The area of the plane region bounded by the curves x + 2y^2 = 0 and [,x + 3y^2 = 1] is equ…

Question

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

Accepted Answer

$$x + 2{y^2} = 0 \Rightarrow {y^2} =  - {x \over 2}$$

$$\left[ {} \right.$$ Left handed parabola with vertex at $$\left( {0,0} \right)$$ $$\left. {} \right]$$

$$x + 3{y^2} = 1 \Rightarrow {y^2} =  - {1 \over 3}\left( {x - 1} \right)$$

$$\left[ {} \right.$$ Left handed parabola with vertex at $$\left( {1,0} \right)$$ $$\left. {} \right]$$

Solving the two equations we get the points of  intersection

as $$\left( { - 2,1} \right),\left( { - 2, - 1} \right)$$

The required area is $ACBDA,$ given by

$$ = \left| {\int\limits_{ - 1}^1 {\left( {1 - 3{y^2} - 2{y^2}} \right)dy} } \right|$$

$$ = \left| {\left[ {y - {{5{y^3}} \over 3}} \right]_{ - 1}^1} \right|$$

$$ = \left| {\left( {1 - {5 \over 3}} \right) - \left( { - 1 + {5 \over 3}} \right)} \right|$$

$$ = 2 \times {2 \over 3} = {4 \over 3}\,\,$$ sq. units.

Solutions · 1