Answer
The area of the region is $4$.
Work Step by Step
$x-y^2=0$ and $x+2y^2=3$
- For $x-y^2=0$, we can rewrite $$x=y^2$$
- For $x+2y^2=3$, we can rewrite $$x=3-2y^2$$
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 $y$.
$$y^2=3-2y^2$$ $$3y^2=3$$ $$y^2=1$$ $$y=-1\hspace{1cm}\text{or}\hspace{1cm}y=1$$
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 $x=3-2y^2$ and on the left by $x=y^2$. Therefore, the area of the region is $$A=\int^1_{-1}(3-2y^2-y^2)dy=\int^1_{-1}(3-3y^2)dy$$ $$A=(3y-y^3)\Big]^1_{-1}$$ $$A=(3-1)-(-3+1)$$ $$A=2-(-2)=4$$