Find the area of the region bounded by the parabola y2=2x and x2=2y
Answers
Answer:
The parabola y
2
=2x opens towards the positive x−axis and its focus is (
2
1
,0)
The straight line x−y=4 passes through (4,0) and (0,−4)
Solving y
2
=2x and x−y=4, we get
y
2
=2(y+4)
⇒y
2
−2y−8=0
⇒(y−4)(y+2)=0
⇒y=4 or y=−2
So, the points of intersection of the given parabola and the line are A(8,4) and B(2,−2)
∴ Required area=Area of the shaded region OABO
=∫
−2
4
x
line
dy−∫
−2
4
x
parabola
dy
=∫
−2
4
(y+4)dy−∫
−2
4
2
y
2
dy
=[
2
(y+4)
2
]
−2
4
−
2
1
[
3
y
3
]
−2
4
=
2
1
(64−4)−
6
1
(64−(−8))
=30−12=18sq.units.
The given curves are y^2=2x and x^2=2y => y=x^2/2
To find the intersection point of two curves,
x^4/4 = 2x
x^4 = 8x
Therefore, the two curves intersect at (0, 0) and (2, 2).
The two curves are reflections of each other in the line y=x. The area bounded by this line and 2y=x² is the integral of (x-x²/2)dx between the limits 0 and 2. This is F(2)-F(0), where F(x)= x²/2-x³/6.
Noting that F(0) vanishes, we find that
F(2)-F(0) = F(2)=2–8/6=2/3 units.
The required area is twice this, or 4/3 units.
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