The area bounded by the curve
Question:

The area bounded by the curve $4 y^{2}=x^{2}(4-x)(x-2)$ is equal to:

1. (1) $\frac{\pi}{8}$

2. (2) $\frac{3 \pi}{8}$

3. (3) $\frac{3 \pi}{2}$

4. (4) $\frac{\pi}{16}$

Correct Option: , 3

Solution:

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

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

$\Rightarrow y_{1}=\frac{x}{2} \sqrt{(4-x)(x-2)}$

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

$\mathrm{D}: \mathrm{x} \in[2,4]$

Required Area

$=\int_{2}^{4}\left(y_{1}-y_{2}\right) d x=\int_{2}^{4} x \sqrt{(4-x)(x-2)} d x \ldots(1)$

Applying $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$

Area $=\int_{2}^{4}(6-x) \sqrt{(4-x)(x-2)} d x \ldots(2)$

$(1)+(2)$

$2 \mathrm{~A}=6 \int_{2}^{4} \sqrt{(4-\mathrm{x})(\mathrm{x}-2)} \mathrm{d} \mathrm{x}$

$A=3 \int_{2}^{4} \sqrt{1-(x-3)^{2}} d x$

$A=3 \cdot \frac{\pi}{2} \cdot 1^{2}=\frac{3 \pi}{2}$