Question

Show that the Area of one loop of the curve y2 = x2n (4-x?)is16÷3

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Answer 1

Answer:

4y

2

=x

2

(4−x

2

)

⇒y=±

2

1

x

2

(4−x

2

)

⇒y=±

2

x

(4−x

2

)

Therefore, Area(A)=4×∫

0

2

2

x

4−x

2

dx

Let 4−x

2

=t⇒−2xdx=dt

⇒A=∫

0

4

t

dt=[

3/2

t

3/2

]

0

4

⇒A=

3

2

64

⇒A=

3

16

sq.units