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}

 

Answers (1)

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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