Question

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

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

Answer 1

The correct option is B $5:8$

$a^4{y}^2=(2a-x)x^5$
$cut\text{off x-axis when y=0}$
$(2a-x)x^5=0$
$A_1=\underset{0}{\overset{2a}{\int }}\frac{\sqrt{(2a-x)}{x}^{5/2}}{a^2}dx$
$putx=2a{\sin }^2\theta$
$dx=4a\sin \theta \cos \theta d\theta$
$\therefore A_1=\underset{0}{\overset{\pi /2}{\int }}\frac{\sqrt{2a}\cos \theta {\left(2a\right)}^{5/2}{\sin }^5\theta \times 4a\sin \theta \cos \theta }{a^2}d\theta$
$=32a^2\underset{0}{\overset{\pi /2}{\int }}{\sin }^6\theta {\cos }^2\theta d\theta$
$=32a^2\frac{\left(\mathrm{5.3.1}\right)\left(1\right)}{\mathrm{8.6.4.2}}\frac{\pi }{2}\left[byWall{i}^{\prime }sformula\right]$
$=\frac{5\pi a^2}{8}$
$Areaofcircle{A}_2=\pi a^2$
$\frac{A_1}{A_2}=\frac{5}{8}$
$A_1:A_2=5:8$