Given curve

$4y^2=x^2(4-x)(x-2)$

So, $y^2=\frac{x^2(4-x)(x-2)}{4}$

Find where the curve exists

Since $y^2\ge 0$, we need $x^2(4-x)(x-2)\ge 0$

Now $x^2\ge 0$ always, so the sign depends on $(4-x)(x-2).$

Let us check intervals:

• For $x<2$: $(4-x)>0$, $(x-2)<0$ so product $<0$
• For $2\le x\le 4$: product $\ge 0$
• For $x>4$: $(4-x)<0$, $(x-2)>0$ so product $<0$

Hence the real curve lies for $2\le x\le 4.$

Write upper half of the curve

$y=\frac{x}{2}\sqrt{(4-x)(x-2)}$

Since the equation is symmetric about the $x$-axis, the total bounded area is $A=2\int_2^4 y\,dx=2\int_2^4 \frac{x}{2}\sqrt{(4-x)(x-2)}\,dx$

Therefore, $A=\int_2^4 x\sqrt{(4-x)(x-2)}\,dx.$

Simplify the radical

Observe: $(4-x)(x-2)= -x^2+6x-8 = 1-(x-3)^2$ because $1-(x-3)^2=1-(x^2-6x+9)=-x^2+6x-8.$

So, $A=\int_2^4 x\sqrt{1-(x-3)^2}\,dx.$

Substitute

Let $u=x-3 \quad\Rightarrow\quad x=u+3,\; dx=du.$

When $x=2$, $u=-1$; when $x=4$, $u=1$.

Thus, $A=\int_{-1}^{1} (u+3)\sqrt{1-u^2}\,du.$

Split the integral: $A=\int_{-1}^{1} u\sqrt{1-u^2}\,du + 3\int_{-1}^{1} \sqrt{1-u^2}\,du.$

Evaluate each part

• The function $u\sqrt{1-u^2}$ is odd, so $\int_{-1}^{1} u\sqrt{1-u^2}\,du=0.$

• Also, $\int_{-1}^{1} \sqrt{1-u^2}\,du$ is the area of a semicircle of radius $1$, hence $\int_{-1}^{1} \sqrt{1-u^2}\,du=\frac{\pi}{2}.$

Therefore, $A=3\cdot \frac{\pi}{2}=\frac{3\pi}{2}.$

Match with options

$\boxed{\frac{3\pi}{2}}$

So the correct option is C.