using green theoroem in a plane to find the finite area enclosed by the parabolas y^2=4ax and x^2=4ay
Hello,
We have, y 2 = 4ax................ (1)
Also, x
2
= 4ay.........................(2)
So, x = y2/4a
Hence, (y 2 /4a) 2 = 4ay
So, y 4 = 64a 3 y
So, y
4
– 64a
3
y = 0
y[y 3 – (4a) 3 ] = 0
Hence, y = 0, 4a
When y = 0, x = 0 and when y = 4a, x = 4a.
The points of intersection of (1) and (2) are (0, 0) and (4a, 4a).
The area of the region between the two curves
= 0 ∫ 4a (y1 – y 2 )dx
=
0
∫
4a
[√(4ax) – x2/4a]dx
= [2√a.(x3/2)/(3/2) – (1/4a)(x
3
/3)]0
4a
= 4/3√a(4a)3/2 – (1/12a)(4a)
3
– 0
= 32/3a
2
– 16/3a
2
= 16/3a 2 sq. units.
Best Wishes.
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