Question

The area of the loop of the curve $y^2=x^2(1-x)$ is ___________.

• A. 15/8
• B. 4/15
• C. 8/15
• D. None of these

Answer 1

The correct option is C 8/15

$\text{g}iven:curve=y^2=x^2(1-x)\text{(i)}$

From (i), $y=\pm x\sqrt{1-x}$

For, AOB(region) we have $y=+x\sqrt{1-x}$ (From the graph which suggests that curve is in $1^{\text{st}}$ quadrant

So, the desired area,

Area $=2{\int }_0^{1}x\sqrt{1-x}dx$

take, $1-x=t^2$

differentiating, $-dx=2$ td $t$

also, limits will change

for $x=1$ ; for $x=0$

$t=0;t=1$

$\text{Area}=4{\int }_1^0(1-t^2)(-tdt)\left(t\right)$

$=4{\int }_1^0(1-t^2)+(-t)dt$

$=4{\int }_0^1(t^2-t^4)dt$

$=4{(\frac{t^3}{3}-\frac{t^5}{5})}_0^1$

$=4(\frac{1}{3}-\frac{1}{5})sq.\text{units}$

$=4\times \frac{2}{15}Sq.\text{units}$

$\text{Area}=\frac{8}{15}sq.units$