SF1626 - Övning 7

Göran Andersson goeran@kth.se, 110413

6.1

(∫_0^1)(∫_0^1)(x y+(y^2))dx dy=(∫_0^1)((∫_0^1)(x y+(y^2))dy )dx

=(∫_0^1)(Subscript[[(1/2)x (y^2)+(1/3)(y^3)], y=0]^1)dx=(∫_0^1)((x (1/2)+(1/3))-0)dx

=(∫_0^1)(x (1/2)+(1/3))dx=(Subscript[[ (1/4)(x^2)+(1/3)x], x=0]^1)=((1/4)+(1/3))-0=(7/12)

Grafik

6.4

△={(x,y)∶0≤x≤1∧-(π/2)≤y≤(π/2)}

(Underscript[∫∫, △])y sin(y+x y)dx dy=(∫_(-(π/2))^(π/2))y ((∫_0^1)sin(y+x y)dx )dy

=(∫_(-(π/2))^(π/2))(Subscript[y [-(1/y)cos(y+x y)], x=0]^1)dy=(∫_(-(π/2))^(π/2))(cos(y)-cos(2y))dy

=(Subscript[[ sin(y)-(1/2)sin(2y)], y=-(π/2)]^(π/2))=(1-(-1))=2

Grafik

6.5

△={(x,y)∶(Abs[x])≤1∧(Abs[y])≤1}

(Underscript[∫∫, △])x (e^(-((x^2)+(y^2))))dx dy=(∫_(-1)^1)(Underscript[(Underscript[x (e^(-(x^2))), ⏟]), udda f(x)])dx(∫_(-1)^1)(e^(-(y^2)))dy=0·(∫_(-1)^1)(e^(-(y^2)))dy=0

Resulatet blir 0 vid integrationen i x-led eftersom integranden är udda m.a.p. x och intervallet är symmetriskt kring 0.

Grafik

6.10

△={(x,y)∶0≤x≤1∧0≤y≤1-x}

(Underscript[∫∫, △])(y/(1+x))dx dy=(∫_0^1)(1/(1+x))((∫_0^(1-x))y dy )dx=(∫_0^1)(Subscript[(1/(1+x))  [(1/2)(y^2)], y=0]^(1-x)) dx

=(∫_0^1)((((1-x))^2)/(2(1+x)))dx(Piecewise[{{t=1+x}, {dt=dx}}])=(∫_1^2)((((2-t))^2)/(2t))dt=(∫_1^2)((2/t)-2+(1/2)t)dt

=(Subscript[[2ln(t)-2t+(1/4)(t^2)], t=1]^2)=2ln(2)-(5/4)

6.11

△={(x,y)∶x≥0∧y≥0∧2x+3y≤6}={(x,y)∶0≤x≤3∧0≤y≤2-2x/3}

(Underscript[∫∫, △])(e^(-2x-3y))dx dy=(∫_0^3)(e^(-2x)) ((∫_0^(2-2x/3))(e^(-3y)) dy )dx=(∫_0^3)(Subscript[(e^(-2x))  [-(1/3)(e^(-3y))], y=0]^(2-2x/3)) dx

=(∫_0^3)(e^(-2x)) ((1/3)-(1/3)(e^(-6+2 x)))dx=(1/3)(∫_0^3) ((e^(-2x))-(e^(-6)))dx=(Subscript[(1/3)  [-(1/2)(e^(-2x))-x (e^(-6))], x=0]^3)=(1/6)(1-7(e^(-6)))

6.12

𝒟={(x,y)∶x≥0∧(x^2)≤y≤1}={(x,y)∶0≤x≤(Sqrt[y])∧0≤y≤1}

(Underscript[∫∫, 𝒟])(x/(1+(y^2)))dx dy=(∫_0^1)(1/(1+(y^2))) ((∫_0^(Sqrt[y]))x dx )dy=(∫_0^1)(Subscript[(1/(1+(y^2)))  [(1/2)(x^2)], x=0]^(Sqrt[y])) dy

=(1/2)(∫_0^1)(y/(1+(y^2)))dy(Piecewise[{{t=1+(y^2)}, {dt=2y dy}}])=(1/2)(∫_1^2) (1/t)((dt)/2)=(Subscript[(1/4)  [ln(t)], t=1]^2)=(1/4)ln(2)

6.14

𝒟={(x,y):(Abs[x])≤y,0≤y≤1}

ℐ=(Underscript[∫∫, 𝒟])(e^(-(y^2)))dx dy=(∫_0^1) (e^(-(y^2)))((∫_(-y)^y)dx )dy=(∫_0^1) (e^(-(y^2)))2y dy

=(∫_(0^2)^(1^2)) (e^(-t)) dt=(Subscript[[-(e^(-t))], 0]^1)=1-(e^(-1))

