find the area between two parabolas x^2=4ay and y^2=4ax
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We have, y2 = 4ax ----------- (1)
x2 = 4ay ------------------------ (2)
(1) and (2) intersects
x = y2/4a (a > 0)
=> (y2/4a)2 = 4ay
=> y4 = 64a3y
=> y4 – 64a3y = 0
=> y[y3 – (4a)3] = 0
=> y = 0, 4a
y = 0, x = 0 and
when y = 4a, x = 4a.
The points of intersection of (1) and (2) are O(0, 0)
and A(4a, 4a).
= Area of the shaded region
= 0∫4a(y1 – y2)dx
= 0∫4a[√(4ax) – x2/4a]dx
= [2√a.(x3/2)/(3/2) – (1/4a)(x3/3)]04a
= 4/3√a(4a)3/2 – (1/12a)(4a)3 – 0
= 32/3a2 – 16/3a2
= 16/3a2 sq. units
x2 = 4ay ------------------------ (2)
(1) and (2) intersects
x = y2/4a (a > 0)
=> (y2/4a)2 = 4ay
=> y4 = 64a3y
=> y4 – 64a3y = 0
=> y[y3 – (4a)3] = 0
=> y = 0, 4a
y = 0, x = 0 and
when y = 4a, x = 4a.
The points of intersection of (1) and (2) are O(0, 0)
and A(4a, 4a).
= Area of the shaded region
= 0∫4a(y1 – y2)dx
= 0∫4a[√(4ax) – x2/4a]dx
= [2√a.(x3/2)/(3/2) – (1/4a)(x3/3)]04a
= 4/3√a(4a)3/2 – (1/12a)(4a)3 – 0
= 32/3a2 – 16/3a2
= 16/3a2 sq. units
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