Question

The area of the region bounded by the curve $a^4{y}^2=(2a-x)x^5$ is to that of the circle whose radius is $a$, is given by the ratio

• A. $4:5$
• B. $5:8$
• C. $2:3$
• D. $3:2$

Answer 1

Answer: (B)

$5:8$

Given curve $a^4{y}^2=(2a-x)x^5$
cut off $x-$axis,when $y=0$
$0=(2a-x)x^5$ is
$A_1={\int }_0^{2a}\frac{\sqrt{(2a-x)}{x}^{\frac{5}{2}}}{a^2}dx$
Put $x=2a{\sin }^2\theta$
$\therefore dx=4a\sin \theta \cos \theta d\theta$
$\therefore A_1={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{2a}\cos \theta {\left(2a\right)}^{\frac{5}{2}}{\sin }^5\theta \times 4a\sin \theta \cos \theta }{a^2}d\theta$
$=32a^2{\int }_0^{\frac{\pi }{2}}{\sin }^6\theta {\cos }^2\theta d\theta$
$=32a^2\frac{\left(\mathrm{5.3.1}\right)\left(1\right)}{\mathrm{8.6.4.2}}.\frac{\pi }{2}$ (by Walli's formula)
$=\frac{5\pi a^2}{8}$
Area of circle,$A_2=\pi a^2$
$\therefore \frac{A_1}{A_2}=\frac{5}{8}$
$\Rightarrow A_1:A_2=5:8$