IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

shakirshafi12

a⁴y²=(2a-x)x⁵

you see if we put y as -y the equation will not change

so the funcition will be symmetric about x axis

y=0 at x=0 and

y=0 at x=2a

so the required area will be

A=2_{[0 ]}

^{[2a ]}

x⁵(2a-x)/a⁴

A=2_{[0 ]}

^{[2a ]}(x²/a²)

x(2a-x)

now put x=2at

A=2_{[0 ]}

^{[1 ]}4t².2a

t(1-t) .2a.dt

A=32a²_{[0 ]}

^{[1 ]}t²

t(1-t)dt

now put t=|sinQ|²

for t>0 and Q>0

t=sin²Q

dt=2cosQ.sinQ t=0 Q=0

t=1 Q=pi/2

A=64a²_{[0 ]}

^{[pi/2 ]} Sin⁶ Q Cos² Q

A=64a²[_{[0 ]}

^{[pi/2 ]} Sin⁶ Q - ₀^{[pi/2 ]} Sin⁸ Q

using the property I_n=(n-1)/n I_n-2

The area of circle=pi*a²