1. The area region bounded by the parabolas y2=4ax and x2=4ay is
1. The area region bounded by the parabolas y2=4ax and x2=4ay is
Sanvi , 8 Years ago
Grade 12th pass
Differential Calculus> 1. The area region bounded by the parabol...
1 Answers
Arun
We have, y2 = 4ax --------------------------- (1)
x2 = 4ay ---------------------------- (2)
(1) and (2) intersects hence
x = y2/4a (a > 0)
=> (y2/4a)2 = 4ay
=> y4 = 64a3y
=> y4 – 64a3y = 0
=> y[y3 – (4a)3] = 0
=> y = 0, 4a
When 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).
The area of the region between the two curves
= 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 hence
x = y2/4a (a > 0)
=> (y2/4a)2 = 4ay
=> y4 = 64a3y
=> y4 – 64a3y = 0
=> y[y3 – (4a)3] = 0
=> y = 0, 4a
When 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).
The area of the region between the two curves
= 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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