Question

y=1/2x与y^2=8-x所围成的平面图形的面积？

y=1/2x与y^2=8-x所围成的平面图形的面积

Answer 1

先求两曲线在第一象限的交点得x=4,化y^2=8-x为y=根号(8-x)(因只取这部分的积分).
这个面积为S(0->4)x/2 dx+S（4->8）根号(8-x)dx=4-2/3·根号(8-x)^3|(4->8)=4+16/3=28/3.

追问

最后是36

追答

嗯，我还算少了横轴下面围成的面积. 道理相同，自己算吧

追问

好的 谢谢您

Answer 2

y=(1/2)x (1)
y^2 = 8-x (2)
from (1) and (2)
(1/4)x^2 =8-x
x^2+4x -32 =0
(x+8)(x-4) =0
x=-8 or 4
A
=∫ y dx
=∫(-8->4) [√(8-x) - (1/2)x ] dx
= [-(2/3)(8-x)^(3/2) - (1/4)x^2]| (-8->4)
=-[(2/3)(8-x)^(3/2) + (1/4)x^2]| (-8->4)
=[ (2/3)(8+8)^(3/2) + (1/4)(-8)^2) ] - [(2/3)(8-4)^(3/2) + (1/4)(4^2)]
=(128/3 + 16 ) - (16/3 + 4)
=112/3 +12
=148/3

追问

可是答案是36

可是答案是36

Answer 3

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谢谢

Answer 4

交点(-8,-4)和(4,2)
S=∫(-4,2)(8-y²-2y)dy
=(8y-1/3y³-y²)|(-4,2)
=8[2-(-4)]-1/3[2³-(-4)³]-[2²-(-4)²]
=48-24+12
=36

追问

请问 这个图像怎么画呀