟[𝕩],{x,-2,2},{y,-2,2},{z,0,1},AxesLabel->Automatic,BoxRatios->Automatic,Mesh->False]

V=(∫_(-∞)^∞)(∫_(-∞)^∞)(∫_(-∞)^∞)Boole[𝒟[𝕩]]\[DifferentialD]z\[DifferentialD]y\[DifferentialD]x