Question

The area enclosed between the curve $y^{2}(2a-x)=x^{3}$ and the line $x = 2$ above the $x$-axis is

• A. $\pi a^{2}$ sq. units
• B. $\frac{3\pi\,a^{2}}{2}$ sq. units
• C. $2 \pi a^{2}$ sq. units
• D. $3 \pi a^{2}$ sq. units

Answer 1

Answer: (B)

The required area $A=\int\limits_{0}^{2a} \sqrt{\frac{x^{3}}{2a-x}}dx$

put $x=2a\,sin^{2}\,\theta$
$\Rightarrow dx=2a\,2sin\,\theta \, cos\,\theta \,d \theta$
$\Rightarrow A=8a^{2} \int\limits_{0}^{\pi} \left(\frac{1-cos\,2\theta}{2}\right)^{2} d\theta $
$=2a^{2} \int\limits_{0}^{\pi}\left(1-2\,cos\,2\theta+cos^{2}\,2\theta\right)d\theta $
$=2a^{2} \int\limits_{0}^{\pi}\left(1-2\,cos\,2\theta+\frac{1+cos\,4\theta}{2}\right)d\theta$
$=\frac{3\pi\,a^{2}}{2}$