SF1626 - Övning 12

Göran Andersson goeran@kth.se, 110506

8.14

a)

En parametrisering av ytan är:

𝔖∶(x,y)(Overscript[→, 𝕣])(x,y,(x^2)+(y^2)), 𝔇={(x,y)∶(x^2)+(y^2)≤1}

𝕟(x,y)=(Subscript[𝕣, x]^′)×(Subscript[𝕣, y]^′)=(1,0,2x)×(0,1,2y)=(Det[({{(OverHat[𝕩]), (OverHat[𝕪]), (OverHat[𝕫])}, {1, 0, 2x}, {0, 1, 2y}})])=(-2 x,-2 y,1)

Normalvektorn kan även erhållas som gradienten till nivåytan f=0, f(x,y,z)=z-((x^2)+(y^2)).

∇f=(-2 x,-2 y,1)

b)

Arean av ytan är

A=(∫_𝔖)dS=(Underscript[∫∫, 𝔇])(Abs[(Subscript[𝕣, x]^′)×(Subscript[𝕣, y]^′)])  dx dy=(Underscript[∫∫, 𝔇])(Sqrt[(((-2 x))^2)+(((-2 y))^2)+(1^2)])  dx dy

=(∫_0^(2π)) dθ (∫_0^1)(Sqrt[4(r^2)+1]) r dr(Piecewise[{{t=4(r^2)+1}, {dt=8r dr}}])=2π(∫_(4·(0^2)+1)^(4·(1^2)+1))(Sqrt[t]) ((dt)/8)

=(π/4)(∫_1^5)(t^(1/2)) dt=(Subscript[(π/4)[((t^(3/2))/(3/2))], t=1]^5)=(π/6) (5 (Sqrt[5])-1)

c)

Massan blir:

m=(∫_𝔖)σ dS=(Underscript[∫∫, 𝔇])(x^2)(Abs[(Subscript[𝕣, x]^′)×(Subscript[𝕣, y]^′)])  dx dy=(Underscript[∫∫, 𝔇])(x^2)(Sqrt[(((-2 x))^2)+(((-2 y))^2)+(1^2)])  dx dy

= (∫_0^(2π))(∫_0^1)(((r cos(θ)))^2)(Sqrt[4(r^2)+1]) r dr dθ

=(∫_0^(2π)) (cos^2)(θ)dθ (∫_0^1)(r^2)(Sqrt[4(r^2)+1]) r dr(Piecewise[{{t=4(r^2)+1}, {dt=8r dr}}])

=(∫_0^(2π)) (1/2) (1+cos(2 θ))dθ (∫_1^5)(1/4)(t-1)(Sqrt[t])  ((dt)/8)

=π·(1/32)(∫_1^5)((t^(3/2))-(t^(1/2)))  dt=(π/120) (1+25 (Sqrt[5]))

8.15

a)

En parametrisering av ytan är:

𝔖∶(s,t)(Overscript[→, 𝕣])(2cos(s),2 sin(s),t), 𝔇={(s,t)∶0≤s≤2π,0≤t≤3}

b)

𝕟(s,t)=(Subscript[𝕣, s]^′)×(Subscript[𝕣, t]^′)=(-2 sin(s),2cos(s),0)×(0,0,1)=(Det[({{(OverHat[𝕩]), (OverHat[𝕪]), (OverHat[𝕫])}, {-2 sin(s), 2cos(s), 0}, {0, 0, 1}})])=(2cos(s),2 sin(s),0)

∴𝕡=(x,y,z)=>𝕟=(x,y,0)

∴𝕡=(1,(Sqrt[3]),2)=>𝕟=(1,(Sqrt[3]),0)

c)

Arean av cylindern blir

A=(∫_𝔖)dS=(Underscript[∫∫, 𝔇])(Abs[(Subscript[𝕣, s]^′)×(Subscript[𝕣, t]^′)])  ds dt=(Underscript[∫∫, 𝔇])(Sqrt[(((2cos(s)))^2)+(((2 sin(s)))^2)+(0^2)])  ds dt

