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The area of the region bounded by the curvea^4 y^2 = 2( 2a -x ) x^5is to that of the circle whose radius is a, is given by the ratio

Option: 1

4: 5 


Option: 2

5 : 4 


Option: 3

2 : 3 


Option: 4

3 : 2 


Answers (1)

best_answer

 

Area between two curves -

If we have two functions f\left ( x \right )\: and\:g \left ( x \right ).Area between two curves are

\int_{a}^{b}\left [ g\left ( x \right )-f\left ( x \right ) \right ]dx

- wherein

 

 

a^4 y^2 = ( 2a - x ) x^5 = 2 a x^5 - x^6 \\\\ here ( 2a -x )\cdot x^5 > 0 \\\\ x \epsilon [ 0,2a ]

\\\\ A = 2/ a^2 \int_{0 }^2a {} \sqrt {a^2 -( x-a )^2 } . x^2 dx \\\\ put \: \: \: \: x- a =a \cos \theta \\\\ A = -2 / a^2 \int_{0 }^\pi {} a \sin \theta a^ 2 ( 1+ \cos \theta )^ 2 ( - a \sin \theta )d \theta

= 2a ^ 2 \int_{0 }^{\pi } \left ( 5/8 - \cos 2 \theta /2 - \cos 4 \theta /8 + 2 \sin ^2 \theta \cos \theta \right ) d \theta \\\\ = 2a ^2 [ 5\theta/8 - \frac{1 \sin 2 \theta }{2*2 }- \frac{\cos 4 \theta }{32}+ 2/3 ( \sin \theta )^3 ] ^{\pi}_0\\\\ A = 2a ^2 [ 5\pi/ 8 ]= 5\pi a^2 /4 \\\\ 5 : 4

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