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The area enclosed between the curves y² = 4x and x² = 4y is
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- 16/3
- 8
- 32/3
- 16
Correct Option: A
Given: y² = 4x
x² = 4y
⇒ | = 4x | |
16 |
or x4 = 64x
or x(x³ – 64) = 0
or x³ = 64
or x = 4
∴ y = 4
∴ Required area = 4∫0 | √4x - | dx | |||
y |
= | 2 × x3/2 × | - | 4 | ||||
3 | 12 | 0 |
= | × (4)x3/2 - | ||
3 | 12 |
= | - | = | |||
3 | 3 | 3 |
Alternately
For point where both parabolas cut each other
y² = 4x, x² = 4y
∴ x² = 4 √4x
or x² = 8√x
or x4 = 16 x
or x³ = 16
∴ x =(4, 0), (–4, 0)
∴ Required area
= 4∫0√4x - 4∫0 | dx = | 2 × | x3/2 - | 4 | = | ||||||
4 | 3 | 12 | 0 | 3 |