=(Underscript[∫∫, 𝔇])2  ds dt=(∫_0^(2π)) 2ds (∫_0^3)dt=4π·3=12π

d)

(x^2)+(y^2)=4(Overscript[⇔, y>0])y=(Sqrt[4-(x^2)])

𝔖∶(x,z)(Overscript[→, 𝕣])(x,(Sqrt[4-(x^2)]),z), 𝔇={(x,z)∶-2≤x≤2,0≤z≤3}

𝕟(x,z)=(Subscript[𝕣, z]^′)×(Subscript[𝕣, x]^′)=(0,0,1)×(1,-(x/(Sqrt[4-(x^2)])),0)=(Det[({{(OverHat[𝕩]), (OverHat[𝕪]), (OverHat[𝕫])}, {0, 0, 1}, {1, -(x/(Sqrt[4-(x^2)])), 0}})])=((x/(Sqrt[4-(x^2)])),1,0)

(x,z)=(1,2)=>𝕟=((1/(Sqrt[4-(1^2)])),1,0)=((1/(Sqrt[3])),1,0)

8.16

a)

En parametrisering av ytan är:

𝔜∶(x,y)(Overscript[→, 𝕣])(x,y,(Sqrt[4-((x^2)+(y^2))])), 𝔇={(x,y)∶(x^2)+(y^2)≤3}

Alternativ:

𝔜∶(θ,ϕ)(Overscript[→, 𝕣])(2sin(θ)cos(ϕ),2sin(θ)sin(ϕ),2cos(θ)), 𝔇={(x,y)∶0≤θ≤(π/3),0≤ϕ≤2π}

b)

Eftersom det är en sfär med centrum i origo är förstås (x,y,z) en normalvektor. Inför areaberäkningen vill vi dock ha en speciell normal nämligen (Subscript[𝕣, x]^′)×(Subscript[𝕣, y]^′).

𝕟(x,y)=(Subscript[𝕣, x]^′)×(Subscript[𝕣, y]^′)=(1,0,-(x/(Sqrt[4-((x^2)+(y^2))])))×(0,1,-(y/(Sqrt[4-((x^2)+(y^2))])))

=((x/(Sqrt[4-((x^2)+(y^2))])),(y/(Sqrt[4-((x^2)+(y^2))])),1)=(1/z) (x,y,z)=>(Abs[𝕟])=(1/z)·2=(2/(Sqrt[4-((x^2)+(y^2))]))

Alternativt:

𝕟(θ,ϕ)=(Subscript[𝕣, θ]^′)×(Subscript[𝕣, ϕ]^′)=(2 cos(θ) cos(ϕ),2 cos(θ) sin(ϕ),-2 sin(θ))×(-2 sin(θ) sin(ϕ),2 sin(θ) cos(ϕ),0)

=(4 (sin^2)(θ) cos(ϕ),4 (sin^2)(θ) sin(ϕ),4 sin(θ) cos(θ) (cos^2)(ϕ)+4 sin(θ) cos(θ) (sin^2)(ϕ))

=2sin(θ)(2sin(θ)cos(ϕ),2sin(θ) sin(ϕ),2cos(θ))=2sin(θ)(x,y,z)

c)

Arean av kalotten blir

A=(∫_𝔜)dS=(Underscript[∫∫, 𝔇])(Abs[(Subscript[𝕣, x]^′)×(Subscript[𝕣, y]^′)])  dx dy=(Underscript[∫∫, 𝔇])(2/(Sqrt[4-((x^2)+(y^2))]))  dx dy

=(∫_0^(2π))dϕ(∫_0^(Sqrt[3]))(2/(Sqrt[4-(r^2)]))  r dr(Piecewise[{{t=4-(r^2)}, {dt=(-2r) dr}}])=2π(∫_4^1)(1/(Sqrt[t]))  (-dt)=4π

Alternativt:

A=(∫_𝔜)dS=(Underscript[∫∫, 𝔇])(Abs[(Subscript[𝕣, θ]^′)×(Subscript[𝕣, ϕ]^′)])  dθ dϕ=(Underscript[∫∫, 𝔇])2sin(θ)·2  dθ dϕ

