Question

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

• A. $\frac{3\mathrm{\pi }}{2}$
• B. $\frac{\mathrm{\pi }}{16}$
• C. $\frac{\mathrm{\pi }}{8}$
• D. $\frac{3\mathrm{\pi }}{8}$

Answer 1

The correct option is A

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

Explanation for the correct option:

Given, curve $4y^2=x^2(4-x)(x-2)$

Taking square root;

$$\begin{gathered} \therefore \left|y\right|=\frac{\left|x\right|}{2}\sqrt{(x-2)(4-x)} \\ \Rightarrow y_1=\frac{x}{2}\sqrt{(4-x)(x-2)} \\ and{y}_2=\frac{-x}{2}\sqrt{(4-x)(x-2)} \end{gathered}$$

$$\begin{gathered} \therefore Area\left(A\right)={\int }_2^4(y_1-y_2)dx \\ ={\int }_2^4\sqrt{(4-x)(x-2)}dx...\left(1\right) \end{gathered}$$

On applying${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$$\begin{gathered} Area={\int }_2^4(6-x)\sqrt{(4-x)(x-2)}dx...\left(2\right) \\ \text{Adding},\left(1\right)\text{and}\left(2\right) \\ \Rightarrow 2A=6{\int }_2^4\sqrt{(4-x)(x-2)}dx \\ \Rightarrow A=3{\int }_2^4\sqrt{1-{(x-3)}^2}dx \\ \Rightarrow A=3\times \frac{\mathrm{\pi }}{2}\times 1^2 \\ \Rightarrow A=\frac{3\mathrm{\pi }}{2} \end{gathered}$$

Hence, the correct answer is option (A).