Question

求由抛物线Y=X²和Y=2-X²所围成图形的面积,并求此图形绕X轴旋转一周所成立体的体积

尽量详细点

Answer 1

y=x^2与y=2-x^2的交点：x^2=2-x^2 ===> x^2=1 ===> x=±1
所以交点是(-1,1)和(1,1)
围成的面积=∫<-1,1>(2-x^2-x^2)dx=2∫<0,1>(2-2x^2)dx
=2*[2x-(2/3)x^3]|<0,1>
=2*[2-(2/3)]
=8/3
旋转体的体积V=∫<-1,1>[π(2-x^2)^2-π(x^2)^2]dx
=2π∫<0,1>(4-4x^2)dx
=8π∫<0,1>(1-x^2)dx
=8π[x-(1/3)x^3]|<0,1>
=8π*[1-(1/3)]
=16π/3

Answer 2

请旋转后看···