求r=3（1-sinβ）所围成的图形面积
1个回答
展开全部
β = 0,r = 3,
β = PI/2,r = 0,
β = PI,r = 3,
β = 3PI/2,r = 0,
β = 2PI,r = 3.
面积 = (1/2)S_{0->2PI}dβ S_{0->3(1-sinβ)}r^2dr
= (1/6)S_{0->2PI}[3(1-sinβ)]^3dβ
= (9/2)S_{0->2PI}[1 - 3sinβ + 3(sinβ)^2 - (sinβ)^3]dβ
= (9/2)S_{0->2PI}[1 + 3(sinβ)^2]dβ
= (9/2)*(2PI) + (9/2)S_{0->2PI}3[1 - cos(2β)]/2dβ
= (9/2)*(2PI) + (9/2)*(3/2)*2PI
= 45PI/2
β = PI/2,r = 0,
β = PI,r = 3,
β = 3PI/2,r = 0,
β = 2PI,r = 3.
面积 = (1/2)S_{0->2PI}dβ S_{0->3(1-sinβ)}r^2dr
= (1/6)S_{0->2PI}[3(1-sinβ)]^3dβ
= (9/2)S_{0->2PI}[1 - 3sinβ + 3(sinβ)^2 - (sinβ)^3]dβ
= (9/2)S_{0->2PI}[1 + 3(sinβ)^2]dβ
= (9/2)*(2PI) + (9/2)S_{0->2PI}3[1 - cos(2β)]/2dβ
= (9/2)*(2PI) + (9/2)*(3/2)*2PI
= 45PI/2
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
