Question

求z=x^2+y^2,z=(x^2+y^2)/2,y=x,y=x^2,所围成的体积

Answer 1

联立y=x,y=x^2
得x=0,y=0或x=1,y=1
所以x的取值范围是[0,1]
所以原式=
∫(0,1)dx∫(x^2,x)dy∫((x^2+y^2)/2,x^2+y^2) dz
=∫(0,1)dx∫(x^2,x)(1/2)(x^2+y^2)dy
=∫(0,1)(1/2)[(x^2 * x+(1/3)x^3)-(x^2*x^2+ (1/3)(x^2)^3]dx
=∫(0,1)[(2/3)x^3 - (1/2)x^4 - (1/6)x^6] dx
=(2/3)*(1/4)-(1/2)*(1/5)-（1/6)*(1/7)
=3/70