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The area of the plane region bounded by the curves x+2y^{2}=0\; and\; x+3y^{2}=1  is equal to

  • Option 1)

    \frac{4}{3}\;

  • Option 2)

    \; \frac{5}{3}\;

  • Option 3)

    \; \frac{1}{3}\;

  • Option 4)

    \; \frac{2}{3}

 

Answers (1)

best_answer

As learnt in concept

Area along y axis -

Let y_{1}= f_{1}(x)\, and \, y_{2}= f_{2}(x) be two curve, then area bounded by the curves and the lines

y = a and y = b is

A=\int_{a}^{b}\left ( x_{2}-x_{1} \right )dy

- wherein

 

x + 2y2=0 and x+3y2= 1

=>2\int_{0}^{+1}(1-3y^{2})-(-2y^{2})dy

=2\int_{-0}^{+1}(1-y^{2})dy

=2\left [ y-\frac{y^{2}}{3} \right ]_{0}^{1}

= 2\times (1-\frac{1}{3})=\frac{4}{3} 


Option 1)

\frac{4}{3}\;

this is correct option

Option 2)

\; \frac{5}{3}\;

this is incorrect option

Option 3)

\; \frac{1}{3}\;

this is incorrect option

Option 4)

\; \frac{2}{3}

this is incorrect option

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perimeter

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