area of
and x^2=4ay
find the
y ² =4ax
Integration.
common two parabola's
using
double integartion
Answers
Answer:
your answer
Step-by-step explanation:
Answer
To Prove that the area enclosed between two parabolas y
2
=4ax and x
2
=4ay is
3
16a
3
Given curves are
y
2
=4ax and
x
2
=4ay
First we have to find the area of Intersection of the two curves
Point of Intersection of the two curves are
(
4a
x
2
)
2
=4ax
(
16a
2
x
4
)=4ax
x
4
=64a
3
x
x
4
−64a
3
x=0
x(x
3
−64a
3
)=0
x=0,x=4a
Also y=0,y=4a
The Point of Intersection of these 2 curves are (0,0) and (4a, 4a )
The Area of the two region between the two curves
= Area of the shaded region
∫
0
4a
[y
2
−y
1
]dx
∫
0
4a
[
4ax
−
4a
x
2
]dx
On Integrating this ,we get
[4a
1/2
3
x
3/2
−
12a
x
3
]
0
4a
=[
3
32a
2
−
3
16a
2
]
=
3
16a
2
Hence , Area =
3
16a
2