Finding area between 2 functions

[tmt1]

1.

I have to find the area between

$x = 2y^2$
and
$x = 1 - y$

I find the intersection points

$ 1 -y = 2y^2$

$2y^2 + y - 1= 0 $

$(2y - 1)(y + 1)= 0$

so y = 1 and -1

However, x = y - 1 is not a vertical line so I am not sure how 1 and -1 can be intersections. Also, when I plug these numbers into the original equation they don't equate, so what am I doing wrong here?

1. I have another problem for finding the area between

$x = 2 - y^2$

and
$x = y^2 - 2$

I found that the points of intersection are y =$ +/- \sqrt{2}$ and the greater function is $ x = 2 - y^2$ so

$\int_{-\sqrt{2}}^{\sqrt{2}} 4 - 2y^2 \,d$

then

$\left[4y - \frac{2y^3}{3}]\right]_1^3$

and

$[4\sqrt{2} - \frac{2\sqrt{8}}{3}] - [- 4 \sqrt{2} + \frac{4 \sqrt{8}}{3}]$

and then

$4\sqrt{2}$

However the answer is $\frac{16\sqrt{2}}{3}$

 

[Prove It]

1.

I have to find the area between

$x = 2y^2$
and
$x = 1 - y$

I find the intersection points

$ 1 -y = 2y^2$

$2y^2 + y - 1= 0 $

$(2y - 1)(y + 1)= 0$

Correct up to here. Now notice that $\displaystyle \begin{align*} 2y - 1 = 0 \implies 2y = 1 \implies y = \frac{1}{2} \end{align*}$. The other solution y = -1 is fine...