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-
A.
$$\frac{5}{3}$$
B.
$$\frac{1}{3}$$
C.
$$\frac{2}{3}$$
D.
$$\frac{4}{3}$$
Answer :
$$\frac{4}{3}$$
Solution :
$$x + 2{y^2} = 0\,\,\, \Rightarrow {y^2} = - \frac{x}{2}$$
[ Left handed parabola with vertex at $$\left( {0,\,0} \right)$$ ]
$$x + 3{y^2} = 1\,\,\, \Rightarrow {y^2} = - \frac{1}{3}\left( {x - 1} \right)$$
[Left handed parabola with vertex at $$\left( {1,\,0} \right)$$ ]
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
$$\eqalign{ & = \left| {\int\limits_{ - 1}^1 {\left( {1 - 3{y^2} - 2{y^2}} \right)dy} } \right| \cr & = \left| {\left[ {y - \frac{{5{y^3}}}{3}} \right]_{ - 1}^1} \right| \cr & = \left| {\left( {1 - \frac{5}{3}} \right) - \left( { - 1 + \frac{5}{3}} \right)} \right| \cr & = 2 \times \frac{2}{3} = \frac{4}{3}\,{\text{sq}}{\text{. units}} \cr} $$
$$x + 2{y^2} = 0\,\,\, \Rightarrow {y^2} = - \frac{x}{2}$$
[ Left handed parabola with vertex at $$\left( {0,\,0} \right)$$ ]
$$x + 3{y^2} = 1\,\,\, \Rightarrow {y^2} = - \frac{1}{3}\left( {x - 1} \right)$$
[Left handed parabola with vertex at $$\left( {1,\,0} \right)$$ ]
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
$$\eqalign{ & = \left| {\int\limits_{ - 1}^1 {\left( {1 - 3{y^2} - 2{y^2}} \right)dy} } \right| \cr & = \left| {\left[ {y - \frac{{5{y^3}}}{3}} \right]_{ - 1}^1} \right| \cr & = \left| {\left( {1 - \frac{5}{3}} \right) - \left( { - 1 + \frac{5}{3}} \right)} \right| \cr & = 2 \times \frac{2}{3} = \frac{4}{3}\,{\text{sq}}{\text{. units}} \cr} $$