微積分質心問題,一階微分方程式

[匿名]

2010-06-21 6:48 am

Find the moment of inertia about the x-axis lx for the circular lamina bounded by x^2+y^2=9 if p(x,y)=k

Find the center of mass of the solid bounded by z=4-x^2-y^2 and above the square with vertices (1,1),(1,-1),(-1,-1)and(-1,1), if the density is p=3

find the solution th the initial value problem :(1+x^2)(1+y^2)=xyy' , y(1)=0

回答 (1)

天助

2010-06-26 10:09 am

✔ 最佳答案

Q1:
moment of inerita about the x-axis
=∫∫_D k dxdy (D: x^2+y^2<=9)
changes to polar coordinate system
=∫[0~2π]∫[0~3] k*(rcosθ)^2 r dr dθ
=∫[0~2π] k*(r^4/4)(cosθ)^2|[0~3] dθ
=(81k/8)∫[0~2π] [1+cos(2θ)] dθ
=(81k/8)*[θ+(1/2)sin(2θ)] for θ=0~2π
=81kπ/4

Q2, Q3:已解過了

2010-06-26 02:13:10 補充：
Sorry!更正:
Q1:moment of inerita about the x-axis(繞x軸旋轉的慣性矩or轉動慣量)
=∫∫_D k dxdy (D: x^2+y^2<=9) (changes to polar coordinate system)
=∫[0~2π]∫[0~3] k*(r sinθ)^2 r dr dθ
=∫[0~2π] k*(r^4/4)(sinθ)^2|[0~3] dθ
=(81k/8)∫[0~2π] [1-cos(2θ)] dθ
=(81k/8)*[θ-(1/2)sin(2θ)] for θ=0~2π
=81kπ/4

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