求r=3(1-sinβ)所围成的图形面积
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求r=3(1-sinβ)所围成的图形面积
![求r=3(1-sinβ)所围成的图形面积](/uploads/image/z/18580939-43-9.jpg?t=%E6%B1%82r%3D3%EF%BC%881-sin%CE%B2%EF%BC%89%E6%89%80%E5%9B%B4%E6%88%90%E7%9A%84%E5%9B%BE%E5%BD%A2%E9%9D%A2%E7%A7%AF)
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
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