(1) View the projection on x-z plane:
A line x=z & both circles: x^2+z^2=r^2 (r=5 & r=6)
Line ∩ Sphere=r*sin45°=r√2/2
V=2π∫[√(r^2-z^2)-z]^2*dz.....z=0~r√2/2
=2π∫[r^2-z^2+z^2-2z√(r^2-z^2)]dz
=2π∫[r^2-2z√(r^2-z^2)]dz
=2πr^2*z+2π∫√(r^2-z^2)]d(r^2-z^2)
=√2πr^3+2π2/3*(r^2-z^2)^1.5
=√2πr^3+4π/3*[(r^2-r^2/2)^1.5-r^3]
=√2πr^3+4π/3*[r^3/8-r^3]
=√2πr^3-4πr^3/3+4π/3*r^3/2√2
=√2πr^3-4πr^3/3+2πr^3/3√2
=√2πr^3-4πr^3/3+√2πr^3/3
=4√2πr^3/3-4πr^3/3
=4(√2-1)πr^3/3
V1=V(6)=4(√2-1)π216/3
V2=V(5)=4(√2-1)π125/3
ΔV=V1-V2
=91*4(√2-1)π/3
=364(√2-1)π/3............ans
(2) Find the maclaurin series for f(x)=Ln|1+x/1-x|and determine its
radius of convergence.use the first four nonzero terms of the series
to approximate ln3.
f(x)=ln|(1+x)/(1-x)|
=Ln|1+x|-Ln|1-x|
=f1(x)-f(2(x)
f1(x)=Ln|1+x|=Ln(1+x) if |x|<1 => f1(0)=0
f2(x)=Ln|1-x|=Ln(1-x) if |x|<1
f1'(x)=1/(1+x) => f'(0)=1
f1"(x)=-1/(1+x)^2 => f"(0)=-1
f1"'(x)=2/(1+x)^3 => f"'(0)=2
f1""(x)=-6/(1+x)^4 => f""(0)=-3!
............................
f1(x)=f(0)/0!+f'(0)x-f"(0)x^2/2+f"'(0)x^3/3-f""(0)x^4/4+.....
=x-x^2/2+x^3/3-x^4/4+......
Similariy, f2(x)=-[x+x^2/2+x^3/3+x^4/4+......]
f(x)=f1(x)-f2(x)
=2[x+x^3/3+x^5/5+x^7/7+......]
Let Ln[(1+x)/(1-x)]=Ln3 => (1+x)/(1-x)=3
1+x=3(1-x) => x=1/2
f(1/2)=2[0.5+0.5^3/3+0.5^5/5+0.5^7/7+]
=1.098065 for four items
Limit <n->∞>|A<n>/A<n-1>|
=|[x^(2n+1)/(2n+1)]/[x^(2n-1)/(2n-1)]|
=|x^2|<1
∴ -1<x<1 for convergence
(3) Find the volumn of solid bound by the graphs of z=Ln(x^2+y^2) z=0,
x^2+y^2=>1.....b and x^2+y^2<=4..........c
z=Ln(x^2+y^2)=Ln(r^2) => r^2=e^z.........a
a∩b(a.b交點): r^2=e^z=1 => z1=Ln(1)=0
a∩c(a.c交點): r^2=e^z=4 => z2=Ln(4)=2*Ln(2)
V1=cylinder(r=1)=πz1=0
V2=cylinder(r=2)=4πz2=8πLn(2)
V3=π∫r^2*dz
=π∫e^z*dz.....<0~z>
=πe^z
=π(e^z-1)
V13=Eq.a&b所包體積
=V1-V3
=π(0-e^z+1)
=π(-1+1)
=0
V23=Eq.a&c所包體積
=V2-V3
=π[8πLn(2)-e^z+1]
=π[8πLn(2)-4+1]
=π[8πLn(2)-3]............ans
(4) 超過2000字無法輸入
文章出自: http://tw.knowledge.yahoo.com/question/question?qid=1612070100098
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