在XOY平面上是以原点为圆心,半径为R的圆区域,
化成极坐标,0<=r<=R,
0<=θ<=2π,x=rcosθ,y=rsinθ,
I=∫ [0,2π]dθ∫ [0,R] [(rcosθ)^2/a^2+(rsinθ)^2/b^2] rdr
=∫ [0,2π] [ (cosθ)^2/a^2+(sinθ)^2/b^2]( r^4/4)[0,R]dθ
=(R^4/4)∫ [0,2π] [1+cos2θ)/(2a^2)]dθ+(R^4/4)∫ [0,2π] [1-cos2θ)/(2b^2)]dθ
=R^4/(8a^2)[ θ+(1/2)sin2θ][0,2π]+R^4/(8b^2)[ θ-(1/2)sin2θ][0,2π]
=R^4/(8a^2)(2π+0)+R^4/(8b^2)[2π-0]
=πR^4/(4a^2)+πR^4/(4b^2)
=(πR^4/4)(1/a^2+1/b^2).
化成极坐标,0<=r<=R,
0<=θ<=2π,x=rcosθ,y=rsinθ,
I=∫ [0,2π]dθ∫ [0,R] [(rcosθ)^2/a^2+(rsinθ)^2/b^2] rdr
=∫ [0,2π] [ (cosθ)^2/a^2+(sinθ)^2/b^2]( r^4/4)[0,R]dθ
=(R^4/4)∫ [0,2π] [1+cos2θ)/(2a^2)]dθ+(R^4/4)∫ [0,2π] [1-cos2θ)/(2b^2)]dθ
=R^4/(8a^2)[ θ+(1/2)sin2θ][0,2π]+R^4/(8b^2)[ θ-(1/2)sin2θ][0,2π]
=R^4/(8a^2)(2π+0)+R^4/(8b^2)[2π-0]
=πR^4/(4a^2)+πR^4/(4b^2)
=(πR^4/4)(1/a^2+1/b^2).
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