Question

Find the area bounded between

${x}^{2} = 4y \: and \: {y}^{2} = 4x$
​

Answer 1

Step-by-step explanation:

Correct option is

B

3

16

x

2

=4y y

2

=4x

x=2

y

=4×2

y

y

2

=8

y

⇒y

4

−8y=0

then y=0,4

x

2

=4y

we get x=4,0

So, the points of intersection are (0,0) & (4,4)

Area =∫

0

4

∫

2

x

x

2

/4

dydx=∫

0

4

[y]

2

x

x

2

/4

dx

=∫

0

4

[

4

x

2

−2

x

]dx

=[

12

x

3

−

3

4

⋅x

3/2

]

0

4

=[

12

4

3

−

3

4

(4)

3/2

]

=

3

16

sq. units.