# Find the area enclosed by the loop in the curve, $4y^{2}= 4x^{2}-x^{3}$ Option 1) $\frac{128}{15}$ Option 2) $\frac{15}{128}$ Option 3) $\frac{130}{17}$ Option 4) $\frac{17}{130}$

put y=0

$0=x(4-x)\\ \therefore x=0$

$x=4$

It means curve makes the loop symmetric about x axis between 0 and 4 $\therefore 2\int_{0}^{4}ydx=\frac{2}{2}\int_{0}^{4}\sqrt{4x^{2}-x^{3}}dx$

$\frac{2}{2}\int_{0}^{4}x\sqrt{4-x}dx$

Now let 4-x=t

-dx=dt

$\int_{4}^{0}(4-t)\sqrt{t}dt=\int_{0}^{4}(4\sqrt{t})-t\sqrt{t})dt$

$\left [ \frac{4.2}{3}t^{\frac{3}{2}}-\frac{2}{5}t^{\frac{5}{2}} \right ]_{0}^{4}$

$\left [ \frac{4.2}{3}4^{\frac{3}{2}}-\frac{1.2}{5}4^{\frac{5}{2}} \right ]$

$\frac{4.2}{3}\times 8-\frac{2}{5} \times 32\\ 64\left [ \frac{1}{3}-\frac{1}{5} \right ]=\frac{128}{15}$

Option 1)

$\frac{128}{15}$

This solution is correct

Option 2)

$\frac{15}{128}$

This solution is incorrect

Option 3)

$\frac{130}{17}$

This solution is incorrect

Option 4)

$\frac{17}{130}$

This solution is incorrect

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