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Mathematics
Mathematics
Problem
The area bounded by the curve
4
y
2
=
x
2
(
4
−
x
)
(
x
−
2
)
4y^2 = x^2(4-x)(x-2)
4
y
2
=
x
2
(
4
−
x
)
(
x
−
2
)
is equal to : (1)
π
8
\frac{\pi}{8}
8
π
(2)
3
π
8
\frac{3\pi}{8}
8
3
π
(3)
3
π
2
\frac{3\pi}{2}
2
3
π
(4)
π
16
\frac{\pi}{16}
16
π
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