Given,

4y² = x²(4 $$-$$ x)(x $$-$$ 2) ..... (1)

Here, Left hand side 4y² is always positive. So Right hand side should also be positive.

In x$$\in$$ [2, 4] Right hand side is positive.

By putting y = $$-$$y in equation (1), equation remains same. So, graph is symmetric about x axis.



Required Area = 2A₁

$$ = 2\int_2^4 y dx$$

$$ = \int_2^4 {x\sqrt {(x - 2)(4 - x)} } dx$$

put $$x = 4{\sin ^2}\theta + 2{\cos ^2}\theta $$

$$ \Rightarrow dx = \left[ {4(2\sin \theta \cos \theta - 4\sin \theta \cos \theta )} \right]d\theta $$

$$ \Rightarrow dx = 4\sin \theta \cos \theta d\theta $$

When lower limit = 2 then

$$2 = 4{\sin ^2}\theta + 2{\cos ^2}\theta $$

$$ \Rightarrow 4(1 - {\cos ^2}\theta ) + 2{\cos ^2}\theta = 2$$

$$ \Rightarrow 4 - 4{\cos ^2}\theta + 2{\cos ^2}\theta = 2$$

$$ \Rightarrow 2{\cos ^2}\theta = 2$$

$$ \Rightarrow \cos \theta = \pm 1$$

$$ \Rightarrow \theta = 0,\pi {} $$

When upper limit = 4 then

$$4 = 4{\sin ^2}\theta + 2{\cos ^2}\theta $$

$$ \Rightarrow - 2{\cos ^2}\theta = 0$$

$$ \Rightarrow \theta = {{\pi {} } \over 2}$$

$$ \therefore $$ Range of $$\theta$$ = 0 to $${{{\pi {} } \over 2}}$$

$$ \therefore $$ Area = $$ \int_0^{{{\pi {} } \over 2}} {(4{{\sin }^2}\theta + 2{{\cos }^2}\theta )\sqrt {(2{{\sin }^2}\theta )(2{{\cos }^2}\theta )} (4\sin \theta \cos \theta )d\theta } $$

$$ = \int_0^{{{\pi {} } \over 2}} {(4{{\sin }^2}\theta + 2{{\cos }^2}\theta )8{{\sin }^2}\theta {{\cos }^2}\theta d\theta } $$

$$ = 32\int_0^{{{\pi {} } \over 2}} {{{\sin }^4}\theta {{\cos }^2}\theta d\theta } + 16\int_0^{{{\pi {} } \over 2}} {{{\sin }^2}\theta {{\cos }^2}\theta d\theta } $$

Using Wallis formula,

$$ = 32.{{3.1.1} \over {6.4.2}}.{{\pi {} } \over 2} + 16.{{1.3.1} \over {6.4.2}}.{{\pi {} } \over 2}$$

$$ = \pi {} + {{\pi {} } \over 2}$$

$$ = {{3\pi {} } \over 2}$$