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The area of the region, enclosed by the circle x^{2}+y^{2}=2 which is not common to the region bounded by the parabola y^{2}=x and the straight line y=x, is :  


Option: 1 \frac{1}{3}(12\pi -1)

Option: 2 \frac{1}{6}(12\pi -1)

Option: 3 \frac{1}{3}(6\pi -1)

Option: 4 \frac{1}{6}(24\pi -1)
 

Answers (1)

best_answer

 

 

 

 

Area Bounded by Curves When Intersects at More Than One Point -

Area bounded by the curves  y = f(x),  y = g(x)  and  intersect each other in the interval [a, b]

First find the point of intersection of these curves  y = f(x) and  y = g(x) , let the point of intersection be x = c

Area of the shaded region  

=\int_{a}^{c}\{f(x)-g(x)\} d x+\int_{c}^{b}\{g(x)-f(x)\} d x

 

When two curves intersects more than one point

rea bounded by the curves  y=f(x),  y=g(x)  and  intersect each other at three points at  x = a, x = b amd x = c.

To find the point of intersection, solve f(x) = g(x).

For x ∈ (a, c), f(x) > g(x) and for x ∈ (c, b), g(x) > f(x).

Area bounded by curves,

\\\mathrm{A=} \int_{a}^{b}\left |f(x)-g(x) \right |dx\\\\\mathrm{\;\;\;\;=} \int_{a}^{c}\left ( f(x)-g(x) \right )dx+\int_{c}^{b}\left ( g(x)-f(x) \right )dx  

 

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\\x^2+y^2=2\Rightarrow r=\sqrt 2\\y^2=x\\y=x\\

Area between parabola and line is A1

\\A1=\int_{0}^{1}(y^2-y)dx\\A1=\int_{0}^{1}(\sqrt x-x)dx=1/6\\\text{Required area }=2\pi-1/6

Correct Option (2)

Posted by

Ritika Jonwal

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