求解答,高数题,如图,谢谢

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求解答,高数题,如图,谢谢求解答,高数题,如图,谢谢求解答,高数题,如图,谢谢这题楼上太繁琐.用极坐标,注意0《r《Rcosθ,代入得:=∫(-π/2,π/2)dθ∫(0,Rcosθ)r√(R^2-r

求解答,高数题,如图,谢谢
求解答,高数题,如图,谢谢

求解答,高数题,如图,谢谢
这题楼上太繁琐.用极坐标,注意0《r《Rcosθ,代入得:
=∫(-π/2,π/2)dθ∫(0,Rcosθ)r√(R^2-r^2)dr
=(1/2)∫(-π/2,π/2)dθ∫(0,Rcosθ)√(R^2-r^2)dr^2
=(-1/3)∫(-π/2,π/2)(R^2-r^2)^(3/2)|(0,Rcosθ)dθ
=(1/3)∫(-π/2,π/2)R^3(1-[(sinθ)^2^(3/2)dθ (偶函数在对称区间的积分为半区间的2倍)
=(2R^3/3)∫(0,π/2)(1-(sinθ)^3)dθ (∫(0,π/2)(sinθ)^3)dθ用公式=2/3)
=(2/3)R^3(π/2-2/3)

用极坐标
x=rcost
y=rsint
围城区域方程为
r^2=R*rcost
r=Rcost
所以r的范围是
0而t的范围由图像可见是(-π/2,π/2)
所以原积分变为
∫<-π/2,π/2>∫<0,Rcost>根号(R^2-r^2) r dr dt
第一层用替换,u=R^2-r^2, ...

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用极坐标
x=rcost
y=rsint
围城区域方程为
r^2=R*rcost
r=Rcost
所以r的范围是
0而t的范围由图像可见是(-π/2,π/2)
所以原积分变为
∫<-π/2,π/2>∫<0,Rcost>根号(R^2-r^2) r dr dt
第一层用替换,u=R^2-r^2, R^2du=-2rdr
rdr=(-1/2)du
=(-1/2)∫<-π/2,π/2>∫ u^(1/2) du dt
=(1/2)∫<-π/2,π/2> (2/3)u^(3/2) | dt
=(R^3/3)∫<-π/2,π/2> [1-|sint|^3] dt
由于对称性
=(2R^3/3)∫<0,π/2> [1-sin^3 t ] dt
=(2R^3/3)∫<0,π/2>[1-(3/4)sint+(1/4)sin3t] dt
=(2R^3/3)[t+(3/4)cost-(1/12)cos3t]|<0,π/2>
=(2R^3/3)[π/2+(3/4)*(-1)-(1/12)(-1)]
=(2R^3/3)[π/2-2/3]

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