=4(∫_0^(π/3)) sin(θ)dθ (∫_0^(2π))dϕ=4·(1/2)·2π=4π

d)

En parametrisering av ytan är:

𝔜∶(x,y)(Overscript[→, 𝕣])(x,y,(Sqrt[(R^2)-((x^2)+(y^2))])), 𝔇={(x,y)∶(x^2)+(y^2)≤(R^2)-(h^2)}

𝕟(x,y)=(Subscript[𝕣, x]^′)×(Subscript[𝕣, y]^′)(Overscript[=, b)])(1/z) (x,y,z)=>(Abs[𝕟])=(1/z)·R=(R/(Sqrt[(R^2)-((x^2)+(y^2))]))

A=(∫_𝔜)dS=(Underscript[∫∫, 𝔇])(Abs[(Subscript[𝕣, x]^′)×(Subscript[𝕣, y]^′)])  dx dy=(Underscript[∫∫, 𝔇])(R/(Sqrt[(R^2)-((x^2)+(y^2))]))  dx dy

=(∫_0^(2π))dϕ(∫_0^(Sqrt[(R^2)-(h^2)]))(R/(Sqrt[(R^2)-(r^2)])) r dr(Piecewise[{{t=(R^2)-(r^2)}, {dt=(-2r) dr}}])=2π R(∫_(R^2)^(h^2))(1/(Sqrt[t]))  (-((dt)/2))=2π R (R-h)

8.18

En parametrisering av ytan är:

𝔖∶(θ,x)(Overscript[→, 𝕣])(x,4cos(θ),4sin(θ)), 𝔇={(θ,x)∶(x^2)+(((4cos(θ)))^2)≤(4^2)}

Normalvektorn blir:

𝕟(θ,x)=(Subscript[𝕣, θ]^′)×(Subscript[𝕣, x]^′)=(0,-4sin(θ),4cos(θ))×(1,0,0)=(0,4 cos(θ),4 sin(θ))

Arean av hålet blir

A=(∫_𝔖)dS=(Underscript[∫∫, 𝔇])(Abs[(Subscript[𝕣, x]^′)×(Subscript[𝕣, y]^′)])  dθ dx=(Underscript[∫∫, 𝔇])4  dθ dx

=4(∫_0^π)(∫_(-z)^z) dx dθ=4(∫_0^π)(∫_(-4sin(θ))^(4sin(θ))) dx dθ=32(∫_0^π)sin(θ)dθ=64

Dvs. arean är 64(cm^2).

8.20

En parametrisering av ytan är:

𝔖∶(r,θ)(Overscript[→, 𝕣])(r cos(θ),r sin(θ),(Sqrt[1-(r^2)])), 𝔇={(r,θ)∶0≤r≤1,0≤θ≤(π/2)}

Normalvektorn blir:

𝕟=(Subscript[𝕣, r]^′)×(Subscript[𝕣, θ]^′)=(cos(θ),sin(θ),-(r/(Sqrt[1-(r^2)])))×(-r sin(θ),r cos(θ),0)

=((((r^2) cos(θ))/(Sqrt[1-(r^2)])),(((r^2) sin(θ))/(Sqrt[1-(r^2)])),r)

∴n=(Abs[𝕟])=(Sqrt[(((r^4)(cos^2)(θ))/(1-(r^2)))+(((r^4)(sin^2)(θ))/(1-(r^2)))+(r^2)])=(r/(Sqrt[1-(r^2)]))

Masstätheten är:

σ=c r z=c r (Sqrt[1-(r^2)])

Massan blir

m=(∫_𝔖)σ dS=(Underscript[∫∫, 𝔇])σ n  dr dθ=(Underscript[∫∫, 𝔇])c r (Sqrt[1-(r^2)])(r/(Sqrt[1-(r^2)]))  dr dθ=c(Underscript[∫∫, 𝔇])(r^2)  dr dθ

=c(∫_0^(π/2))dθ(∫_0^1) (r^2)dr =c·(π/2)·(1/3)=c (π/6)

Dvs. massan är c (π/6).