Grafik

6.15

𝒟={(x,y):-y≤x≤2 y∧0≤y≤1}

(Underscript[∫∫, 𝒟])(1/(1+(((x-2y))^2)))dx dy=(∫_0^1) ((∫_(-y)^(2y))(1/(1+(((x-2y))^2)))dx )dy

=(∫_0^1)(Subscript[[arctan(x-2y)], x=-y]^(2y)) dy=(∫_0^1) arctan(3y) dy

=(∫_0^1)1· arctan(3y) dy=(Subscript[[y arctan(3y)], y=0]^1)-(∫_0^1) ((3y)/(1+(((3y))^2))) dy(Piecewise[{{t=1+(((3y))^2)}, {dt=18y dy}}])

=arctan(3)-(1/6)(∫_1^10) (1/t) dt=arctan(3)-(1/6)ln(10)

Grafik

6.17

a)

(Subscript[𝒟, a])={(x,y):0≤y≤x∧0≤x≤1}

(Subscript[ℐ, a])=(Underscript[∫∫, (Subscript[𝒟, a])])x y (Sqrt[(x^2)+(y^2)])dx dy=(∫_0^1)x  ((∫_0^x) (Sqrt[(x^2)+(y^2)])y dy )dx(Piecewise[{{t=(x^2)+(y^2)}, {dt=2y dy}}])

(∫_0^1)x  ((∫_(x^2)^(2(x^2))) (Sqrt[t])((dt)/2) )dx=(∫_0^1)(Subscript[x  [((t^(3/2))/3)], t=(x^2)]^(2(x^2)))dx=(1/3) (2 (Sqrt[2])-1)(∫_0^1) (x^4) dt=(1/15) (2 (Sqrt[2])-1)

b)

(Subscript[𝒟, b])={(x,y):0≤y≤1∧0≤x≤1}

På grund av symmetri blir

(Underscript[∫∫, (Subscript[𝒟, b])])x y (Sqrt[(x^2)+(y^2)])dx dy=2(Subscript[ℐ, a])

6.19

𝒟={(x,y):(Abs[x])+(Abs[y])≤1}={(x,y):-1≤x+y≤1∧-1≤x-y≤1}

Vi använder koordinattransformationen:

(Piecewise[{{u=x+y}, {v=x-y}}])=>((d(u,v))/(d(x,y)))=(Det[({{1, 1}, {1, -1}})])=-2=>((d(x,y))/(d(u,v)))=(1/(-2))=-(1/2)

𝒟'={(u,v):-1≤u≤1,-1≤v≤1}

ℐ=(Underscript[∫∫, 𝒟])((((x^2)-(y^2)))^10) dx dy=(Underscript[∫∫, 𝒟])((((x+y)(x-y)))^10) dx dy=(Underscript[∫∫, 𝒟'])(((u v))^10) (Abs[((d(x,y))/(d(u,v)))])du dv

=(Underscript[∫∫, 𝒟'])(((u v))^10) (1/2)du dv=(1/2)(∫_(-1)^1) (u^10)du (∫_(-1)^1)(v^10) dv=2(∫_0^1) (u^10)du (∫_0^1)(v^10) dv=2 (1/11)(1/11)=(2/121)

6.21

Vi använder polära koordinater:

(Piecewise[{{x=r cos(θ)}, {y=r sin(θ)}}])=>((d(x,y))/(d(r,θ)))=(Det[({{cos(θ), -r sin(θ)}, {sin(θ), r cos(θ)}})])=r

𝒟={(x,y):1≤(x^2)+(y^2)≤2}=>𝒟'={(r,θ):1≤r≤(Sqrt[2]),0≤θ≤2π}

ℐ=(Underscript[∫∫, 𝒟])ln((x^2)+(y^2)+1) dx dy=(Underscript[∫∫, 𝒟'])ln((r^2)+1) (Abs[((d(x,y))/(d(r,θ)))])dr dθ=(Underscript[∫∫, 𝒟'])ln((r^2)+1) r dr dθ

=(∫_0^(2π)) dθ (∫_1^(Sqrt[2]))(1/2) ln((r^2)+1)2 r dr=2π·(1/2)(∫_1^2) ln(t+1) dt

=π(∫_1^2) ln(t+1) dt=π(∫_2^3) ln(u) du=π ( (Subscript[[u ln(u)], u=2]^3)-(∫_2^3) du)=π (3ln(3)-2ln(2)-1)

==π (3ln(3)-2ln(2)-1)=π (ln((27/4))-1)

Grafik

6.22

Vi använder polära koordinater:

