求解这题旋转体面积
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Vx=∫(0,π)πy^2dx
=π∫(0,π)(1+sinx)^2dx
=π∫(0,π)(1+2sinx+sin^2x)dx
=π∫(0,π)[1+2sinx+1/2(1-cos2x)]dx
=π[x-2cosx+1/2x-1/4sin2x]|(0,π)
=π[3/2x-2cosx-1/4sin2x]|(0,π)
=π[3/2(π-0)-2(cosπ-cos0)-1/4(sin2π-sin0)]
=π[3π/2-2(-1-1)-1/4(0-0)]
=π(3π/2+4)
=(3/2)π^2+4π
注:^2——表示平方
=π∫(0,π)(1+sinx)^2dx
=π∫(0,π)(1+2sinx+sin^2x)dx
=π∫(0,π)[1+2sinx+1/2(1-cos2x)]dx
=π[x-2cosx+1/2x-1/4sin2x]|(0,π)
=π[3/2x-2cosx-1/4sin2x]|(0,π)
=π[3/2(π-0)-2(cosπ-cos0)-1/4(sin2π-sin0)]
=π[3π/2-2(-1-1)-1/4(0-0)]
=π(3π/2+4)
=(3/2)π^2+4π
注:^2——表示平方
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