Answer
The area of the region is $8/3$.
Work Step by Step
$x+y^2=0$ and $x+3y^2=2$
- For $x+y^2=0$, we can rewrite $$x=-y^2$$
- For $x+3y^2=2$, we can rewrite $$x=2-3y^2$$
1) Draft the graph:
The graph is drafted below. We would use integration with respect to y here.
2) Find the limits of integration:
To find the limits of integration, we find intersection points with respect to y.
$$-y^2=2-3y^2$$ $$2y^2=2$$ $$y^2=1$$ $$y=\pm1$$
So the upper limit is $1$ and the lower limit is $-1$.
3) Find the area:
The region from $y=-1$ to $y=1$ is bounded on the right by $x=2-3y^2$ and on the left by $x=-y^2$. Therefore, the area of the region is
$$A=\int^1_{-1}[2-3y^2-(-y^2)]dy=\int^1_{-1}(2-2y^2)dy$$ $$A=2y-\frac{2y^3}{3}\Big]^1_{-1}$$ $$A=\Big(2-\frac{2}{3}\Big)-\Big(-2+\frac{2}{3}\Big)$$ $$A=2\Big(2-\frac{2}{3}\Big)$$ $$A=2\times\frac{4}{3}=\frac{8}{3}$$
