The area of the region bounded by the curves x+3 y^2=0 and x+4 y^2=1 is equal to :

Question

The area of the region bounded by the curves $x+3 y^2=0$ and $x+4 y^2=1$ is equal to :

Accepted Answer

Given curves are

$x+3{y}^{2}=0\, \, x+4{y}^{2}=1$

First, write them in the form $x=$:

$x=-3{y}^{2}$

and

$x=1-4{y}^{2}$

To find the bounded area, we first find their points of intersection.

Set the two values of $x$ equal:

$-3{y}^{2}=1-4{y}^{2}$

${y}^{2}=1$

$y=\pm 1$

So the curves intersect at $y=-1$ and $y=1$.

Now, for a fixed value of $y$, the right curve is

$x=1-4{y}^{2}$

and the left curve is

$x=-3{y}^{2}$

So the horizontal width of the region is

$(1-4{y}^{2})-(-3{y}^{2})=1-{y}^{2}$

Hence, area of the enclosed region is

$A={\int }_{1}^{-1}(1-{y}^{2})\, dy$

Now evaluate:

$A={[y-\frac{{y}^{3}}{3}]}_{1}^{-1}$

$A=(1-\frac{1}{3})-(-1+\frac{1}{3})$

$A=\frac{2}{3}-(-\frac{2}{3})$

$A=\frac{4}{3}$

Therefore, the required area is

$\frac{4}{3}$

So, the correct option is:

Solutions · 1