(Piecewise[{{x=r cos(θ)}, {y=r sin(θ)}}])=>((d(x,y))/(d(r,θ)))=(Det[({{cos(θ), -r sin(θ)}, {sin(θ), r cos(θ)}})])=r

𝒟={(x,y):(x^2)+(y^2)≤25∧y≥(Abs[x])}=>𝒟'={(r,θ):0≤r≤5,(π/4)≤θ≤((3π)/4)}

ℐ=(Underscript[∫∫, 𝒟])(x^2) (e^((x^2)+(y^2))) dx dy=(Underscript[∫∫, 𝒟'])(r^2) (cos^2)(θ)(e^(r^2)) (Abs[((d(x,y))/(d(r,θ)))])dr dθ=(Underscript[∫∫, 𝒟'])(r^2) (cos^2)(θ)(e^(r^2)) r dr dθ

=(∫_(π/4)^((3π)/4))(cos^2)(θ) dθ (∫_0^5)(r^2) (e^(r^2)) r dr(Piecewise[{{t=(r^2)}, {dt=2r dr}}])=(∫_(π/4)^((3π)/4))(1/2) (1+cos(2 θ)) dθ (∫_0^25)(1/2)t (e^t)dt

=((Subscript[[(1/2)θ+(1/4)sin(2θ)], θ=(π/4)]^((3π)/4)))((Subscript[[(1/2)t (e^t)], t=0]^25)-(∫_0^25)(1/2) (e^t)dt)=((π/4)-(1/2))(12 (e^25)+(1/2))

6.23

Vi använder polära koordinater:

(Piecewise[{{x=r cos(θ)}, {y=r sin(θ)}}])=>((d(x,y))/(d(r,θ)))=(Det[({{cos(θ), -r sin(θ)}, {sin(θ), r cos(θ)}})])=r

𝒟={(x,y):(x^2)+(y^2)≤1∧-x≤y≤x∧x≥0}=>𝒟'={(r,θ):0≤r≤1,-(π/4)≤θ≤(π/4)}

(Underscript[∫∫, 𝒟])((x^2)-(y^2)) (e^(x y)) dx dy=(Underscript[∫∫, 𝒟'])(r^2) ((cos^2)(θ)-(sin^2)(θ))(e^((r^2)cos(θ)sin(θ))) r dr dθ=(Underscript[∫∫, 𝒟'])(r^3) cos(2 θ)(e^((1/2) (r^2) sin(2 θ))) dr dθ

=(∫_0^1)(r^3) ((∫_(-(π/4))^(π/4))(e^((1/2) (r^2) sin(2 θ))) cos(2 θ)dθ )dr=(Piecewise[{{t=(1/2)sin(2 θ)}, {dt=cos(2 θ)dθ}}])=(∫_0^1)(r^3) ((∫_(-(1/2))^(1/2))(e^((r^2) t)) dt )dr

=(∫_0^1)(r^3) ((Subscript[[((e^((r^2) t))/(r^2))], t=-(1/2)]^(1/2)) )dr=(∫_0^1)r ((e^((1/2)(r^2)))- (e^(-(1/2)(r^2))))dr(Piecewise[{{u=(1/2)(r^2)}, {du=r dr}}])=(∫_0^(1/2))((e^u)- (e^(-u)))du

=2(∫_0^(1/2))sinh(u)du=(Subscript[2[cosh(u)], u=0]^(1/2))=2(cosh((1/2))-1)=4 (sinh^2)((1/4))

6.24

𝒟={(x,y):(x^2)-2 x+(y^2)≤0}={(x,y):(((x-1))^2)+(y^2)≤1}

Vi använder polära koordinater:

(Piecewise[{{x-1=r cos(θ)}, {y=r sin(θ)}}])⇔(Piecewise[{{x=1+r cos(θ)}, {y=r sin(θ)}}])=>((d(x,y))/(d(r,θ)))=(Det[({{cos(θ), -r sin(θ)}, {sin(θ), r cos(θ)}})])=r

=>𝒟'={(r,θ):0≤r≤1,0≤θ≤2π}

(Underscript[∫∫, 𝒟])((x^2)+(y^2))  dx dy=(Underscript[∫∫, 𝒟'])((((1+r cos(θ)))^2)+(((r sin(θ)))^2)) r dr dθ=(Underscript[∫∫, 𝒟'])(1+(r^2)+2 r cos(θ)) r dr dθ

=(∫_0^(2π)) ((∫_0^1)(r+(r^3)+2 (r^2) cos(θ))dr )dθ=(∫_0^(2π)) ((Subscript[[((r^2)/2)+((r^4)/4)+((2(r^3))/3)cos(θ)], r=0]^1) )dθ

=(∫_0^(2π)) ((2/3) cos(θ)+(3/4) )dθ=((3π)/2)