Question

求抛物线y^2=2x与直线y=4-x围成平面图形的面积?

Answer 1

1.先求抛物线与直线的交点

y^2=2x y=4-x

(4-x)^2=2x

x^2-10x+16=0

x1=2 y1=4-2=2 点(2,2)

x2=8 y2=4-8=-4 点(8,-4)

2.再求积分 y积分范围从-4到2(上2,下-4,下同)

y^2=2x x=y^2/2

y=4-x x=4-y

∫(-4,2)(4-y-y^2/2)dy

=(4y-1/2y^2-y^3/6)|(-4,2)

=(8-2-4/3)-(-16-8+32/3)

=30-12

=18