• 2022-06-10
    求由曲线[tex=2.857x1.286]SX6Mf6VLzor8G12z5cl4Ag==[/tex]与[tex=5.286x1.286]MyVth5cTUdXAzTISg8ZCbQ==[/tex]所围图形的面积。
  • 知识点:平面图形面积思路:由于所围图形表达为[tex=1.5x1.286]O60CWzH2K4cRIAlsvRByrg==[/tex]型时解法较简单,所以用[tex=1.5x1.286]O60CWzH2K4cRIAlsvRByrg==[/tex]型做解:见图6-2-3[img=412x330]1781bb38e0f47e2.png[/img][tex=0.714x1.286]wbdAxWgHFhoV9XdVGDcK2w==[/tex]两条曲线的交点:[tex=14.429x3.357]M/Yeox5bOq02SPK7XRukbwf9cF9ir9sB6qew+OiRmbOmtP7Q8h4TOzWycF6NKWnMyVgHeNYOA7MwGcnuebSCkXT74Ruix2500mRI1qjKIWppzWiRH+ud50giBgYME8p9RSneYUEJQ2TowmOQX68Rz2FoQuNiB6TnWXSx87aeCVqsKzsEvZAddNaHz+Y+rqwp[/tex],[tex=0.714x1.286]Mjp1ERIg12NQkOrp1BseMg==[/tex]所围区域[tex=0.857x1.286]s+r8LBAs3scxfl88DGExcg==[/tex]表达为[tex=1.5x1.286]O60CWzH2K4cRIAlsvRByrg==[/tex]型:[tex=8.571x3.357]M/Yeox5bOq02SPK7XRukb30ebviEFlAGNvFAnNoDgBqh1BoRBpY4a1geavWHmiicA7G589QWq3BJOente1fiE9M0fm7HKKBydKmMnKJaD6psHDSOnErPWXglF1VabwOs[/tex],[tex=0.714x1.286]Mjp1ERIg12NQkOrp1BseMg==[/tex][tex=12.786x2.643]ew5a4y33kS23MV98UziWwWFHBrOEdhtKmDSRBf02gX2C0JlRODijaWDYq8u83JtdfWYb0fmjhH2H2/WMZq4Ew0ZY3KOfCMcFXEEu/JBjOlQ=[/tex][tex=11.214x2.857]o9xqxRkX+a6tHfKoOsf4588k8yvjWs4qs7W+i2fJjKEDhw5ITb9zAQExzAE2SvS+kUmhOtfQVgOyRrfTWv0mMTgtmjlF7fhlmIt4zkSLcLMceEmNjxuiGlcviuWmUonUWaSaLYbRreP96OI6cVSb4w==[/tex](由于图形关于[tex=0.571x1.286]XubEW9+1+hkJqH7jXe5MrA==[/tex]轴对称,所以也可以解为:[tex=13.286x2.571]EqKwyjvsOeHmZTqlOP6x/KC8FLSqF5VORwNkUuL2z/+0acMxfx9RiRjg6pM2jv9KDYF5woqvW0MASqufiPMA3w==[/tex][tex=11.286x2.786]tetFgXjuGwM6PzwDJzmlfUa+KHRkTGarhBJp+/ywXlJPeJTnK89RMgiqwzuor5egJ5kFZKjwgWGPrdLaCDL9kHQrB9XPMYoEt2mxQdgNN4hlVAR6ZGu7DZCo6RnGs5lG[/tex])

    内容

    • 0

      求由下列曲线所围图形的面积:[tex=2.286x1.429]f0KdLlH9l+9WWJPSEUUoew==[/tex],[tex=3.571x1.429]IfSo6jez8EmoVEIxPb9lsA==[/tex]

    • 1

      求由下列曲线所围图形的面积:[tex=7.143x1.357]t0KoOPQWCkvuCRcn7bUYKg==[/tex],[tex=1.786x1.214]/1Hc3IEqjvG22LyL7cBWzg==[/tex]

    • 2

      画出旋转抛物面[tex=4.929x1.286]4S08oViEap2mcmnzdJxBs2dRuZpwGIilf376wcz90AM=[/tex],柱面[tex=2.857x1.286]SX6Mf6VLzor8G12z5cl4Ag==[/tex],平面[tex=2.286x1.286]JLs9PeQldj+slOTItz+PvA==[/tex]及 [tex=2.357x1.286]jgIRiGqlkdCMqO92sJAASg==[/tex]所围立体的图形。

    • 3

      求由曲线 [tex=11.143x1.571]kS5/E8RlvrILx5ClC5INNhGjb6BOLSCuFCyTHlpRO4ucE02vqyrBfIz87YjxQrRS[/tex] 所围图形的面积

    • 4

      求由下列曲线所围图形的面积:[tex=5.357x1.429]JK/gWvp9RATL+tJmFtfGJA==[/tex],[tex=3.571x1.214]wr4IRpvyhFZN38hrYxCmgA==[/tex]