數學挑戰題 (13) -- 積分應用 / 面積 (中) (20分)

Question

求以下曲線 / 軸 所包圍的面積
1.
y = x ^2 , √x + √y = 2 , Y 軸
2.
√x + √y = 1 , y =√x , X 軸

Answer 1

1)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Crazyarea1.jpg

2)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Crazyarea2.jpg

Answer 2

(1)
y=x^2,√x+ √y =2,y axis

x=√y---(1)
x=4-4√y+y---(2)
√y=4-4√y+y
Let u=√y
u=4-4u+u^2
u^2-5u+4=0
(u-1)(u-4)=0
u=1 or u=4
y=1 or y=16
所以取值範圍是1≤y≤16
面積=∫(from 16 to 1) [ √y-(4-4√y+y)] dy
=∫(from 16 to 1) [ √y-4+4√y-y)] dy
=(2/3)y^(3/2)|(from 16 to 1)-4y|(from 16 to 1)+(8/3)y^(3/2)|(from 16 to 1)-(y^2)/2|(from 16 to 1)
=[(128/3)-(2/3)]-(64-4)+[(512/3)-(8/3)]-(128-1/2)
=42-60+168-255/2
=22.5

(2)
√x + √y = 1 , y =√x , x axis
y=x-2√x+1---(1)
y=√x---(2)
Let u=√x
u^2-2u+1=u
u^2-3u+1=0
u=(3士√3)/2
x=6+3√3 or x=6-3√3
所以取值範圍是6-3√3≤x≤6+3√3
面積=∫(from 6+3√3 to from 6-3√3) (x-2√x+1)-√x
=∫(from 6+3√3 to from 6-3√3) x-3√x+1
=(x^2)/2 |(from 6+3√3 to from 6-3√3)-3(2/3)x^(3/2)|(from 6+3√3 to from 6-3√3)+x|(from 6+3√3 to from 6-3√3)
=36√2-2(6+3√3)^(3/2)+2(6-3√3)^(3/2)+6√3