Question

一道高数题，定积分应用求面积和体积

Answer 1

先求抛物线方程：y=(x-1)(x-3)=x²-4x+3=(x-2)²-1；
①.设两坐标轴与抛物线所围图形的面积微S₁; x轴与抛物线所围图形的面积为S₂;
那么S₁=∫<0,1>[(x-2)²-1]dx=[(1/3)(x-2)³-x]<0,1>=[-(1/3)-1]-(-8/3)=4/3;
S₂=∣∫<1,3>[(x-2)²-1]dx∣=∣(1/3)(x-2)³-x∣<1,3>=∣[(1/3)-3]-[-(1/3)-1]∣=4/3；
∴S₁=S₂;
②.设面积S₁绕x轴旋转所得旋转体的体积为V₁; 面积S₂绕x轴旋转所得旋转体的体积为V₂;
于是V₁=π∫<0,1>[(x-2)²-1]²dx=π∫<0,1>[(x-2)^4-2(x-2)²+1]dx
=π[(1/5)(x-2)^5-(2/3)(x-2)³+x]<0,1>=π[-(1/5)+(2/3)+1]-π[-(32/5)+(16/3)]=38π/15;
V₂=π∫<1,3>[(x-2)²-1]²dx=π∫<1,3>[(x-2)^4-2(x-2)²+1]dx
=π[(1/5)(x-2)^5-(2/3)(x-2)³+x]<1,3>=π[(1/5)-(2/3)+3]-π[-(1/5)+(2/3)+1]=16π/15;
故V₁:V₂=(38π/15) : (16π/15)=19 : 8;