由y=sinx得：
x1=arcsiny,x1∈(0,π/2),y∈(0,1)
x2=π-arcsiny,x2∈(π/2,π),y∈(0,1)
∴V=∫(0,1)π[(x2)²-(x1)²]dy
=π∫(0,1)[(π-arcsiny)²-(arcsiny)²]dy
=π∫(0,1)[π(π-2arcsiny)dy
=π²[πy|(0,1)-2∫(0,1)arcsinydy]
=π²{π-2[yarcsiny|(0,1)-∫(0,1)ydy/√(1-y²)]}
=π³-2π²[π/2+1/2·∫(0,1)d(1-y²)/√(1-y²)]
=π³-2π²[π/2+1/2·2√(1-y²)|(0,1)]
=π³-2π²[π/2+(-1)]
=2π²