1个回答
展开全部
(5) x^2+y^2 = Rx, 化极坐标即 r = Rcost
I = ∫<-π/2, π/2>dt∫<0, Rcost>√(R^2-r^2) rdr
= (-1/2)∫<-π/2, π/2>dt∫<0, Rcost>√(R^2-r^2) d(R^2-r^2)
= (-1/2)∫<-π/2, π/2>dt (2/3)[(R^2-r^2)^(3/2)]<0, Rcost>
= (R^3/3)∫<-π/2, π/2>[1-(sint)^3]dt
= (R^3/3){∫<-π/2, π/2>dt + ∫<-π/2, π/2>[1-(cost)^2]dcost}
= (R^3/3){ π + [cost-(1/3)(cost)^3]<-π/2, π/2>} = πR^3/3
I = ∫<-π/2, π/2>dt∫<0, Rcost>√(R^2-r^2) rdr
= (-1/2)∫<-π/2, π/2>dt∫<0, Rcost>√(R^2-r^2) d(R^2-r^2)
= (-1/2)∫<-π/2, π/2>dt (2/3)[(R^2-r^2)^(3/2)]<0, Rcost>
= (R^3/3)∫<-π/2, π/2>[1-(sint)^3]dt
= (R^3/3){∫<-π/2, π/2>dt + ∫<-π/2, π/2>[1-(cost)^2]dcost}
= (R^3/3){ π + [cost-(1/3)(cost)^3]<-π/2, π/2>} = πR^3/3
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
