Question

Find the area of the loop of the curve $x=3t^2,y=3t-t^3$

• A. $\frac{72}{5}\sqrt{3}sq.units$
• B. $\frac{76}{3}\sqrt{5}sq.units$
• C. $\frac{78}{5}\sqrt{3}sq.units$
• D. $\frac{70}{3}\sqrt{5}sq.units$

Answer 1

The correct option is A $\frac{72}{5}\sqrt{3}sq.units$

Given curve is $x=3t^2$

$\Rightarrow t=\sqrt{\frac{x}{3}}$
From second curve,
$y=3.\sqrt{\frac{x}{3}}-\frac{x}{3}\sqrt{\frac{x}{3}}$
Now $y=0$ at $x=0$ and $x=9$.
Hence the area enclosed by $y$ between $x=0$ and $x=9$ is
$\frac{1}{\sqrt{3}}{\int }_0^{9}3\sqrt{x}-\frac{(x{)}^{\frac{3}{2}}}{3}.dx$
$=\frac{1}{\sqrt{3}}\left[2\right(x{)}^{\frac{3}{2}}-\frac{2}{15}(x{)}^{\frac{5}{2}}{]}_0^9$
$=\frac{1}{\sqrt{3}}\left(2\right(27)-\frac{486}{15})$
$=\frac{1}{\sqrt{3}}[54-\frac{162}{5}]$
$=\frac{27}{\sqrt{3}}(2-\frac{6}{5})$
$=9\sqrt{3}.\frac{4}{5}$
$=\frac{36\sqrt{3}}{5}$

As the curve form a loop, the required area will be twice of the above.

Hence, the required area is $\frac{72\sqrt{3}}{5}$