Question

Mohon dibantu : luas daerah yang dibatasi parabola 5y^2=16x dan y^2=8x-24

Answer 1

X konstan

5y^2 = 16x
x = (5/16)y^2

y^2 = 8x - 24
8x = y^2 + 24
x = (1/8)y^2 + 3

tentukan titik potong

x = x

(5/16)y^2 = (1/8)y^2 + 3
(5/16)y^2 = (2/16)y^2 + 3
(3/16)y^2 = 3
3y^2 = 48
y^2 = 16
y = +-4

y = 4 dan y = -4

4
∫ (1/8)y^2 + 3 - (5/16)y^2 dy
-4

...4
= ∫ -(3/16)y^2 + 3 dy
..-4

........................4
= -(1/16)y^3 + 3y |
.......................-4

= -(1/16)(4)^3 + 3(4) - (-(1/16)(-4)^3 + 3(-4))
= 8 - (-8)
= 16 satuan luas.